Probability

When selecting three numbers from the set $\{1, 2, 3, \ldots, 50\}$, we want to find the probability that they form an arithmetic progression (AP).

First, determine the total number of ways to choose three numbers from the set:

$ { }^{50}C_{3} = \frac{50 \times 49 \times 48}{6} $

For the numbers $a$, $b$, and $c$ to be in AP, $2b = a + c$, which implies that $a + c$ must be even. Therefore, $a$ and $c$ must both be even or both be odd.

The set $\{1, 2, 3, \ldots, 50\}$ consists of 25 odd numbers and 25 even numbers.

The number of ways to choose two even numbers is:

$ { }^{25}C_{2} = \frac{25 \times 24}{2} $

Similarly, the number of ways to choose two odd numbers is also:

$ { }^{25}C_{2} = \frac{25 \times 24}{2} $

Thus, the total number of favorable cases (either two evens or two odds) is:

$ 2 \times { }^{25}C_{2} = 2 \times \frac{25 \times 24}{2} $

Therefore, the probability that the selected numbers are in arithmetic progression is:

$ \text{Probability} = \frac{2 \times \frac{25 \times 24}{2}}{\frac{50 \times 49 \times 48}{6}} = \frac{3}{98} $

$ p $ represent the probability of getting head in a toss of a fair coin, so

$ \begin{array}{l} p=\frac{1}{2} \\\\ q=\frac{1}{2} \end{array} $

$ \Rightarrow p(X=r)={ }^{6} C_{r}\left(\frac{1}{2}\right)^{r}\left(\frac{1}{2}\right)^{6-r} $

Probability of getting 3 head $ r=3 $

$ \Rightarrow \quad p(X=3)={ }^{6} C_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{3}=\frac{5}{16} $

To solve the problem, we need to determine the probability that the word written by the student was "PROBABILITY," given that the letters "LI" were left after erasing the rest.

Definitions

Event B: "LI" is the sequence of letters left.

Event A: The student wrote "PROBABILITY."

We need to find $ P(A|B) $, which represents the probability of event A given event B.

Calculations

Probability of Event $ A \cap B $:

Event $ A \cap B $ happens when the student writes "PROBABILITY" and "LI" is the part remaining after erasing. In "PROBABILITY," "LI" can appear in exactly one position. Hence, the probability of choosing "PROBABILITY" and having "LI" is:

$ P(A \cap B) = \frac{1}{4} \times \frac{1}{10} $

$\frac{1}{4}$ represents the probability of choosing "PROBABILITY" from the four words.

$\frac{1}{10}$ represents the probability of choosing "LI" from the 10 possible pairs of consecutive letters in "PROBABILITY."

Similar Probability Calculations for Other Words:

For "ABILITY": "LI" can be one of 6 pairs.

$ P(\text{ABILITY}) = \frac{1}{4} \times \frac{1}{6} $

For "FACILITY": "LI" can be one of 7 pairs.

$ P(\text{FACILITY}) = \frac{1}{4} \times \frac{1}{7} $

For "MOBILITY": "LI" can be one of 7 pairs.

$ P(\text{MOBILITY}) = \frac{1}{4} \times \frac{1}{7} $

Probability of Event B:

$ P(B) = P(\text{PROBABILITY} \cap B) + P(\text{ABILITY} \cap B) + P(\text{FACILITY} \cap B) + P(\text{MOBILITY} \cap B) $

Conditional Probability $ P(A|B) $:

$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4} \times \frac{1}{10}}{\frac{1}{4} \times \frac{1}{10} + \frac{1}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{1}{7}} $

Final Calculation:

The probability is simplified to:

$ \boxed{\frac{21}{116}} $

In a standard deck of 52 cards, there are 4 aces. Therefore, the probability of drawing an ace on a single trial is given by:

$ \text{Probability} = \frac{4}{52} = \frac{1}{13} $

When drawing two cards with replacement, the situation can be modeled using a binomial distribution where the number of trials is 2 and the probability of success (drawing an ace) is $\frac{1}{13}$.

The mean (or expected value) of a binomial distribution is calculated as:

$ \text{Mean} = \text{Number of trials} \times \text{Probability of success} = 2 \times \frac{1}{13} = \frac{2}{13} $

Therefore, the mean of the probability distribution of $X$, where $X$ denotes the number of aces drawn, is $\frac{2}{13}$.

Total number of outcomes when 2 dice are thrown $ =36 $

Number of co-prime numbers on each die are

$ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) $,

$ (5,3),(5,4),(5,6),(6,1),(6,5) $

Number of favourable outcomes $ =23 $

$ \therefore $ Required probability $ =\frac{23}{36} $

Total number of outcomes

$ ={ }^{52} C_{2}=1326 $

Let $ A=\{4,8,10\}, B=\{6,9\} $ and $ C=\{3\} $ $ A^{\prime}= $ Cards which have the number either 4 or 8 or 10

$ B^{\prime}= $ Cards which have the number either 6 or 9

$ C^{\prime}= $ Cards which have the number 3

Now,

$ \begin{array}{l} n\left(A^{\prime}\right)=3 \times 4=12 \\\\ n\left(B^{\prime}\right)=2 \times 4=8 \\\\ n\left(C^{\prime}\right)=1 \times 4=4 \end{array} $

Favourable cases

Case I 0 cards from $ A^{\prime}, 1 $ cards from $ B^{\prime} $ and 1 card from $ C^{\prime} $.

Case II 0 cards from $ A^{\prime}, 2 $ cards from $ B^{\prime} $ and 0 cards from $ C^{\prime} $

Case III 1 cards from $ A^{\prime}, 1 $ card from $ B^{\prime} $ and 0 cards from $ C^{\prime} $

Case IV 1 cards from $ A^{\prime}, 0 $ cards from $ B^{\prime} $ and 1 card from $ C^{\prime} $

Total number of cases

$ \begin{array}{l} ={ }^{12} C_{0} \times{ }^{8} C_{1} \times{ }^{4} C_{1}+{ }^{12} C_{0} \times{ }^{8} C_{2} \times{ }^{4} C_{0} \\\\ +{ }^{12} C_{1} \times{ }^{8} C_{1} \times{ }^{4} C_{0}+{ }^{12} C_{1} \times{ }^{8} C_{0} \times{ }^{4} C_{1} \\\\ =1 \times 8 \times 4+1 \times 28 \times 1+12 \times 8 \times 1+12 \times 1 \times 4 \\\\ =32+28+96+48=204 \\\\ \therefore \text { Required probability }=\frac{204}{1326}=\frac{102}{663} \end{array} $

Bag $ P $ contains 3 white, 2 red, and 5 blue balls, while bag $ Q $ contains 2 white, 3 red, and 5 blue balls. We are interested in finding the probability of drawing a red ball from bag $ Q $ after transferring one ball from bag $ P $ to bag $ Q $.

First, define $ E_1 $, $ E_2 $, and $ E_3 $ as the events of drawing a white, red, and blue ball from bag $ P $, respectively. The probabilities are calculated as follows:

$ P(E_1) = \frac{3}{10} $ (probability of drawing a white ball from $ P $)

$ P(E_2) = \frac{2}{10} $ (probability of drawing a red ball from $ P $)

$ P(E_3) = \frac{5}{10} $ (probability of drawing a blue ball from $ P $)

Now, consider the composition of bag $ Q $ after transferring one ball from bag $ P $:

If $ E_1 $ occurs (a white ball is transferred), bag $ Q $ will have 3 white balls, 3 red balls, and 5 blue balls. The probability of drawing a red ball from $ Q $ becomes $ \frac{3}{11} $.

If $ E_2 $ occurs (a red ball is transferred), bag $ Q $ will have 2 white balls, 4 red balls, and 5 blue balls. The probability of drawing a red ball from $ Q $ becomes $ \frac{4}{11} $.

If $ E_3 $ occurs (a blue ball is transferred), bag $ Q $ will have 2 white balls, 3 red balls, and 6 blue balls. The probability of drawing a red ball from $ Q $ becomes $ \frac{3}{11} $.

The overall probability $ P(E) $ of drawing a red ball from bag $ Q $ is computed using the law of total probability:

$ P(E) = P(E_1) \cdot \frac{3}{11} + P(E_2) \cdot \frac{4}{11} + P(E_3) \cdot \frac{3}{11} $

Substitute the values:

$ P(E) = \left(\frac{3}{10} \times \frac{3}{11}\right) + \left(\frac{2}{10} \times \frac{4}{11}\right) + \left(\frac{5}{10} \times \frac{3}{11}\right) $

$ P(E) = \frac{9}{110} + \frac{8}{110} + \frac{15}{110} = \frac{32}{110} = \frac{16}{55} $

The probability that a ball chosen from bag $ Q $ is red after transfer is $ \frac{16}{55} $.

$ \begin{aligned} & K+2 K+3 K^2+K^2=1\\\\ &\begin{aligned} & 3 K+4 K^2=1 \Rightarrow 4 K^2+3 K-1=0 \\\\ & 4 K^2+4 K-K-1=0 \\\\ & (4 K-1)(K+1)=0 \\\\ & K=\frac{1}{4}(K \neq-1) \\\\ & \text { Mean }(E(X))=2 K+6 K+15 K^2+9 K^2 \\\\ & =8 K+24 K^2 \\\\ & =2+24 \times \frac{1}{16}=2+\frac{3}{2}=\frac{7}{2} \\\\ & E\left(X^2\right)=4 K+18 K+75 K^2+81 K^2 \\\\ & =22 K+156 K^2 \\\\ & =\frac{22}{4}+156 \times \frac{1}{16}=\frac{22}{4}+\frac{39}{4}=\frac{61}{4} \\\\ & \text { Variance }=E\left(X^2\right)-(E(X))^2 \\\\ & =\frac{61}{4}-\frac{49}{4}=\frac{12}{4}=3 \end{aligned} \end{aligned} $

The given problem is same as the probability of de-arrangement of $n!$ objects.

$\therefore$ Required probability

$ =\frac{5!\left[1-1!+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right]}{5!} $

$ \begin{aligned} & =1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120} \\\\ & =\frac{60-20+5-1}{120}=\frac{44}{120}=\frac{11}{30} \end{aligned} $

Total possible outcomes,

$ n(S)=6 \times 6 \times 6=216 $

$\therefore$ Required probability $=\frac{27}{216}=\frac{1}{8}$

$ \begin{aligned} P(A) & =P(B)=P(C)=\frac{1}{3} \\\\ P(R / A) & =\frac{2}{5} \Rightarrow P(R / B)=\frac{3}{5} \\\\ P(R / C) & =1 / 5 \\\\ P(C / R) & =\frac{P(C) \times P(R / C)}{P(A) \times P(R / A)+P(B) \times P(R / B)} \\\\ & =\frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3} \times \frac{2}{5}+\frac{1}{3} \times \frac{3}{5}+\frac{1}{3} \times \frac{1}{5}} \\\\ & =\frac{1}{2+3+1}=\frac{1}{6} \end{aligned} $

Given,

We know that $\sum p\left(x_i\right)=1$

$ \begin{array}{rr} \therefore & 2 k^2+k+k+k^2=1 \\\\ \Rightarrow & 3 k^2+2 k-1=0 \\\\ \Rightarrow & 3 k^2+3 k-k-1=0 \\\\ \Rightarrow & 3 k(k+1)-1(k+1)=0 \\\\ \Rightarrow & -(3 k-1)(k+1)=0 \\\\ \Rightarrow & k=\frac{1}{3} \end{array} $

[ $\because$ probability cannot be negative]

Mean of $X=\sum x_i p\left(x_i\right)$

$ \begin{aligned} & =(1)\left(2 k^2\right)+2(k)+3(k)+5\left(k^2\right) \\\\ & =7 k^2+5 k \\\\ & =7\left(\frac{1}{3}\right)^2+5\left(\frac{1}{3}\right)=\frac{7}{9}+\frac{5}{3}=\frac{22}{9} \end{aligned} $

Let the coin be tossed $n$ times.

It is a binomial distribution problem Here, $p=P(H)=\frac{1}{2}, q=P(T)=\frac{1}{2}$

According to the question,

$ \begin{aligned} & { }^n C_5 p^5 q^{n-5}={ }^n C_4 p^4 q^{n-4} \\\\ & \Rightarrow{ }^n C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{n-5}={ }^n C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^{n-4} \\\\ & \Rightarrow{ }^n C_5={ }^n C_4 \Rightarrow n=5+4=9 \\\\ & \therefore \text { Required probability }={ }^9 C_6\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^{9-6} \\\\ & \quad=\frac{9 \times 8 \times 7}{3!}\left(\frac{1}{2}\right)^9=\frac{21}{128} \end{aligned} $

Possible outcomes are

$ \begin{aligned} & (1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5) \end{aligned} $

$\therefore$ Desired outcomes are $(1,6),(2,3),(3,2),(3,4),(4,3),(5,6)$, $(6,1),(6,5)$

$\therefore$ Required probability $=\frac{8}{15}$

Since, we can choose any face card (3 options) and any prime card ( 4 options) in a suit

$\Rightarrow 3 \times 4=12$ favourable combination in one suit.

$\because$ This situation can happen in any of the 4 suits

$\Rightarrow 12 \times 4=48$ overall favourable combination

and when drawing 2 cards from the same suit, then total possible ways

$ ={ }^{13} c_2=78 $

Since, there are 4 suits, then total ways $=78 \times 4$

Hence, required probability $=\frac{48}{78 \times 4}=\frac{2}{13}$

Let $E_1, E_2$ and $E_3$ denote the stock from $C_1, C_2$ and $C_3$ respectively. Let $A$ be event that refrigerator sold at random is found to be defective by a customer.

$ \begin{gathered} P\left(E_1\right)=\frac{25}{100}, P\left(E_2\right)=\frac{35}{100}, P\left(E_3\right)=\frac{40}{100} \\\\ P\left(\frac{A}{E_1}\right)=\frac{3}{100}, P\left(\frac{A}{E_2}\right)=\frac{2}{100} \Rightarrow P\left(\frac{A}{E_3}\right)=\frac{1}{100} \end{gathered} $

$ \begin{aligned} &\text { Using Baye's theorem, we get }\\\\ &\begin{aligned} A\left(\Sigma_2 / A\right) & =\frac{Y\left(E_2\right) \cdot P\left(A / E_2\right)}{\Sigma P\left(E_i\right) \cdot P\left(A / E_i\right)}=\frac{\frac{35}{100} \cdot \frac{2}{100}}{\frac{25}{100} \cdot \frac{3}{100}+\frac{35}{100} \cdot \frac{2}{100}+\frac{40}{100} \cdot \frac{1}{100}} \\\\ & =\frac{70}{5(15+14+8)}=\frac{14}{37} \end{aligned} \end{aligned} $

To determine the probability of having exactly two students who are good at Mathematics in a group of 8 students. We use Binomial probability formula,

$ p(X=k)={ }^n C_k p^k(1-p)^{n-k} $

Here, $n=8, k=2, p=0.6=\frac{6}{10}$

Then, required probability

$ \begin{aligned} & =P(X=2)={ }^8 C_2 \times\left(\frac{6}{10}\right)^2 \times\left(\frac{4}{10}\right)^6 \\\\ & =28 \times \frac{3^2}{5^2} \times \frac{2^6}{5^6}=\frac{7 \times 3^2 \times 2^8}{5^8} \end{aligned} $

To solve this problem, we use poisson distribution.

The Poisson probability formula is $P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}$ Here,

$\lambda=$ average number of events $=4$

$K=$ number of events (customer)

So, probability that more than 2 customers visit the shop in a specific hour $=P(X>2)$

$ \begin{aligned} & =1-P(X \leq 2)=1-[P(X=0)+P(X=1)+P(X=2)] \\\\ & =1-\left[\frac{4^0 e^{-4}}{0!}+\frac{4^1 e^{-4}}{1!}+\frac{4^2 e^{-4}}{2!}\right]=1-\left[e^{-4}+4 e^{-4}+8 e^{-4}\right] \\\\ & =1-13 e^{-4}=1-\frac{13}{e^4}=\frac{e^4-13}{e^4} \end{aligned} $

When two dice are rolled, the total number of possible outcomes is $6 \times 6 = 36$.

We need to find the probability that the sum of the numbers on the two dice is either 10 or 11.

Here are the combinations that result in a sum of 10 or 11:

For a sum of 10: $(4,6), (6,4), (5,5)$

For a sum of 11: $(6,5), (5,6)$

This gives us a total of 5 favorable outcomes: $(4,6), (6,4), (5,5), (6,5), (5,6)$.

Therefore, the probability of getting a sum of 10 or 11 is:

$ \frac{5}{36} $

Given the probabilities: $ P(A) = \frac{1}{4} $, $ P(A \mid B) = \frac{1}{2} $, and $ P(B \mid A) = \frac{2}{3} $, we need to find $ P(B) $.

First, consider the formula for conditional probability:

$ P(B \mid A) = \frac{P(A \cap B)}{P(A)} $

Inserting the known values:

$ \frac{P(A \cap B)}{P(A)} = \frac{2}{3} $

This implies:

$ P(A \cap B) = \frac{2}{3} \times P(A) $

Substituting $ P(A) = \frac{1}{4} $:

$ P(A \cap B) = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6} $

Next, for $ P(A \mid B) $:

$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} $

Given $ P(A \mid B) = \frac{1}{2} $, we have:

$ \frac{P(A \cap B)}{P(B)} = \frac{1}{2} $

Substituting $ P(A \cap B) = \frac{1}{6} $:

$ \frac{1}{6} = \frac{1}{2} \times P(B) $

Solving for $ P(B) $:

$ P(B) = \frac{1}{3} $

Given the problem, we have the following probabilities:

The probability that person A speaks the truth is 75%, which can be expressed as $ P(A) = \frac{3}{4} $.

The probability that person B speaks the truth is 80%, which can be expressed as $ P(B) = \frac{4}{5} $.

From these, we can find the probabilities that they do not speak the truth:

$ P(A') = 1 - P(A) = \frac{1}{4} $

$ P(B') = 1 - P(B) = \frac{1}{5} $

The probability that A and B contradict each other happens in two scenarios:

A speaks the truth, and B does not. This probability is given by:

$ P(A) \cdot P(B') = \frac{3}{4} \times \frac{1}{5} = \frac{3}{20} $

A does not speak the truth, and B does. This probability is given by:

$ P(A') \cdot P(B) = \frac{1}{4} \times \frac{4}{5} = \frac{4}{20} = \frac{1}{5} $

The total probability that they contradict each other is the sum of these probabilities:

$ P(\text{contradiction}) = \frac{3}{20} + \frac{4}{20} = \frac{7}{20} $

Let $E_1$ be the event of choosing bag A. $E_2$ be the event of choosing bag $B$, and $B$ be the event of drawing a red ball

$ \begin{aligned} & P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2} \\ & P\left(\frac{E}{E_1}\right)=\frac{3}{5}, P\left(\frac{E}{E_2}\right)=\frac{5}{9} \end{aligned} $

$ \begin{aligned} \therefore P\left(\frac{E_2}{E}\right) & =\frac{P\left(\frac{E}{E_2}\right) P\left(E_2\right)}{P\left(E_1\right) P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right)} \\ & =\frac{\frac{5}{9} \times \frac{1}{2}}{\frac{1}{2} \times \frac{3}{5}+\frac{5}{9} \times \frac{1}{2}}=\frac{\frac{5}{9}}{\frac{3}{5}+\frac{5}{9}} \\ & =\frac{\frac{5}{9}}{\frac{(27+25)}{45}}=\frac{5}{9} \times \frac{45}{52}=\frac{25}{52} \end{aligned} $

To find the value of $ k $, we start by considering the probability distribution table of the random variable $ X $:

$ \begin{array}{c|l|l|l|l} \hline X=x & 1 & 2 & 3 & 4 \\ \hline P(X=x) & 2k & 4k & 3k & k \\ \hline \end{array} $

The probabilities must sum up to 1, as they represent a complete probability distribution. Therefore, we have:

$ 2k + 4k + 3k + k = 1 $

Simplify the equation:

$ 10k = 1 $

Solve for $ k $:

$ k = \frac{1}{10} $

Hence, the value of $ k $ is $\frac{1}{10}$.

Given the conditions for a binomial distribution $ B(n, p) $:

The sum of the mean and the variance is 5.

The product of the mean and the variance is 6.

Starting with the formulas for the mean and variance:

Mean $ = np $

Variance $ = npq $, where $ q = 1 - p $.

Using these expressions:

Sum equation:

$ np + npq = 5 \quad \Rightarrow \quad np(1+q) = 5 \quad \text{(Equation 1)} $

Product equation:

$ np \times npq = 6 \quad \Rightarrow \quad (np)^2 \times q = 6 $

Substitute from Equation 1:

$ \left(\frac{5}{1+q}\right)^2 \times q = 6 $

Simplify this to:

$ 25q = 6(1 + q)^2 $

Expanding and rearranging:

$ 25q = 6(1 + q^2 + 2q) $

$ 25q = 6 + 6q^2 + 12q $

$ 6q^2 - 13q + 6 = 0 $

Solving the quadratic equation:

$ (3q - 2)(2q - 3) = 0 $

Solving gives possible values for $ q $:

$ q = \frac{2}{3} \quad \text{(choose this as} \, q \neq \frac{3}{2}) $

From $ q = \frac{2}{3} $, calculate $ p $ and $ n $:

$ p = 1 - q = \frac{1}{3} $

From Equation 1:

$ np(1+q) = 5 $

$ n \cdot \frac{1}{3} \cdot \left(1 + \frac{2}{3}\right) = 5 $

$ n \times \frac{1}{3} \times \frac{5}{3} = 5 $

$ n = 9 $

Finally, calculate $ 6(n + p - q) $:

$ 6(n + p - q) = 6\left(9 + \frac{1}{3} - \frac{2}{3}\right) $

$ = 6\left(9 - \frac{1}{3}\right) $

$ = 6 \times \frac{26}{3} = 52 $

Given that the coefficients $ a, b, $ and $ c $ in the quadratic equation $ ax^2 + bx + c = 0 $ are determined by rolling a die, we need to find the probability that the equation will have equal roots.

For a quadratic equation to have equal roots, the discriminant must be zero. Therefore:

$ b^2 - 4ac = 0 $

Rewriting, we get:

$ \left(\frac{b}{2}\right)^2 = ac $

Each coefficient $ a, b, $ and $ c $ is an integer between 1 and 6, inclusive. We examine possible integer values for equal roots by considering different values of $ b $.

If $ b = 1 $:

$ \left(\frac{1}{2}\right)^2 = ac \Rightarrow \frac{1}{4} = ac $

No integer values of $ a $ and $ c $ satisfy this condition.

If $ b = 2 $:

$ 1 = ac \Rightarrow a = 1, c = 1 \quad \text{so possible combination is } (1, 2, 1) $

If $ b = 3 $:

$ \left(\frac{3}{2}\right)^2 = ac \Rightarrow \frac{9}{4} = ac $

No integer values of $ a $ and $ c $ satisfy this condition.

If $ b = 4 $:

$ 4 = ac $

Possible pairs $(a, c)$ are:

$ a = 4, c = 1 $ leading to $(4, 4, 1)$

$ a = 1, c = 4 $ leading to $(1, 4, 4)$

$ a = 2, c = 2 $ leading to $(2, 4, 2)$

If $ b = 5 $:

$ \left(\frac{5}{2}\right)^2 = ac \Rightarrow \frac{25}{4} = ac $

No integer values of $ a $ and $ c $ satisfy this condition.

If $ b = 6 $:

$ 9 = ac $

Possible pair $(a, c)$:

$ a = 3, c = 3 $ leading to $(3, 6, 3)$

In total, these configurations give us five favorable outcomes:

$(1, 2, 1)$

$(4, 4, 1)$

$(1, 4, 4)$

$(2, 4, 2)$

$(3, 6, 3)$

The total number of possible equations is:

$ 6 \times 6 \times 6 = 216 $

Each coefficient can have any of the 6 possible values from a die roll.

Therefore, the required probability is:

$ \frac{5}{216} $

To solve for the probability that player $A$ wins the game, we need to calculate the probabilities and consider the alternating turns of $A$ and $B$, where $A$ needs to roll a sum of 6 before $B$ rolls a sum of 7.

Calculate the probability of sums:

Sum of 6: The outcomes that sum to 6 are (1,5), (5,1), (2,4), (4,2), and (3,3). Therefore, there are 5 outcomes.

Probability of rolling a 6: $ P(\text{Sum } 6) = \frac{5}{36} $

Sum of 7: The outcomes that sum to 7 are (1,6), (6,1), (2,5), (5,2), (3,4), and (4,3). This gives us 6 outcomes.

Probability of rolling a 7: $ P(\text{Sum } 7) = \frac{6}{36} = \frac{1}{6} $

Calculate the probability of $A$ winning:

Since $A$ rolls first, for $A$ to win, the following sequence can occur:

$A$ rolls a 6 on the first turn.

If $A$ does not roll a 6, with probability $\frac{31}{36}$ (the complementary probability of not rolling 6), and $B$ does not roll a 7, with probability $\frac{5}{6}$, then the game is in the same state again after one complete cycle (both $A$ and $B$ have thrown once each), and we recalculate.

This forms a geometric series where the success probability is $\frac{5}{36}$, and the continuation state probability is $\frac{31}{36} \times \frac{5}{6}$.

Use the formula for geometric series to find the probability $A$ wins:

$ P(A \text{ wins}) = \frac{\frac{5}{36}}{1 - \frac{31}{36} \times \frac{5}{6}} $

Solving this evaluates to:

$ P(A \text{ wins}) = \frac{30}{61} $

Therefore, the probability that $A$ wins is $\frac{30}{61}$.

Given:

$ P(E_1) = \frac{1}{2} $

$ P(E_1 \cup E_2) = \frac{2}{3} $

Since $ E_1 $ and $ E_2 $ are independent events:

$ P(E_1 \cap E_2) = P(E_1) \cdot P(E_2) $

Using the formula for the union of two events:

$ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) $

Substituting the given values:

$ \frac{2}{3} = \frac{1}{2} + P(E_2) - P(E_1) \cdot P(E_2) $

Simplifying:

$ \frac{2}{3} = \frac{1}{2} + P(E_2) - \frac{1}{2}P(E_2) $

$ \frac{2}{3} = \frac{1}{2} + \left(1 - \frac{1}{2}\right)P(E_2) $

$ \frac{2}{3} = \frac{1}{2} + \frac{1}{2}P(E_2) $

Rearranging terms:

$ \frac{2}{3} - \frac{1}{2} = \frac{1}{2}P(E_2) $

Calculate $\frac{2}{3} - \frac{1}{2}$:

$ \frac{4}{6} - \frac{3}{6} = \frac{1}{6} $

Substitute back:

$ \frac{1}{6} = \frac{1}{2}P(E_2) $

Solving for $P(E_2)$:

$ P(E_2) = \frac{1}{3} $

For the conditional probability $P(E_1 \mid E_2)$:

$ P(E_1 \mid E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{P(E_1) \cdot P(E_2)}{P(E_2)} = P(E1) = \frac{1}{2} $

Additional probability calculations:

$ P(\bar{E}_2 \mid E_1) = \frac{2}{3} $

$ P(\bar{E}_1 \cup \bar{E}_2) = \frac{5}{6} $

In summary, the matches are:

$ P(E_2) = \frac{1}{3} $

$ P(E_1 \mid E_2) = \frac{1}{2} $

Additional calculated probabilities align with the conditions provided.

$ \begin{aligned} & P_1\left(R_1\right)=\frac{4}{9} \quad P_2\left(R_2\right)=\frac{3}{9} \\ & P_1\left(B_1\right)=\frac{5}{9} \quad P_2\left(B_2\right)=\frac{6}{9} \\ & 1 R \text { and } 2 B \text { Ball }=R B B+B R B+B B R \\ & =P\left(R_1\right) P\left(B_2\right) P\left(B_2\right)+P\left(B_1\right) P\left(R_2\right) P\left(B_2\right) \\ & +P\left(B_1\right) P\left(B_2\right) P\left(R_2\right) \\ & =\frac{4}{9} \times \frac{6}{9} \times \frac{5}{8}+\frac{5}{9} \times \frac{3}{9} \times \frac{6}{8}+\frac{5}{9} \times \frac{6}{9} \times \frac{3}{8} \\ & =\frac{5}{27}+\frac{5}{36}+\frac{5}{36}=\frac{5}{27}+\frac{10}{36} \\ & =\frac{5}{27}+\frac{5}{18}=\frac{10+15}{54}=\frac{25}{54} \end{aligned} $

To determine the variance of the random variable $ X $ with the given probability distribution, we first need to find the value of $ k $ by ensuring the probabilities sum to 1.

Step 1: Solve for $ k $

Given:

$ 0.05 + 0.1 + 2k + 0 + 0.3 + k + 0.1 = 1 $

Simplifying, we find:

$ 3k = 0.45 $

$ k = 0.15 $

Step 2: Construct the probability distribution with the calculated $ k $

The probability distribution is:

$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline X = x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X = x) & 0.05 & 0.1 & 0.3 & 0 & 0.3 & 0.15 & 0.1 \\ \hline \end{array} $

Step 3: Calculate the mean $\mu$

The mean $\mu$ is calculated as:

$ \mu = \sum x_i P(x_i) = (-3)(0.05) + (-2)(0.1) + (-1)(0.3) + (0)(0) + (1)(0.3) + (2)(0.15) + (3)(0.1) $

$ \mu = -0.15 - 0.2 - 0.3 + 0 + 0.3 + 0.3 + 0.3 = 0.25 $

Step 4: Calculate the variance $\sigma^2$

Variance is calculated using:

$ \sigma^2 = \sum x_i^2 P(x_i) - \mu^2 $

Calculating $\sum x_i^2 P(x_i)$:

$ =(-3)^2(0.05) + (-2)^2(0.1) + (-1)^2(0.3) + (0)^2(0) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.1) $

$ =0.45 + 0.4 + 0.3 + 0 + 0.3 + 0.6 + 0.9 $

$ =2.95 $

Thus, variance $\sigma^2$ is:

$ \sigma^2 = 2.95 - (0.25)^2 $

$ \sigma^2 = 2.95 - 0.0625 $

$ \sigma^2 = 2.8875 $

Therefore, the variance of the random variable $ X $ is approximately 2.8875.

To find the probability that a radar system cannot detect an enemy plane at least two times in four consecutive scans, follow these steps:

Probability of Detection:

$ P(\text{Detection}) = P(D) = \frac{1}{10} $

Probability of Not Detecting:

$ P(\text{Not Detection}) = P(ND) = \frac{9}{10} $

Event Description: We are looking for the probability that the radar fails to detect the enemy plane at least two times (which includes all combinations where the plane is detected fewer than two times in four consecutive scans).

Calculate Complement Events:

The event we're interested in is the complement of detecting the plane all four times or detecting it exactly three times. First, calculate those probabilities:

All Four Detections:

$ P(\text{All Four Detections}) = \left(\frac{1}{10}\right)^4 $

Exactly Three Detections:

There are 4 combinations (combinations of choosing 3 detections out of 4), and one scenario where detection doesn't happen:

$ P(\text{Exactly Three Detections}) = \binom{4}{3} \left(\frac{1}{10}\right)^3 \left(\frac{9}{10}\right) $

Calculate and Subtract from 1:

$ = 1 - \left(\left(\frac{1}{10}\right)^4 + 4 \times \left(\frac{1}{10}\right)^3 \times \frac{9}{10}\right) $

Result:

$ = 1 - \left(\frac{1}{10000} + \frac{36}{10000}\right) = \frac{10000 - 523}{10000} = \frac{9477}{10000} = 0.9477 $

Thus, the probability that the radar cannot detect the plane at least two times in four consecutive scans is 0.9477.

Let the modular arithmetic modulo 3. Each number from 1 to 20 can be represented as either 0,1 or 2 modulo 3. Specifically numbers divisible by 3 are congruent to 0 , numbers leaving remainder 1 are congruent to 1 and number having remainder 2 are congruent to 2 .

The valid combinations of the sum of three numbers to be 0 modulo are There 0 's, three 1's, three 2 's, one 0 and two numbers that are 1 and 2 or vice-versa.

Total number to choose 3 numbers from 20.

$ ={ }^{20} C_3=\frac{20!}{17!3!}=1140 $

All three number are congruent to $0 \mathrm{~m}^3$.

$ \binom{6}{3}=20 $

All three numbers are congruent to

$ 1 m^3=\binom{7}{3}=35 $

All three number are congruent to

$ 2 m^3=\binom{7}{3}=35 $

One numbers congruent to 0,1 and 2

$ \begin{aligned} =6 \times 7 \times 7 & =294 \\ \text { On adding } & =20+35+35+294 \\ & =384 \end{aligned} $

Then, probability $=\frac{384}{1140}=\frac{32}{95}$

The possible sum values for three dice range from $3$ (when all dice show 1) to $18$ (when all dice show 6).

If person $A$ gets a sum of 13, we need to calculate the number of ways person $B$ can get a sum higher than 13, which would be sums of 14, 15, 16, 17, or 18.

Here is the breakdown of the number of ways to achieve each of these sums with three dice:

Sum of 14: There are 15 ways.

Sum of 15: There are 10 ways.

Sum of 16: There are 6 ways.

Sum of 17: There are 3 ways.

Sum of 18: There is 1 way.

Therefore, the total number of favorable outcomes for $B$ to get a sum higher than 13 is:

$ 15 + 10 + 6 + 3 + 1 = 35 \text{ ways} $

The total possible outcomes when rolling three dice is:

$ 6^3 = 216 $

Thus, the probability that person $B$ gets a sum higher than 13 is:

$ \frac{35}{216} $

Total possible arrangements

12 people ( 8 teacher +4 students). total number of possible arrangement is $=(12-1)=111$

Favourable arrangements

8 teachers can be arrange in a circle in $(8-1)!=7!$ ways

Now, there are 8 gaps between teachers where students can be seated such that no two students are adjacent.

We need to choose 4 out of 8 gaps $=\binom{8}{4}$

$ \begin{aligned} & =\frac{8!}{4!4!}=\frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!}=7 \times 2 \times 5 \\ & =70 \end{aligned} $

Total favorable arrangements

$ \begin{aligned} & =7!\times 70 \times 4! \\ \text { Probability } & =\frac{\text { Favourable arrangement }}{\text { Total arrangement }} \\ & =\frac{7}{33} \end{aligned} $

Let $E=$ Drawn balls be green

$A=3$ green balls in the bag

$B=4$ green balls in the bag

$C=5$ green balls in the bag

$D=6$ green balls in the bag

So, $P(A)=P(B)=P(C)=P(D)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned}So,\,\,\,\, & P\left(\frac{E}{A}\right)=\frac{{ }^3 C_3}{{ }^6 C_3}=\frac{1}{\frac{6!}{3!3!}}=\frac{1}{20} \\ & P\left(\frac{E}{B}\right)=\frac{{ }^4 C_3}{{ }^6 C_3}=\frac{4}{20} \\ & P\left(\frac{E}{C}\right)=\frac{{ }^5 C_3}{{ }^6 C_3}=\frac{\frac{5!}{3!2!}}{\frac{6!}{3!3!}}=\frac{10}{20} \\ & P\left(\frac{E}{D}\right)=\frac{{ }^6 C_3}{{ }^6 C_3}=1=\frac{20}{20} \end{aligned} $

Thus, $P\left(\frac{C}{E}\right)$

$ =\frac{P(B) \cdot P\left(\frac{E}{C}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) P\left(\frac{E}{B}\right)+P(C) P\left(\frac{E}{C}\right)} +P(D) P\left(\frac{E}{D}\right)$

$ =\frac{P(B) P\left(\frac{E}{B}\right)}{P(B) P\left(\frac{E}{A}\right)+P(B) P\left(\frac{E}{B}\right)+P(B) P\left(\frac{E}{C}\right)} +P(B) P\left(\frac{E}{D}\right)\\$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\, \text { [from Eq. (i)] } $

$ \begin{aligned} & =\frac{P\left(\frac{E}{B}\right)}{P\left(\frac{E}{A}\right)+P\left(\frac{E}{B}\right)+P\left(\frac{E}{C}\right)+P\left(\frac{E}{D}\right)} \\ & =\frac{\frac{10}{20}}{\frac{1}{20}+\frac{4}{20}+\frac{10}{20}+\frac{20}{20}} \\ & =\frac{10}{1+4+10+20}=\frac{2}{7} \end{aligned} $

In a binomial distribution, the mean ($\mu$) is given by $np$, and the standard deviation ($\sigma$) is defined as:

$ \sigma = \sqrt{np(1-p)} $

Given:

$ \mu - \sigma = 3 \quad \text{(i)} $

$ \mu^2 - \sigma^2 = 21 \quad \text{(ii)} $

From equation (ii), using the identity $(a-b)(a+b) = a^2 - b^2$, we get:

$ (\mu - \sigma)(\mu + \sigma) = 21 $

Substituting the value from equation (i):

$ 3(\mu + \sigma) = 21 $

$ \mu + \sigma = 7 \quad \text{(iii)} $

Adding equations (i) and (iii):

$ 2\mu = 10 \Rightarrow \mu = 5 $

Thus:

$ 5 + \sigma = 7 \quad \Rightarrow \quad \sigma = 2 $

Now, with $\mu = np = 5$ and using the expression for standard deviation:

$ \sigma = \sqrt{np(1-p)} = 2 $

Squaring both sides gives:

$ 5(1-p) = 4 \quad \Rightarrow \quad 1-p = \frac{4}{5} \quad \Rightarrow \quad p = \frac{1}{5} $

$ np = 5 \quad \Rightarrow \quad n = \frac{5}{p} = \frac{5}{1/5} = 25 $

For the binomial distribution, the probability of $x = k$ is:

$ P(x = k) = \binom{n}{k} p^k (1-p)^{n-k} $

Calculate $P(x = 1)$ and $P(x = 2)$:

For $k = 1$:

$ P(x=1) = \binom{25}{1} \left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^{24} = 25 \cdot \frac{1}{5} \cdot \left(\frac{4}{5}\right)^{24} = 5\left(\frac{4}{5}\right)^{24} $

For $k = 2$:

$ P(x=2) = \binom{25}{2} \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^{23} = 300 \cdot \frac{1}{25} \cdot \left(\frac{4}{5}\right)^{23} = 12\left(\frac{4}{5}\right)^{23} $

Thus, the ratio $P(x=1) : P(x=2)$ is:

$ \frac{5\left(\frac{4}{5}\right)^{24}}{12\left(\frac{4}{5}\right)^{23}} = \frac{5 \cdot \frac{4}{5}}{12} = \frac{4}{12} = \frac{1}{3} $

Therefore:

$ P(x=1) : P(x=2) = 1 : 3 $

Lets denote $P$ as the normalization constant.

Since, $k$ can take values $1,2,3,4,5$ and 6 we have the probabilities $P(x=k)=k^2 p$.

$ \begin{aligned} & \sum_{k=1}^6 p(x=k)=1 \\ & p \sum_{k=1}^6 k^2=1 \\ & p\left(l^2+2^2+3^2+4^2+5^2+6^2\right)=1 \\ & p \cdot 91=1 \\ & \Rightarrow \quad p=\frac{1}{91} \end{aligned} $

$ \begin{aligned} &\text { Mean of } x\\ &\begin{aligned} E(X) & \left.=\sum_{k=1}^6 k[P(x=k))\right.] \\ & =\sum_{k=1}^6 k \cdot k^2 P=\sum_{k=1}^6 k^3 p \\ E(x) & =P \sum_{k=1}^6 k^3 \\ & =P\left(l^3+2^3+3^3+4^3+5^3+6^3\right) \\ & =P(1+8+27+64+125+216) \\ & =P \cdot 441=\frac{1}{91} 441 \\ E(x) & =\frac{441}{91} \end{aligned} \end{aligned} $

The word S E N S E L E S S N E S

$S$ have 13 letters

$ S=6, E=4, N=2, L=1 $

The total number of distinct arrangement

$ =\frac{13!}{6!4!2!1!} $

Let all the E's as a single unit, then E E E $\mathrm{E}=1$

$\therefore$ The number of distinct arrangements of these 10 units is

$ \frac{10!}{6!!2!1!} $

So, Probability

$ \begin{aligned} & =\frac{\text { Number of favourable arrangements }}{\text { Total number of arrangement }} \\ & =\frac{\frac{10!}{6!21!!1!}}{\frac{13!}{6!4!2!1!}}=\frac{10!4!}{13!}=\frac{10!4!}{13 \cdot 12 \cdot 11 \cdot 10!} \\ & =\frac{4 \cdot 3 \cdot 2 \cdot 1}{13 \cdot 12 \cdot 11}=\frac{2}{143} \end{aligned} $

First we calculate the square of the numbers from 1 to 10 modules 6

$ \begin{aligned} & 1^2 \equiv 1(\bmod 6) \\ & 2^2 \equiv 4(\bmod 6) \\ & 3^2 \equiv 9 \equiv 3(\bmod 6) \\ & 4^2 \equiv 16 \equiv 4(\bmod 6) \\ & 5^2 \equiv 25 \equiv 1(\bmod 6) \\ & 6^2 \equiv 36 \equiv 0(\bmod 6) \\ & 7^2 \equiv 49 \equiv 1(\bmod 6) \\ & 8^2 \equiv 64 \equiv 4(\bmod 6) \\ & 9^2 \equiv 81 \equiv 3(\bmod 6) \\ & 10^2 \equiv 100 \equiv 4(\bmod 6) \end{aligned} $

Next we determine possible values of $x^2+y^2$ modulo 6 which are $0,1,3$ and 4 consider,

$ \begin{aligned} & 0-0 \equiv 0(\bmod 6) \\ & 1-1 \equiv 0(\bmod 6) \\ & 3-3=0(\bmod 6) \\ & 4-4 \equiv 0(\bmod 6) \\ & 1-4 \equiv-3 \equiv 3(\bmod 6) \\ & 4-1 \equiv 3(\bmod 6) \\ & 1-3 \equiv-2 \equiv 4(\bmod 6) \\ & 3-1 \equiv 2(\bmod 6) \\ & 4-3 \equiv 1(\bmod 6) \\ & 3-4 \equiv-1 \equiv 5(\bmod 6) \end{aligned} $

$ \begin{aligned} & 0-1 \equiv-1 \equiv 5(\bmod 6) \\ & 0-3 \equiv-3 \equiv 3(\bmod 6) \\ & 0-4 \equiv-4 \equiv 2(\bmod 6) \\ & 1-0 \equiv 1(\bmod 6) \\ & 4-2 \equiv 2(\bmod 6) \\ & 2-4 \equiv-2 \equiv 4(\bmod 6) \end{aligned} $

From the above only cases where $\left|x^2-y^2\right|$ is divisible by 6 occur, when $x^2 \equiv y^2$ (mod 6), these pairs are

$ \begin{aligned} & x^2 \equiv y^2 \equiv 0(\bmod 6) \\ & x^2 \equiv y^2 \equiv 1(\bmod 6) \\ & x^2 \equiv y^2 \equiv 3(\bmod 6) \\ & x^2 \equiv y^2 \equiv 4(\bmod 6) \end{aligned} $

Now, counting the number of possible pairs $(x, y)$ that satisfy each condition.

$0(\bmod 6)$ is satisfied by 6 .

$1(\bmod 6)$ is satisfied by $1,5,7$.

$3(\bmod 6)$ is satisfied by 3,9 .

$4(\bmod 6)$ is satisfied by $2,4,8,10$

The total number of favorable outcomes

0 has 1 pair $(6,6)$

1 has $3 \times 3=9$ pairs $(\{1,5,7\} \times\{1,5,7\})$

3 has $2 \times 2=4$ pairs $(\{3,9\} \times\{3,9\})$

4 has $4 \times 4=16$ pairs

$ (\{2,4,8,10\} \times\{2,4,8,10\}) $

Total number of favorable outcomes

$ =1+9+4+16=30 $

Total possible outcomes $=10 \times 10=100$

Thus, the probability that $\left|x^2-y^2\right|$ is divisible by 6 is

$ \begin{aligned} & =\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }} \\ & =\frac{30}{100}=\frac{3}{10} \end{aligned} $

Here's the step-by-step explanation:

We have three bags:

Bag A contains 3 white balls and 4 red balls.

Bag B contains 4 white balls and 5 red balls.

Bag C contains 5 white balls and 6 red balls.

We are drawing one ball from each of these bags. We need to find the probability of drawing exactly one white ball and two red balls from these bags.

First, calculate the probability of drawing a white ball from each bag:

From Bag A: $ P(W_A) = \frac{3}{7} $

From Bag B: $ P(W_B) = \frac{4}{9} $

From Bag C: $ P(W_C) = \frac{5}{11} $

Next, calculate the probability of drawing a red ball from each bag:

From Bag A: $ P(R_A) = \frac{4}{7} $

From Bag B: $ P(R_B) = \frac{5}{9} $

From Bag C: $ P(R_C) = \frac{6}{11} $

Now, consider the combinations of drawing one white ball and two red balls:

White from A, Red from B and C:

$ P(W_A) \times P(R_B) \times P(R_C) = \frac{3}{7} \times \frac{5}{9} \times \frac{6}{11} = \frac{90}{693} $

Red from A, White from B, Red from C:

$ P(R_A) \times P(W_B) \times P(R_C) = \frac{4}{7} \times \frac{4}{9} \times \frac{6}{11} = \frac{96}{693} $

Red from A and B, White from C:

$ P(R_A) \times P(R_B) \times P(W_C) = \frac{4}{7} \times \frac{5}{9} \times \frac{5}{11} = \frac{100}{693} $

Add up these probabilities to find the total probability of getting one white and two red balls:

$ \frac{90}{693} + \frac{96}{693} + \frac{100}{693} = \frac{286}{693} = \frac{26}{63} $

Thus, the probability of drawing one white ball and two red balls from the bags is $\frac{26}{63}$.

The possible outcomes to get sum of 4 with a pair of dice are $(1,3),(2,2),(3,1)$

Favourable outcomes $=3$

So, the probability $P$ of getting a sum of 4 in a single throw $=\frac{3}{36}=\frac{1}{12}$

Now, probability (not getting a sum of 4)

$ \bar{P}=1-P=1-\frac{1}{12}=\frac{11}{12} $

If $A$ starts the game, the sequence of throw is $A, B, A, B$ and so on.

$\therefore$ The probability that $B$ wins the game on his first throw.

$ P(\text { B wins })=\frac{11}{12} \times \frac{1}{12}=\frac{11}{144} $

If both $A$ and $B$ fail on their first throw, the same sequence repeats.

Therefore, $P$ (B wins) $=\left(\frac{11}{12}\right)^2 \times P(\mathrm{~B}$ wins)

This gives us a recursive equation

$ \begin{aligned} & P(\mathrm{~B} \text { wins })=\frac{11}{144}+\left(\frac{11}{12}\right)^2 \times P(\mathrm{~B} \text { wins }) \\ & \Rightarrow P(\mathrm{~B} \text { wins })=\frac{11}{144}+\frac{121}{144} \times P \text { (B wins) } \\ & \Rightarrow P \text { (B wins) }\left(1-\frac{121}{144}\right)=\frac{11}{144} \\ & \Rightarrow P \text { (B wins) }\left(\frac{23}{144}\right)=\frac{11}{144} \\ & \Rightarrow P \text { (B wins) }=\frac{11}{23} \end{aligned} $

Hence, the probability that $B$ wins the game is $\frac{11}{23}$.

Given, number of black balls $=3$

and number of red balls $=5$

$\therefore$ Total number of balls $=5+3=8$

Probability of black ball $=\frac{3}{8}$

and Probability of real ball $=\frac{5}{8}$

Now, probability distribution of number of red balls

Given that $ X \sim B(5, p) $ is a binomial random variable and $ P(X=3) = P(X=4) $, we can equate the probabilities:

$ \Rightarrow { }^5C_3 p^3 (1-p)^2 = { }^5C_4 p^4 (1-p) $

Calculating each term:

The binomial coefficient for $ X=3 $ is ${ }^5C_3 = \frac{5!}{3!2!} = 10$

The binomial coefficient for $ X=4 $ is ${ }^5C_4 = \frac{5!}{4!1!} = 5$

Equating and simplifying the expression:

$ 10 p^3 (1-p)^2 = 5 p^4 (1-p) $

$ \Rightarrow 2(1-p) = p \Rightarrow 2 - 2p = p \Rightarrow 3p = 2 \Rightarrow p = \frac{2}{3} $

Thus, $ p = \frac{2}{3} $ and $ q = 1 - p = \frac{1}{3} $.

Now calculate $ P(|X-3|<2) $, which is equivalent to $ P(1 < X < 5) = P(X=2) + P(X=3) + P(X=4) $:

$ P(X=2) = { }^5C_2 \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^3 = 10 \cdot \frac{4}{9} \cdot \frac{1}{27} = \frac{40}{243} $

$ P(X=3) = { }^5C_3 \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = 10 \cdot \frac{8}{27} \cdot \frac{1}{9} = \frac{80}{243} $

$ P(X=4) = { }^5C_4 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right) = 5 \cdot \frac{16}{81} \cdot \frac{1}{3} = \frac{80}{243} $

Adding these probabilities together:

$ P(|X-3|<2) = \frac{40}{243} + \frac{80}{243} + \frac{80}{243} = \frac{200}{243} $

Therefore, $ P(|X-3|<2) = \frac{200}{243} $.

To find the probability that a multiple of 3 does not appear on any of the 12 dice when they are thrown, follow these steps:

Identify Possible Outcomes: When you roll a single die, you have 6 possible outcomes: 1, 2, 3, 4, 5, and 6.

Identify Multiples of 3: The multiples of 3 on a single die are 3 and 6. Therefore, there are 2 favorable outcomes.

Calculate Probability of Getting a Multiple of 3: Use the formula for probability:

$ P(\text{A}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{6} = \frac{1}{3} $

Determine Probability of Not Getting a Multiple of 3: The probability of not getting a multiple of 3 on a single die is:

$ 1 - \frac{1}{3} = \frac{2}{3} $

Compute Probability for 12 Dice: Since each die throw is independent, the probability that none of the 12 dice shows a multiple of 3 is:

$ \left(\frac{2}{3}\right)^{12} $

Therefore, the probability that a multiple of 3 does not appear on any of the 12 dice is:

$ \left(\frac{2}{3}\right)^{12} $

Given:

Total number of boys = 40

Total number of girls = 30

First, calculate the number of girls who are good at Mathematics. To do this, multiply the total number of girls by the percentage of girls who are good at Mathematics:

$ \text{Number of girls good at Mathematics} = 30 \times 40\% = 12 \text{ girls} $

Next, find the number of girls who are not good at Mathematics by subtracting the number of girls who are good at Mathematics from the total number of girls:

$ 30 - 12 = 18 $

Now, calculate the probability that a girl selected at random is not good at Mathematics. Divide the number of girls who are not good at Mathematics by the total number of girls:

$ \frac{18}{30} = \frac{3}{5} $

Therefore, the probability that a randomly selected girl is not good at Mathematics is $\frac{3}{5}$.

Total ways to choose 3 apples from 12.

$ { }^{12} C_3=\frac{12!}{3!(12-3)!}=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220 $

Now, favourable outcome (atmost one rotten apples)

$ \begin{aligned} \text { Zero rotten apples }\left({ }^9 C_3\right) & =\frac{9!}{2!(9-2)!} \\ & =\frac{9 \times 8 \times 7}{3 \times 2 \times 1}=84 \end{aligned} $

One rotten apples ${ }^3 \mathrm{C}_1 \times{ }^9 \mathrm{C}_2=108$

Total favourable outcomes $=84+108$

$ =192 $

$\therefore$ Probability of getting atmost one rotten apples $=\frac{192}{220}=\frac{48}{55}$

Therefore, the probability of drawing atmost one rotten apple when 3 apples are drawn from the basket is $\frac{48}{55}$

Find the mean of the binomial distribution since, then are 2 possible out comes for each coin toss, the probability of getting heads is $P=\frac{1}{2}$. On substitute $P=\frac{1}{2}$ and $n=7$ into formula for the mean of a binomial distribution.

$ \mu=n P=7 \cdot \frac{1}{2}=\frac{7}{2} $

Now, find the variance of the bin 0 mind distribution substitute $P=\frac{1}{2}$ and $n \approx ?$ into formula

$ \begin{aligned} \sigma^2 & =n P(1-P) \\ & =\frac{7}{2}\left(1-\frac{1}{2}\right)=\frac{7}{4} \\ \text { or } \quad \sigma & = \pm \frac{\sqrt{7}}{2} \end{aligned} $

Find the probability of getting 3 heads

$ \begin{aligned} P(X=K) & ={ }^n C_k P^k(1-P)^{n-k} \\ P(3 \text { heads }) & ={ }^7 C_3\left(\frac{1}{2}\right)^3\left(1-\frac{1}{2}\right)^{7-3}=\frac{35}{128} \end{aligned} $

Now, find the required

On substituting $\mu=\frac{7}{2}, \sigma^2=\frac{7}{4}$ and

$ P(3 \text { head })=\frac{35}{128} $

into the given expression $\frac{\frac{7}{2} \times \frac{7}{4}}{\frac{35}{128}}=\frac{112}{5}$

Therefore, the value of $\frac{\mu \sigma^2}{P(X=3)}$ is $\frac{112}{5}$.

$1 \%$ of its product are defective. $n=300$ items

On using binomial distribution $n=300$

$p=0.01$ (probability of an item defectire) and $x$ be the number of defective items PMF (probability mass function) of $X$

$ \begin{aligned} & P(X=k)=\binom{n}{k} p^k q^{n-k}=\binom{n}{k} p^k(1-p)^{n-k} \\ & P(X \leq 1)=P(X=0)+P(X=1) \\ & =\binom{300}{0}(0.01)^0(1-0.01)^{300}+\binom{300}{1}(0.01)^1(1-0.01)^{299} \\ & =\frac{300!}{0!300!} 1 \times(0.99)^{300}+\frac{300!}{299!1!} \times 0.01 \\ & \times(0.99)^{299} \\ & =1 \times 0.0490+300 \times 0.01 \times 0.0495 \\ & =0.0490+0.486687=0.1976087 \\ & \text { So, check option (1) 3/e }=3 / 20.0855 \\ & =0.14936 \end{aligned} $

(2) $2 / e^2=2 / 7.38905=0.270670$

(3) $3 / e^2=3 / 7.38905=0.406006$

(4) $4 / e^3=4 / 20.0855=0.199148$

$ = 0.1976087 $

To find the probability that a randomly chosen five-digit number, formed from the digits 0, 1, 2, 3, and 4 (each used exactly once), is divisible by 4, follow these steps:

First, calculate the total number of five-digit numbers you can form from these digits:

Choose the first digit (to avoid starting with 0): 4 choices (1, 2, 3, or 4).

Choose the second digit: 4 remaining choices.

Choose the third digit: 3 remaining choices.

Choose the fourth digit: 2 remaining choices.

The last digit is the final remaining digit.

Thus, the total number of possible numbers is:

$ 4 \times 4 \times 3 \times 2 \times 1 = 96 $

For a number to be divisible by 4, its last two digits must form a number divisible by 4. From the set of digits {0, 1, 2, 3, 4}, the pairs that are divisible by 4 are: 12, 20, 24, 32, and 40.

Let's consider the formation of numbers with each of these pairs as the last two digits:

Pairs 20, 40:

The first digit (non-zero) has 3 choices.

The second digit has 2 remaining choices.

The third digit is the last remaining digit.

$ 3 \times 2 \times 1 = 6 \text{ arrangements per pair} $

Pairs 12, 24, 32:

The first digit (non-zero) has 2 choices.

The second digit has 2 remaining choices.

The third digit is the last remaining digit.

$ 2 \times 2 \times 1 = 4 \text{ arrangements per pair} $

Now, combine these results:

Two pairs (20, 40) with 6 arrangements each contribute:

$ 2 \times 6 = 12 $

Three pairs (12, 24, 32) with 4 arrangements each contribute:

$ 3 \times 4 = 12 $

Hence, the total number of numbers divisible by 4 is:

$ 18 + 12 = 30 $

Therefore, the probability that a randomly chosen number is divisible by 4 is:

$ \frac{30}{96} = \frac{5}{16} $

Total natural number of cases obtained by taking multiplication of only two numbers out of $100={ }^{100} C_2$

$ n(U)=\frac{100 \times 99}{2}=4950 $

Now, $A$ be the event that product is an even number,

$ \begin{aligned} \therefore \quad n(A) & ={ }^{50} C_1 \times{ }^{50} C_1+{ }^{50} C_2 \\ & =50 \times 50+1225=3725 \end{aligned} $

and out of 100 , there are the numbers 4 , $8, \ldots, 100$ which is divisible by 4 .

Which are 25 in numbers such that when any one of them is multiplied with any of remaining 75 numbers.

Then, the pairs of numbers whose product is divisible by

$ \begin{aligned} 4 & ={ }^{25} C_1 \times{ }^{75} C_1+{ }^{25} C_2 \\ & =25 \times 75+300=1875+300 \\ & =2175 \end{aligned} $

$n(B)=2175$

Since, all the number divisible by 4 is also an even number.

$ \begin{aligned} n(A \cap B) & =2475 \\ n(A \cap \bar{B}) & =n(A)-n(A \cap B) \\ & =3725-2475=1250 \end{aligned} $

$ \begin{aligned} \text { Thus, required probability } & =\frac{n(A \cap \bar{B})}{n(U)} \\ & =\frac{1250}{4950}=\frac{25}{99} \end{aligned} $

In Box $P$

Red Balls 3, White Balls 1, Black balls 2 In Box $Q$

Red Balls 3, White Balls 2, Black balls 4 Total Balls in Box $P=3+1+2=6$

Total Balls in Box $Q=9$

Thus, required probability $=$ Red from $P \times$ White from $Q+$ Red from $P \times$ Black from $Q+$ White from $P \times$ Red from $Q+$ White from $P \times$ Black from $Q+$ Black from $P \times$ Red from $Q+$ Black from $P \times$ White from $Q$

$ \begin{aligned} & { }^3 C_1 \times{ }^2 C_1+{ }^3 C_1 \times{ }^4 C_1+{ }^1 C_1 \times{ }^3 C_1 \\ = & \frac{+{ }^1 C_1 \times{ }^4 C_1+{ }^2 C_1 \times{ }^3 C_1+{ }^2 C_1 \times{ }^2 C_1}{{ }^6 C_1 \times{ }^9 C_1} \\ = & \frac{6+12+3+4+6+4}{54}=\frac{35}{54} \end{aligned} $