Probability

$E_1$ : Card drawn is king

$E_2$ : Card drawn is not king

$A$ : Man reports it is a king

We have,

$ \begin{aligned} P\left(E_1\right) & =\frac{4}{52}=\frac{1}{13} \\ P\left(E_2\right) & =\frac{12}{13} \\ P\left(\frac{A}{E_1}\right) & =\frac{3}{4} \\ \text { and } P\left(\frac{A}{E_2}\right) & =\frac{1}{4} \end{aligned} $

By Baye's rule,

$ \begin{aligned} P\left(\frac{E_1}{A}\right) & =\frac{P\left(E_1\right) P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)} \\ & =\frac{\left(\frac{1}{13}\right)\left(\frac{3}{4}\right)}{\left(\frac{1}{13}\right)\left(\frac{3}{4}\right)+\left(\frac{12}{13}\right)\left(\frac{1}{4}\right)} \\ & =\frac{3}{3+12}=\frac{3}{15}=\frac{1}{5} \end{aligned} $

Mean - Variance $=1$

$ \begin{aligned} & & n p-n p q & =1 \\ \Rightarrow & & n p(1-q) & =1 \quad \ldots \text { (i) } \\ \Rightarrow & & n p^2 & =1 \end{aligned} $

$ \begin{array}{l} \text { Also, } & & n^2 p^2-n^2 p^2 q^2 & =11 \\ \Rightarrow & & n^2 p^2\left(1-q^2\right) & =11 \\ \Rightarrow & & n^2 p^2(1-q)(1+q) & =11 \\ \Rightarrow & & n^2 p^2 \cdot p \cdot(1+q) & =11 \\ \Rightarrow & & n p^2 \cdot n p(1+q) & =11 \\ \Rightarrow & & n p(1+q) & =11 \\ \Rightarrow & & n p+n p q & =11 \quad \ldots \text { (ii) } \end{array} $

From Eqs. (i) and (ii), we get

$ n p=6 $

$ \text { and } \quad n p q=5 $

Now, $n p^2=1$

$ \begin{array}{ll} \Rightarrow & 6 \cdot p=1 \Rightarrow p=\frac{1}{6} \\ \therefore & q=\frac{5}{6} \text { and } n=36 \end{array} $

Also, $p(x=2)=m\left(\frac{5}{6}\right)^n$

$ \begin{aligned} & \Rightarrow \quad{ }^{36} C_2\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{34}=m\left(\frac{5}{6}\right)^n \\ & \Rightarrow \quad m=\frac{126}{5} \\ & \Rightarrow \quad m: n=7: 10 \end{aligned} $

Probability to hit a target $(p)=\frac{1}{3}$

and not hit the target $(q)=\frac{2}{3}$

Total times he fires $=4$

$\therefore$ Probability of hit the target atleast thrice

$ \begin{aligned} & =P(X \geq 3) \\ & =P(X=3)+P(X=4) \\ & ={ }^4 C_3\left(\frac{2}{3}\right)^3 \cdot\left(\frac{1}{3}\right)+{ }^4 C_4\left(\frac{2}{3}\right)^4 \cdot\left(\frac{1}{3}\right)^0 \\ & =\frac{4}{3^3} \times \frac{2^3}{3}+\frac{2^4}{3^4} \\ & =\frac{32}{3^4}+\frac{16}{3^4}=\frac{48}{81}=\frac{16}{27} \end{aligned} $

Explanation

(I) Mutually Exclusive Events

Since $A$ and $B$ are mutually exclusive, they have no outcomes in common. This implies:

$ A \cap B = \emptyset $

Thus, we have:

$ P(A \cap B) = 0 $

Therefore:

$ P(A \cup B) = P(A) + P(B) $

(II) Independent Events

For independent events $A$ and $B$, we have:

$ P(A \cap B) = P(A) \cdot P(B) $

The probability of the complement of $A$ given $B$ is:

$ P\left(\frac{\bar{A}}{B}\right) = 1 - P\left(\frac{A}{B}\right) $

Calculating further:

$ P\left(\frac{\bar{A}}{B}\right) = 1 - \frac{P(A \cap B)}{P(B)} $

$ P\left(\frac{\bar{A}}{B}\right) = 1 - \frac{P(A) \cdot P(B)}{P(B)} $

Thus:

$ P\left(\frac{\bar{A}}{B}\right) = 1 - P(A) $

(III) Event Intersection Equal to A

Given:

$ A \cap B = A $

This implies that all outcomes of $A$ also occur in $B$. Therefore, the probability satisfies:

$ P(A) \leq P(B) $

(IV) Union Equals Sample Space

Since $A \cup B = S$, the entire sample space is covered:

$ P(A \cup B) = P(S) = 1 $

Using the formula for the probability of a union:

$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $

Equating this to the total probability, we have:

$ 1 = P(A) + P(B) - P(A \cap B) $

Rearranging yields:

$ P(A \cap B) = P(A) + P(B) - 1 $

Substituting the complement:

$ P(A \cap B) = P(B) - (1 - P(A)) $

Finally:

$ P(A \cap B) = P(B) - P(\bar{A}) $

To solve this problem, we need to calculate $P(A \mid A \cap B) + P(B \mid A \cap B)$.

Step 1: Determine $P(A \mid A \cap B)$.

Using the formula for conditional probability, we have:

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

Since $A \cap (A \cap B) = A \cap B$, it follows that:

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

Step 2: Determine $P(B \mid A \cap B)$.

Similarly, using the conditional probability formula:

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

Since $B \cap (A \cap B) = A \cap B$, it follows that:

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

Conclusion:

Adding the two results:

$ P(A \mid A \cap B) + P(B \mid A \cap B) = 1 + 1 = 2 $

We have,

$ S=\{1,2,3,4,5,6,7,8,9\} $

For even sum $\rightarrow$ both number are even

$\rightarrow$ both number are odd

even numbers $=2,4,6,8$

odd numbers $=1,3,5,7,9$

Then, $P(B / A)=\frac{P(B \cap A)}{P(A)}$

Where $A$ is an event when sum is evenand $B$ is an event when numbers are odd.

$ \begin{aligned} P(B \cap A) & =\frac{{ }^5 C_2}{{ }^9 C_2} \\ P(A) & =\frac{{ }^5 C_2+{ }^4 C_2}{{ }^9 C_2} \end{aligned} $

$ \begin{aligned} P(B / A) & =\frac{P(B \cap A)}{P(A)}=\frac{{ }^5 C_2 /{ }^9 C_2}{\frac{{ }^5 C_2+{ }^4 C_2}{{ }^9 C_2}} \\ & =\frac{{ }^5 C_2}{{ }^5 C_2+{ }^4 C_2}=\frac{10}{10+6}=\frac{10}{16}=\frac{5}{8} \end{aligned} $

We are given that $A$, $B$, and $C$ are mutually exclusive and exhaustive events of a random experiment, and $E$ is an event that occurs together with either $A$, $B$, or $C$. The conditional probabilities are provided as follows:

$P(E \mid A) = 0.6$

$P(E \mid B) = 0.3$

$P(E \mid C) = 0.1$

Along with the probabilities:

$P(A) = 0.3$

$P(B) = 0.5$

These conditions lead us to find $P(C/E)$.

Since $A$, $B$, and $C$ are mutually exclusive and exhaustive, $P(C) = 1 - P(A) - P(B) = 1 - 0.3 - 0.5 = 0.2$.

To find $P(E)$, we use the law of total probability:

$ P(E) = P(E \mid A) \cdot P(A) + P(E \mid B) \cdot P(B) + P(E \mid C) \cdot P(C) $

Substituting in the given values:

$ P(E) = (0.6)(0.3) + (0.3)(0.5) + (0.1)(0.2) $

Calculating each term:

$0.6 \cdot 0.3 = 0.18$

$0.3 \cdot 0.5 = 0.15$

$0.1 \cdot 0.2 = 0.02$

Adding these results:

$ P(E) = 0.18 + 0.15 + 0.02 = 0.35 $

Now, to find $P(C \mid E)$, we use Bayes' theorem:

$ P(C \mid E) = \frac{P(E \mid C) \cdot P(C)}{P(E)} $

Substituting the known values:

$ P(C \mid E) = \frac{(0.1)(0.2)}{0.35} = \frac{0.02}{0.35} = \frac{2}{35} $

To find the mean of the discrete random variable $ X $, we start with the given probability distribution:

$ \begin{array}{c|c|c|c|c|c|c} X = x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X = x) & \frac{1}{10} & \frac{K+2}{10} & \frac{K+3}{10} & \frac{K+3}{10} & \frac{K+4}{10} & \frac{K+2}{10} \\ \end{array} $

Step 1: Determine $ K $

The sum of all probabilities should be equal to 1:

$ \frac{1}{10} + \frac{K+2}{10} + \frac{K+3}{10} + \frac{K+3}{10} + \frac{K+4}{10} + \frac{K+2}{10} = 1 $

Simplifying the equation:

$ \frac{1 + 5K + 14}{10} = 1 $

$ 5K + 15 = 10 $

$ 5K = -5 \quad \Rightarrow \quad K = -1 $

Step 2: Calculate Individual Probabilities

Using $ K = -1 $:

$ P(X = -1) = \frac{-1 + 2}{10} = \frac{1}{10} $

$ P(X = 0) = \frac{-1 + 3}{10} = \frac{2}{10} $

$ P(X = 1) = \frac{-1 + 3}{10} = \frac{2}{10} $

$ P(X = 2) = \frac{-1 + 4}{10} = \frac{3}{10} $

$ P(X = 3) = \frac{-1 + 2}{10} = \frac{1}{10} $

Step 3: Find Mean $ \mu = E(X) $

$ \mu = E(X) = \sum x_i \cdot P(X = x_i) $

$ = (-2) \cdot \frac{1}{10} + (-1) \cdot \frac{1}{10} + 0 \cdot \frac{2}{10} + 1 \cdot \frac{2}{10} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{10} $

$ = -\frac{2}{10} - \frac{1}{10} + 0 + \frac{2}{10} + \frac{6}{10} + \frac{3}{10} $

$ = \frac{8}{10} = \frac{4}{5} $

Therefore, the mean $ \mu $ of the random variable $ X $ is $\frac{4}{5}$.

To find the probability ratio $ P_1 : P_2 $ where $ P_1 $ is the probability of obtaining a sum of 6 at least once in 9 throws of two dice, and $ P_2 $ is the probability of obtaining a sum of 8 at least once in the same number of throws, follow these steps:

Probability Calculation

Total Possible Outcomes: Each die has 6 faces, so there are a total of $ 6 \times 6 = 36 $ possible outcomes when throwing two dice.

Outcomes for Sum of 6:

Possible pairs: $ (1,5), (2,4), (3,3), (4,2), (5,1) $.

Number of outcomes: 5

Probability of not getting a sum of 6 in one throw is:

$ \frac{36 - 5}{36} = \frac{31}{36} $

Probability of Not Getting Sum of 6 in 9 Throws:

$ \left(\frac{31}{36}\right)^9 $

Probability of Getting Sum of 6 at Least Once ($ P_1 $):

$ P_1 = 1 - \left(\frac{31}{36}\right)^9 $

Outcomes for Sum of 8:

Possible pairs: $ (2,6), (3,5), (4,4), (5,3), (6,2) $.

Number of outcomes: 5

Probability of not getting a sum of 8 in one throw is:

$ \frac{36 - 5}{36} = \frac{31}{36} $

Probability of Not Getting Sum of 8 in 9 Throws:

$ \left(\frac{31}{36}\right)^9 $

Probability of Getting Sum of 8 at Least Once ($ P_2 $):

$ P_2 = 1 - \left(\frac{31}{36}\right)^9 $

Ratio Calculation

Since the expressions for $ P_1 $ and $ P_2 $ are identical:

$ P_1 : P_2 = \frac{1 - \left(\frac{31}{36}\right)^9}{1 - \left(\frac{31}{36}\right)^9} = 1:1 $

Thus, the probability ratio is $ 1:1 $.

Total number of ways in which 7 balls can be placed in 4 boxes $=4^7$

Out of 7 balls, 3 balls are choosen for the first box in ${ }^7 C_3$ ways.

Now, remaining 4 balls can be placed in remaining 3 boxes in $3^4$ ways.

$ \begin{aligned} & \therefore \text { Required probability }=\frac{{ }^7 C_3 \times 3^4}{4^7} \\ & =\frac{7 \times 6 \times 5}{1 \times 2 \times 3} \times \frac{3^4}{4^7}=\frac{35}{64}\left(\frac{3}{4}\right)^4 \end{aligned} $

Let the first 5 consecutive natural numbers be $1,2,3,4,5$

Total number of pairs of $(x, y)={ }^5 C_2=10$

We need to check whether $x^4-y^4$ is divisible by 5 .

$ \begin{aligned} & \because \quad 1^4=1 \equiv 1 \bmod 5 \\ & 2^4=16 \equiv 1 \bmod 5 \\ & 3^4=81 \equiv 1 \mathrm{mod} 5 \\ & 4^4=256 \equiv 1 \mathrm{mod} 5 \\ & 5^4=625 \equiv 0 \bmod 5 \end{aligned} $

Both $x$ and $y$ are chosen from $\{1,2,3$, 4).

$\therefore$ Favourable pairs $={ }^4 C_2=6$

$\therefore$ Required probability

$ \begin{aligned} & =\frac{\text { Number of favourable pairs }}{\text { Total number of pairs }} \\ & =\frac{6}{10}=\frac{3}{5} \end{aligned} $

To find the probability that the last ball drawn from the bag is red, consider the following:

The bag contains 2 white (W), 3 green (G), and 5 red (R) balls.

We need to calculate the probability of different combinations of drawing three balls where the last ball is red.

The possible successful sequences with a red ball at the end are:

Drawing a white, another white, and then a red (WWR)

Drawing two green balls and then a red (GGR)

Drawing a white, then a green, and then a red (WGR)

Drawing a green, then a white, and then a red (GWR)

Drawing one white, then two reds (WRR)

Drawing one green, then two reds (GRR)

Drawing a red first, then a white, then a red (RWR)

Drawing a red first, then a green, then a red (RGR)

All three drawn balls are red (RRR)

The probability calculation is as follows:

$ \begin{aligned} P(\text{last ball is red}) &= P(WWR) + P(GGR) + P(WGR) + P(GWR) \\ &\quad + P(WRR) + P(GRR) + P(RWR) + P(RGR) \\ &\quad + P(RRR) \\ &= \frac{2 \times 1 \times 5}{10 \times 9 \times 8} + \frac{3 \times 2 \times 5}{10 \times 9 \times 8} + \frac{2 \times 3 \times 5}{10 \times 9 \times 8} \\ &\quad + \frac{3 \times 2 \times 5}{10 \times 9 \times 8} + \frac{2 \times 5 \times 4}{10 \times 9 \times 8} + \frac{3 \times 5 \times 4}{10 \times 9 \times 8} \\ &\quad + \frac{5 \times 2 \times 4}{10 \times 9 \times 8} + \frac{5 \times 3 \times 4}{10 \times 9 \times 8} + \frac{5 \times 4 \times 3}{10 \times 9 \times 8} \\ &= \frac{10 + 30 + 30 + 30 + 40 + 60 + 40 + 60 + 60}{10 \times 9 \times 8} \\ &= \frac{360}{720} \\ &= \frac{1}{2} \end{aligned} $

Therefore, the probability that the last ball drawn is red is $\frac{1}{2}$.

To determine the probability that a black ball is from the first set of bags, let's define the following:

B: Probability of drawing a black ball.

$B_1$: Probability associated with drawing from the first set of bags, where each bag contains 3 white and 5 black balls.

$B_2$: Probability associated with drawing from the second set of bags, where each bag contains 6 white and 4 black balls.

Given:

Probability of choosing a bag from the first set ($P(B_1)$) is $ \frac{1}{2} $.

Probability of choosing a bag from the second set ($P(B_2)$) is also $ \frac{1}{2} $ since there are two bags in the first set and four bags in the second set but we simplify by considering overall probabilities.

Probabilities of drawing a black ball from each type of bag:

$P(B \mid B_1) = \frac{5}{8}$ for Bag 1.

$P(B \mid B_2) = \frac{4}{10} = \frac{2}{5}$ for Bag 2.

Calculation:

The total probability of drawing a black ball ($P(B)$) is calculated by adding the probabilities of drawing a black ball from each type of bag, weighted by the probability of choosing each type of bag:

$ \begin{aligned} P(B) & = P(B_1) \cdot P(B \mid B_1) + P(B_2) \cdot P(B \mid B_2) \\ & = \frac{1}{2} \times \frac{5}{8} + \frac{1}{2} \times \frac{2}{5} \\ & = \frac{5}{16} + \frac{1}{5} \\ & = \frac{41}{80} \end{aligned} $

Final Probability:

Using Bayes' Theorem, we find the probability that a black ball is from the first set of bags:

$ \begin{aligned} P(B_1 \mid B) & = \frac{P(B \mid B_1) \cdot P(B_1)}{P(B)} \\ & = \frac{\frac{5}{8} \times \frac{1}{2}}{\frac{41}{80}} \\ & = \frac{5}{16} \times \frac{80}{41} \\ & = \frac{25}{41} \end{aligned} $

Thus, the probability that the black ball is from the first set of bags is $ \frac{25}{41} $.

let $X$ be the random variable representing the number of kings drawn.

$ \begin{aligned} & P(X=0)=\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{188}{221} \\ & P(X=1)=\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4 \times 48 \times 2}{52 \times 51}=\frac{32}{221} \\ & P(X=2)=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221} \\ & E(X)=p_i x_i=0+\frac{32}{221}+\frac{2}{221}=\frac{34}{221}=\frac{2}{13} \end{aligned} $

P(defective articles)

$ \begin{aligned} &=\frac{3}{15}=\frac{1}{5}=p(let) \\ & q=1-p=1-\frac{1}{5}=\frac{4}{5} \end{aligned} $

Here, $n=5$

$ P(X=2)={ }^5 C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^{5-2}=\frac{10 \times 64}{3125}=\frac{128}{625} $

To determine the probability that at least one letter is placed in the wrong envelope, consider the following:

Total Arrangements: There are 5 letters, each meant for a specific envelope. The total number of ways to place the letters is calculated using permutations because each letter can go into any of the 5 envelopes. This number is given by $5! = 120$.

Correct Arrangement: There is only 1 way to place all letters in the correctly addressed envelopes, which means every letter is in its intended envelope.

At Least One Mistake: To find the probability that at least one letter is in the wrong envelope, we subtract the probability that all letters are correctly placed from 1. The probability that all letters are correctly placed is $\frac{1}{120}$.

Therefore, the probability that at least one letter is placed in the wrong envelope is:

$ 1 - \frac{1}{120} = \frac{119}{120} $

Here, $x=8, P=\frac{1}{2}$ and $q=\frac{1}{2}$

frobabllity that he will pass

$ \begin{aligned} & =F X=\theta+P(X=7)+P(X=8 \\ & ={ }^8 C_4\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2+{ }^8 C_7\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)+{ }^8 C_3\left(\frac{1}{2}\right)^2 \\ & =\left(\frac{1}{2}\right)^2 P^3 C_2+{ }^3 C_1+{ }^8 C_4 1=\frac{37}{256} \end{aligned} $

Probability that he will fail is the cxamination $ =1-\frac{37}{256}=\frac{219}{256} $

To find the probability that a person traveled to college by car, given that they arrived on time, let's denote the events as follows:

$ E_1 $: The event that a person travels by car.

$ E_2 $: The event that a person travels by bus.

$ E_3 $: The event that a person travels by train.

The probabilities for choosing each mode of transport are:

$ P(E_1) = \frac{1}{5}, \quad P(E_2) = \frac{2}{5}, \quad P(E_3) = \frac{3}{5} $

Let $ A $ be the event of reaching the college on time. The given probabilities of being late for each mode of transport are:

$ P(\bar{A} \mid E_1) = \frac{2}{7} $ leading to $ P(A \mid E_1) = 1 - \frac{2}{7} = \frac{5}{7} $

$ P(\bar{A} \mid E_2) = \frac{4}{7} $ leading to $ P(A \mid E_2) = 1 - \frac{4}{7} = \frac{3}{7} $

$ P(\bar{A} \mid E_3) = \frac{1}{7} $ leading to $ P(A \mid E_3) = 1 - \frac{1}{7} = \frac{6}{7} $

Using Bayes' Theorem, we calculate $ P(E_1 \mid A) $, the probability that the person traveled by car, given they arrived on time:

$ P(E_1 \mid A) = \frac{P(A \mid E_1) \cdot P(E_1)}{P(A \mid E_1) \cdot P(E_1) + P(A \mid E_2) \cdot P(E_2) + P(A \mid E_3) \cdot P(E_3)} $

Substituting the known values:

$ = \frac{\left(\frac{5}{7}\right) \times \left(\frac{1}{5}\right)}{\left(\frac{5}{7} \times \frac{1}{5}\right) + \left(\frac{3}{7} \times \frac{2}{5}\right) + \left(\frac{6}{7} \times \frac{3}{5}\right)} $

Calculate the numerator and the denominator:

$ = \frac{\frac{1}{5} \times \frac{5}{7}}{\frac{1}{5} \times \frac{5}{7} + \frac{2}{5} \times \frac{3}{7} + \frac{3}{5} \times \frac{6}{7}} $

Simplify:

$ = \frac{\frac{5}{35}}{\frac{5}{35} + \frac{6}{35} + \frac{18}{35}} = \frac{\frac{5}{35}}{\frac{29}{35}} = \frac{5}{29} $

Thus, the probability that the person traveled by car, given they arrived on time, is $\frac{5}{29}$.

To find the probability that the target is hit by $P$ or $Q$ but not by $R$, let's define the events:

$E_1$: $P$ hits the target.

$E_2$: $Q$ hits the target.

$E_3$: $R$ hits the target.

The given probabilities are:

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

$P(E_2) = \frac{3}{5}$

$P(E_3) = \frac{5}{7}$

These events are independent, so:

The probability that $P$ hits and both $Q$ and $R$ miss is:

$ P(E_1 \cap \overline{E_2} \cap \overline{E_3}) = P(E_1) \cdot (1 - P(E_2)) \cdot (1 - P(E_3)) = \frac{2}{3} \cdot \frac{2}{5} \cdot \frac{2}{7} $

The probability that $Q$ hits and both $P$ and $R$ miss is:

$ P(\overline{E_1} \cap E_2 \cap \overline{E_3}) = (1 - P(E_1)) \cdot P(E_2) \cdot (1 - P(E_3)) = \frac{1}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} $

The probability that both $P$ and $Q$ hit but $R$ misses is:

$ P(E_1 \cap E_2 \cap \overline{E_3}) = P(E_1) \cdot P(E_2) \cdot (1 - P(E_3)) = \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} $

Now, summing these probabilities gives the required probability that the target is hit by $P$ or $Q$, but not by $R$:

$ \begin{align*} P(\text{hit by } P \text{ or } Q \text{ but not } R) &= P(E_1 \cap \overline{E_2} \cap \overline{E_3}) + P(\overline{E_1} \cap E_2 \cap \overline{E_3}) + P(E_1 \cap E_2 \cap \overline{E_3}) \\ &= \frac{2}{3} \cdot \frac{2}{5} \cdot \frac{2}{7} + \frac{1}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} + \frac{2}{3} \cdot \frac{3}{5} \cdot \frac{2}{7} \\ &= \frac{8}{105} + \frac{6}{105} + \frac{12}{105} \\ &= \frac{26}{105} \end{align*} $

Thus, the probability that the target is hit by $P$ or $Q$ but not by $R$ is $\frac{26}{105}$.

To solve for the probability that exactly 3 out of 5 bulbs chosen are defective, given that 20% of the bulbs are defective, we can use the binomial probability formula. Here's how it's done:

Define the variables:

Total number of trials, $ n = 5 $

Probability of a defective bulb, $ p = 20\% = \frac{1}{5} $

Probability of a non-defective bulb, $ q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5} $

Calculate the probability:

We need to find the probability of exactly 3 defective bulbs, $ P(X=3) $.

Apply the binomial probability formula:

$ P(X = 3) = {^5C_3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2 $

Calculate the individual components:

The coefficient, $ ^5C_3 $, is the number of ways to choose 3 defective bulbs from 5. This is calculated as:

$ ^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = 10 $

Plug in the values:

$ \left(\frac{1}{5}\right)^3 = \frac{1}{125} $

$ \left(\frac{4}{5}\right)^2 = \frac{16}{25} $

Calculate the probability:

$ P(X = 3) = 10 \times \frac{1}{125} \times \frac{16}{25} = 10 \times \frac{16}{3125} = \frac{160}{3125} = \frac{32}{625} $

Therefore, the probability that exactly 3 out of the 5 bulbs chosen are defective is $\frac{32}{625}$.

To find the probability that a random variable $ X $, which follows a Poisson distribution with a mean ($\lambda$) of 5, is less than 3, we calculate $ P(X < 3) $.

This is equivalent to finding the sum of probabilities $ P(X = 0) $, $ P(X = 1) $, and $ P(X = 2) $:

$ \begin{aligned} P(X < 3) &= P(X = 0) + P(X = 1) + P(X = 2) \\ &= \frac{\lambda^0 e^{-\lambda}}{0!} + \frac{\lambda^1 e^{-\lambda}}{1!} + \frac{\lambda^2 e^{-\lambda}}{2!}. \end{aligned} $

Substituting $\lambda = 5$, we have:

$ \begin{aligned} P(X < 3) &= e^{-5} \left( 1 + 5 + \frac{25}{2} \right) \\ &= e^{-5} \left( \frac{2}{2} + \frac{10}{2} + \frac{25}{2} \right) \\ &= e^{-5} \times \frac{37}{2} \\ &= \frac{37}{2} e^{-5}. \end{aligned} $

Let $X$ be the number rolled on the die, and let $W$ be the event that all balls drawn are white. We want to find the probability $P(W)$, which can be calculated using the law of total probability as follows :

$P(W) = \sum\limits_{x=1}^{6} P(W|X=x)P(X=x)$

The probability of rolling any number from 1 to 6 on the die is equal, so $P(X=x) = \frac{1}{6}$ for all $x \in \{1, 2, 3, 4, 5, 6\}$.

Now let's calculate the conditional probabilities $P(W|X=x)$ for each possible value of $x$ :

1. $P(W|X=1) = {{{}^6{C_1}} \over {{}^{10}{C_1}}}= \frac{6}{10} = \frac{3}{5}$, since there are 6 white balls out of a total of 10 balls.

2. $P(W|X=2) = {{{}^6{C_2}} \over {{}^{10}{C_2}}} = \frac{15}{45} = \frac{1}{3}$, since there are 15 ways to choose 2 white balls out of 6, and 45 ways to choose 2 balls out of 10.

3. $P(W|X=3) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = \frac{20}{120} = \frac{1}{6}$, since there are 20 ways to choose 3 white balls out of 6, and 120 ways to choose 3 balls out of 10.

4. $P(W|X=4) = {{{}^6{C_4}} \over {{}^{10}{C_4}}} = \frac{15}{210} = \frac{1}{14}$, since there are 15 ways to choose 4 white balls out of 6, and 210 ways to choose 4 balls out of 10.

5. $P(W|X=5) = {{{}^6{C_5}} \over {{}^{10}{C_5}}} = \frac{6}{252} = \frac{1}{42}$, since there are 6 ways to choose 5 white balls out of 6, and 252 ways to choose 5 balls out of 10.

6. $P(W|X=6) = {{{}^6{C_6}} \over {{}^{10}{C_6}}} = \frac{1}{210}$, since there are 1 ways to choose 6 white balls out of 6, and 210 ways to choose 6 balls out of 10.

Using the law of total probability, we have :

$P(W) = \frac{1}{6} \left(P(W|X=1) + P(W|X=2) + P(W|X=3) + P(W|X=4) + P(W|X=5) + P(W|X=6)\right)$

$P(W) = \frac{1}{6} \left(\frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}\right)$

To simplify this expression, find a common denominator :

$P(W) = \frac{1}{6} \left(\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210}\right)$

Add the fractions :

$P(W) = \frac{1}{6}\left(\frac{126+70+35+15+5+1}{210}\right)=\frac{42}{210}=\frac{1}{5}$

$ \begin{aligned} & n p-n p q=1 \\\\ \Rightarrow & n p(1-q)=1 \\\\ \Rightarrow & n p^2=1 \end{aligned} $

$ \begin{aligned} & 2 P(X=2)=3 P(X=1) \\\\ & 2 \cdot{ }^n C_2 p^2 q{ }^{n-2}=3 \cdot{ }^n C_1 p \cdot q^{n-1} \\\\ \Rightarrow & 2 \cdot \frac{n \cdot(n-1)}{2} \cdot p=3 \cdot n \cdot q \\\\ \Rightarrow & (n-1) p=3(1-p) \\\\ \Rightarrow & \left(\frac{1}{p^2}-1\right) p=3(1-p) \\\\ \Rightarrow & \frac{(1-p)(1+p)}{p}=3(1-p) \\\\ \Rightarrow & 1+p=3 p \\\\ \Rightarrow & p=\frac{1}{2} \\\\ \therefore & n=4 \end{aligned} $

$ \begin{aligned} n^2 P(x>1) & =n^2(1-P(x=1)-P(x=0)) \\\\ & =16\left(1-{ }^4 C_1 \cdot\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4\right)=11 \end{aligned} $

The given probabilities for getting a head (H) and a tail (T) are as follows:

$ P(H) = \frac{3}{4}, \quad P(T) = \frac{1}{4} $

The random variable X can take the values 1, 2, or 3. These correspond to the following events:

- X = 1 : A head is obtained on the first toss. This happens with probability $P(H) = \frac{3}{4}$.

- X = 2 : A tail is obtained on the first toss and a head on the second. This happens with probability $P(T)P(H) = \frac{1}{4} \times \frac{3}{4}$.

- X = 3 : Either two tails and then a head are obtained, or three tails are obtained.
This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3$.

Now, we calculate the mean (expected value) of X:

$ \begin{aligned} E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\ & = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\ & = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\ & = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\ & = 3 \cdot \left(\frac{7}{16}\right) \\\\ & = \frac{21}{16}. \end{aligned} $

$ \begin{array}{|c|l|c|} \hline \alpha-\beta & {\text { Case }} & \mathbf{P} \\ \hline 5 & (6,1) & 1 / 36 \\ \hline 4 & (6,2)(5,1) & 2 / 36 \\ \hline 3 & (6,3)(5,2)(4,1) & 3 / 36 \\ \hline 2 & (6,4)(5,3)(4,3)(3,1) & 4 / 36 \\ \hline 1 & (6,5)(5,4)(4,3)(3,2)(2,1) & 5 / 36 \\ \hline 0 & (6,6)(5,5) \ldots \ldots(1,1) & 6 / 36 \\ \hline-1 & ----- & 5 / 36 \\ \hline-2 & -----& 4 / 36 \\ \hline-3 & ----- & 3 / 36 \\ \hline-4 & (2,6)(1,5) & 2 / 36 \\ \hline-5 & (1,6) & 1 / 36 \\ \hline \end{array} $

$E[X^2] = \sum\limits_{i=-5}^{5} (x_i)^2 P(x_i)$

Substituting the values from table :

$E[X^2] = 2\left[\frac{25}{36}+\frac{32}{36}+\frac{27}{36}+\frac{16}{36}+\frac{5}{36}\right] = \frac{105}{18} = \frac{35}{6}$

Next, we calculate the expected value of the differences. The expected value is calculated as the sum of the products of each outcome and its corresponding probability. Given that the table is symmetric around 0, the expected value is 0.

$E[X] = \sum\limits_{i=-5}^{5} x_i P(x_i) = 0$

Now, we can calculate the variance, which is the expected value of the squared differences minus the square of the expected value of the differences :

$Var[X] = E[X^2] - (E[X])^2 = \frac{35}{6} - 0^2 = \frac{35}{6}$

Here, $p = 35$ and $q = 6$, and they are co-prime.

The positive divisors of 35 are 1, 5, 7, and 35. The sum of these divisors is $1 + 5 + 7 + 35 = 48$.

We have, $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$

Let $M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, where $a, b, c, d \in\{0,1,2\}$

$ n(s)=3^4=81 $

If $A$ is invertible, then $|A| \neq 0$

Now, if $|A|=0$, then $|M|=0$

$\therefore a d-b c=0$ or $a d=b c$

Case I : When $a d=b c=0$, then

There are five ways when $a d=0$ i.e.,

$(a, d)=(0,0),(0,1),(0,2),(1,0),(2,0)$

Similarly, there are again five ways, when $b c=0$

$\therefore$ There are total $5 \times 5=25$ ways, when $a d=b c=0$

Case II : When $a d=b c=1$

There is only one way, when $a d=b c=1$

$ \text { i.e. } \quad a=b=c=d=1 $

Case III : When $a d=b c=2$

There are two ways, when $a d=2$, i.e.

$ (a, d)=(1,2) \text { or }(2,1) $

Similarly, there are two ways

when $b c=2$ i.e., $(b, c)=(1,2)$ or $(2,1)$

Case IV : When $a d-b c=4$

There is only way, when $a d=b c=4$

$ \text { i.e., } a=b=c=d=2 $

$\therefore$ Total number of ways, when

$ \begin{aligned} & (\bar{A})=\frac{31}{81}|A|=0 \text { is } 25+1+4+1=31 \\\\ & \text { Hence, } P(A)=1-P(\bar{A})=1-\frac{31}{81}=\frac{50}{81} \end{aligned} $

Given that,

$\mathrm{P}($ odd number seven times $)=\mathrm{P}($ Odd number nine times)

$ \begin{aligned} & \Rightarrow{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{n-7}={ }^n \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{n-9} \\\\ & \Rightarrow{ }^n \mathrm{C}_7={ }^n \mathrm{C}_9 \\\\ & \Rightarrow n=7+9=16 \end{aligned} $

$ \begin{aligned} & \text { Hence, } \mathrm{P}(\text { Even number twice })=\frac{k}{2^{15}} \\\\ & \Rightarrow{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2}=\frac{k}{2^{15}} \\\\ & \Rightarrow \frac{16 \times 15}{2} \times \frac{1}{2^{16}}=\frac{k}{2^{15}} \Rightarrow k=60 \end{aligned} $

$N$ denote the sum of the numbers obtained when two dice are rolled.

Such that $2^N < N$!

$ \text { i.e., } 4 \leq N \leq 12 \text { i.e., } N \in\{4,5,6, \ldots 12\} $

Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$

So, required probability $=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$

$ 4 m-3 n=4 \times 11-3 \times 12=44-36=8 $

As, we know that sum of all the probabilities $=1$

$ \begin{aligned} & \text { So, } \sum_{x=1}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\\\ & \Rightarrow k\left[1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+\ldots . \infty\right]=1 \end{aligned} $

$ \begin{aligned} & \text { Let } S=1+\frac{2}{3}+\frac{3}{3^2}+\ldots .+\infty \\\\ & \Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots .+\infty \end{aligned} $

On subtracting, we get

$ \begin{aligned} & \frac{2 S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots .+\infty \\\\ & \Rightarrow \frac{2 S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}} \\\\ & \Rightarrow \frac{2 S}{3}=\frac{3}{2} \\\\ & \Rightarrow S=\frac{9}{4} \end{aligned} $

$ \begin{aligned} & \text { So, } k \times \frac{9}{4}=1 \Rightarrow k=\frac{4}{9} \\\\ & \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{X}<2) \\\\ & =1-\mathrm{P}(\mathrm{X}=0)-\mathrm{P}(\mathrm{X}=1) \\\\ & =1-\frac{4}{9}(1)-\frac{4}{9} \times \frac{2}{3} \\\\ & =1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27} \end{aligned} $

Given : $P(A)=\frac{20}{100}=\frac{2}{10}$

$ P(B)=\frac{30}{100}=\frac{3}{10} $

$ P(C)=\frac{50}{100}=\frac{5}{10} $

Let $\mathrm{E} \rightarrow$ Event that the bolt is defective.

$ \text { So, } P(E / A)=\frac{3}{100}, $

$P\left(\frac{E}{B}\right)=\frac{4}{100}, P\left(\frac{E}{C}\right)=\frac{2}{100} $

So, $\mathrm{P}(\mathrm{C} / \mathrm{E})$

$ \begin{aligned} & =\frac{P\left(\frac{E}{C}\right) \times P(C)}{P\left(\frac{E}{A}\right) \times P(A)+P\left(\frac{E}{B}\right) \times P(B)+P\left(\frac{E}{C}\right) \times P(C)} \\\\ & =\frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10}} \\\\ & =\frac{10}{6+12+10}=\frac{10}{28}=\frac{5}{14} \end{aligned} $

Total number of outcomes $=6 \times 6 \times 6=216$

Number of outcomes getting different numbers on the three dice are ${ }^6 P_3=\frac{6 !}{3 !}=120$

$\therefore$ Required probability $=\frac{120}{216}=\frac{5}{9}$

$\therefore p=5$ and $q=9$

$\therefore q-p=9-5=4$

Given, a pair of dice is thrown once, then number of outcomes $=6 \times 6=36$

A total of 5 are $\{(1,4),(2,3),(4,1),(3,2)\}$

$ \begin{aligned} & \therefore p =p(\text { success })=\frac{4}{36}=\frac{1}{9} \\\\ & q =1-\frac{1}{9}=\frac{8}{9} \end{aligned} $

Let $X$ represent the number of success

$ \begin{aligned} & \therefore P(\text { at least } 4 \text { success }) \\\\ &=P(X=4)+P(X=5) \\\\ &={ }^5 C_4 p^4 q+{ }^5 C_5 p^5 q^{5-5} \\\\ &=5\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right)+\left(\frac{1}{9}\right)^5 \\\\ &=\frac{40+1}{\left(3^2\right)^5}=\frac{41}{310} \\\\ &=\frac{41 \times 3}{3^{11}}=\frac{123}{3^{11}} \\\\ & \therefore k=123 \end{aligned} $

$\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\\\ & n(A)=15 \\\\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\\\ & n(B)=9 \\\\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\\\ & n(C)=9\end{aligned}$

$ (4,5) \in A \text { and }(4,5) \in B $

$\therefore A$ and $B$ are not exclusive events

$ \begin{aligned} & n((A \cup B) \cap C)=n(A \cap C)+n(B \cap C)-n(A \cap B \cap C) \\\\ = & 3+3-0 \\\\ = & 6 \end{aligned} $

Option (D) is correct.

$ \begin{aligned} & n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\\\ & \Rightarrow n(B) \cdot n(C) \neq n(B \cap C) \\\\ & \therefore B \text { and } C \text { are not independent. } \end{aligned} $

$ \begin{aligned} & \text { Given } \\\\ & \mathrm{np}+\mathrm{npq}=5 \\\\ & \Rightarrow \mathrm{np}(1+\mathrm{q})=5 ........(i) \\\\ & \text { and (np) (npq) }=6 \\\\ & \Rightarrow \mathrm{n}^2 \mathrm{p}^2 \mathrm{q}=6 ........(ii) \\\\ & (\mathrm{i})^2 \div(\mathrm{ii}) \\\\ & \frac{(1+q)^2}{9}=\frac{25}{6} \\\\ & \Rightarrow 6 \mathrm{q}^2-13 \mathrm{q}+6=0 \\\\ & \Rightarrow \mathrm{q}=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\\\ & \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\\\ & \frac{n}{3}\left(1+\frac{2}{3}\right)=5 \\\\ & \Rightarrow \mathrm{n}=9 \\\\ & 6(\mathrm{n}+\mathrm{p}-\mathrm{q})=52 \end{aligned} $

Let $E_{i} \rightarrow$ Bag have at least $i$ black balls

$E \rightarrow 2$ balls are drawn & both black

$ \begin{aligned} & \therefore P\left(\frac{E_{5} \text { or } E_{6}}{E}\right)=\frac{P\left(\frac{E}{E_{5}}\right)+P\left(\frac{E}{E_{6}}\right)}{\sum\limits_{i=1}^{6} P\left(\frac{E}{E_{i}}\right)} \\\\ & =\frac{\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}}{0+\frac{{ }^{2} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{3} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{4} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}} \\\\ & =\frac{10+15}{1+3+6+10+15}=\frac{25}{35}=\frac{5}{7} \end{aligned} $

${}^5{C_0} \times {3^5} = 243$

${}^5{C_2} \times {2^2} \times {3^3} = 1080$

${}^5{C_4} \times {2^4}\,.\,3 = 240$

$\therefore$ required probability

$ = {{243 + 1080 + 240} \over {6 \times 6 \times 6 \times 6 \times 6}} = {{521} \over {2592}}$

$P({w_1}) + {{P({w_1})} \over 2} + {{P({w_1})} \over {{2^2}}}\, + \,..... = 1$

$\therefore$ $P({w_1}) = {1 \over 2}$

Hence, $P({w_n}) = {1 \over {{2^n}}}$

Every number except 1, 2, 3, 4, 6 is representable in the form

$2k + 3l$ where $k,l \in N$.

$\therefore$ $P(B) = 1 - P({w_1}) - P({w_2}) - P({w_3}) - P({w_4}) - P({w_6})$

$ = {3 \over {64}}$

Required probability $=1-\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$

Taking $\mathrm{D}_{(15)}$ as $\frac{15 !}{e}$

$\mathrm{D}_{(14)}$ as $\frac{14 \text { ! }}{e}$

$\mathrm{D}_{(13)}$ as $\frac{13 \text { ! }}{e}$

$ \text { We get, } \text { Required probability } = 1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right) $

$ =1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08 $

$n-2, \sqrt{3 n}, n+2 \rightarrow$ G.P.

$3 n=n^{2}-4$

$\Rightarrow n^{2}-3 n-4=0$

$\Rightarrow n=4,-1$ (rejected)

$P(S=4)=\frac{3}{36}=\frac{1}{12}=\frac{4}{48}$

$\therefore k=4$

$x+y=66$

$ \begin{aligned} & \frac{x+y}{2} \geq \sqrt{x y} \\\\ \Rightarrow & 33 \geq \sqrt{x y} \\\\ \Rightarrow & x y \leq 1089 \\\\ \therefore & M=1089 \\\\ S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\\\ & 66 x-x^{2} \geq 605 \\\\ \Rightarrow & x^{2}-66 x+605 \leq 0 \end{aligned} $

$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$

Therefore $S=\{11,12,13 \ldots 55\} $

$\Rightarrow n(S)=45$

Elements of $S$ which are multiple of 3 are

$ \begin{aligned} & 12+(n-1) 3=54 \Rightarrow 3(n-1)=42 \Rightarrow n=15 \\\\ & n(A)=15 \end{aligned} $

$\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}$

For unique solution $\Delta \neq 0$

i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$

$\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$

$\Rightarrow N^{2}-5 N+6 \neq 0$

$\therefore N \neq 2$ and $N \neq 3$

$\therefore $ Probability of not getting 2 or 3 in a throw of dice $=\frac{2}{3}$

As given $\frac{2}{3}=\frac{k}{6} \Rightarrow k=4$

$\therefore$ Required value $=1+4+5+6+4=20$

$\Omega=$ sample space

$\mathrm{A}=$ be an event

$ \Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$

$\mathrm{A}=\left\{\frac{1}{2}\right\}$ = Selecting a point on the wire which is at $\left\{\frac{1} {2}\right\}$ or 0.5

As wire is an 1-D object so from geometrical probability

$P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$

Here total length of wire = 1 unit

and point has zero length so point A at $\left\{\frac{1} {2}\right\}$ or 0.5 has length = 0.

$ \therefore $ Favorable length = 0

$ \therefore $ $\mathrm{P}(\mathrm{A})=0 {\text { but }} \mathrm{A} \neq \phi$

Now $\overline{\mathrm{A}}$ = $[0,1]$ - $\left\{\frac{1} {2}\right\}$

So, length of $\overline{\mathrm{A}}$ = Length of entire wire - Length of point A = 1

$ \therefore $ $\mathrm{P}(\overline{\mathrm{A}})=1 {\text { but }} \overline{\mathrm{A}} \neq \Omega$.

Then both statements are false.

Attention : According to NTA option A is correct. Which is wrong. That is proven here using geometrical probability.

Note :

Geometrical probability :

1. For 1-D object, $P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$

2. For 2-D object, $P(A)=\frac{\text { Favourable Area }}{\text { Total Area }}$

3. For 2-D object, $P(A)=\frac{\text { Favourable volume }}{\text { Total volume }}$

We are given a biased coin with the probability of getting heads (H) as $\frac{1}{3}$ and thus, the probability of getting tails (T) is $\frac{2}{3}$ (since the total probability of getting either heads or tails should sum to 1).

The experiment involves tossing this coin repeatedly until we obtain two identical outcomes in a row. We are interested in the case where the experiment ends with two consecutive heads.

Let's consider two different scenarios for the sequences of coin tosses :

1. Sequences that begin with a head and end with HH (like HH, HTHH, HTHTHH, and so on).
2. Sequences that begin with a tail and end with HH (like THH, THTHH, THTHTHH, and so on).

Both types of sequences will result in the experiment ending, so we need to consider both in our final probability calculation.

Scenario 1 : Sequences beginning with H

The simplest sequence is getting HH right away. The probability of this event is $(\frac{1}{3})^2 = \frac{1}{9}$.

The next sequence is getting a head, then a tail, and then HH (i.e., HTHH). The probability of this sequence is $\frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{81}$.

We can see a pattern in these sequences. Each sequence in this scenario can be described as getting one head, followed by some number of "tail, head" pairs, and ending with HH.

Hence, the sequences in this scenario form a geometric series, where the ratio between consecutive terms is $r = P(HT) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$.

The sum of an infinite geometric series is given by the formula $\frac{a}{1 - r}$, where $a$ is the first term of the series and $r$ is the common ratio.

In this scenario, the first term is the probability of getting HH immediately, which is $\frac{1}{9}$. So, the sum of this series is :

$S_1 = \frac{1/9}{1 - 2/9} = \frac{1/9}{7/9} = \frac{1}{7}.$

Scenario 2 : Sequences beginning with T

In this scenario, we first get a tail, and then we follow the same pattern as in the first scenario.

The simplest sequence is THH, with a probability of $\frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{2}{27}$.

The next sequence is getting a tail, a head, another tail, and then HH (i.e., THTHH). The probability of this sequence is $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot (\frac{1}{3})^2 = \frac{4}{243}$.

As before, these sequences form a geometric series, with the ratio between consecutive terms is $r = P(TH) = \frac{2}{9}$.

In this scenario, the first term is the probability of getting THH, which is $\frac{2}{27}$. So, the sum of this series is :

$S_2 = \frac{2/27}{1 - 2/9} = \frac{2/27}{7/9} = \frac{2}{21}.$

Finally, to calculate the total probability of the experiment ending with two consecutive heads, we need to sum up the probabilities calculated in both scenarios:

$P(\text{HH}) = S_1 + S_2 = \frac{1}{7} + \frac{2}{21} = \frac{5}{21}.$

So, the answer to the problem is $\frac{5}{21}$ (Option B).

$ \frac{x^2}{8}+\frac{y^2}{20}<1 \text { and } y^2<5 x $

Let $ \frac{x^2}{8}+\frac{y^2}{20}=1$ ........(1)

and $ y^2=5 x $ .........(2)

On solving (1) and (2), we get

$ \begin{aligned} & \frac{x^2}{8}+\frac{5 x}{20}=1 \\\\ & \frac{x^2}{8}+\frac{x}{4}=1 \end{aligned} $

$\begin{aligned} x^2+2 x =8 \\\\ \Rightarrow x^2+2 x-8 =0 \\ \\\Rightarrow (x+4)(x-2) =0 \\\\ \Rightarrow x =-4, 2 \\\\ \Rightarrow x =2(-4 \text { is not possible }) \\\\ \Rightarrow y^2 =10 \\\\ \Rightarrow y = \pm \sqrt{10}\end{aligned}$


$\begin{aligned} X=\{(1,1),(1,0),(1,-1),(1,2),(1,-2),(2,1),(2-1), \\ (2,3),(2,3),(2,-3),(2,-2),(2,2),(2,0)\}\end{aligned}$

Let $S$ be the sample space and $E$ be the event $n(S)={ }^{12} C_3$

For E :

Selecting 3 points in which 2 points are either or $\mathrm{x}=1$ and $ \mathrm{x}=2$ but distance between them is even.

Triangles with base 2 :

$ =3 \times 7+5 \times 5=46 $

Triangles with base 4 :

$ =1 \times 7+3 \times 5=22 $

Triangles with base 6 :

$ =1 \times 5=5 $

$P(E)=\frac{46+22+5}{{ }^{12} C_3}=\frac{73}{220}$