Probability

Total number of function

$ =5 \times 5 \times 5=125 $

Numbers of one-one function

$ ={ }^5 C_3 \times 3! $

∴ Probability of one-one function

$ =\frac{10 \times 6}{125}=\frac{12}{25} $

$ \begin{aligned} & \text { } P(B)=0.4 \\ & \Rightarrow P(A \cap \bar{B})=P(A)-P(A \cap B)=0.5 \\ & \Rightarrow P(A \cup B)+\frac{P(B \cap A \cup \bar{B})}{P(A \cup \bar{B})}=1.15 \end{aligned} $

$ \begin{aligned} & P(A \cup B)+\frac{P(A \cap B)}{P(A)+1-P(A \cup B)}=1.15 \\ & \text { Let } P(A)=x \\ & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & =0.4+0.5=0.9 \\ & \Rightarrow 0.9+\frac{x-0.5}{x+1-0.9}=1.15 \\ & \Rightarrow \frac{x-0.5}{x+0.1}=0.25 \Rightarrow \frac{10 x-5}{10 x+1}=\frac{1}{4} \\ & \Rightarrow 40 x-20=10 x+1 \\ & \Rightarrow 30 x=21 \Rightarrow x=0.7 \end{aligned} $

Probability of ball to be black

$ \begin{aligned} & =\frac{1}{2} \times \frac{{ }^x C_1}{{ }^{10} C_1}+\frac{1}{2} \times \frac{{ }^y C_1}{{ }^{10} C_1} \\ & =\frac{x+y}{20} \end{aligned} $

Probability of ball to be black and drawn from box 2

$ \begin{aligned} & =\frac{\frac{1}{2} \times \frac{{ }^y C_1}{{ }^{10} C_1}}{\frac{x+y}{20}}=\frac{1}{5} \\ \Rightarrow \quad \frac{\frac{y}{20}}{\frac{x+y}{20}} & =\frac{1}{5} \Rightarrow 5 y=x+y \\ \Rightarrow \quad x & =4 y \end{aligned} $

Number of black balls in box 1 will be 4 or 8 .

Here, $n=3$

$ p=\frac{3}{5} $

It is a binomial distribution

$ \begin{aligned} \therefore \text { Mean } & =n p \\ & =3 \times \frac{3}{5}=\frac{9}{5} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} P(X=k) & =\frac{e^{-\lambda} \cdot \lambda^k}{k!} \\ P(X=5) & =\frac{e^{-\lambda} \cdot \lambda^5}{5!} \\ P(X=2) & =\frac{e^{-\lambda} \cdot \lambda^2}{2!} \\ \frac{P(X=5)}{P(X=2)} & =\frac{\lambda^3}{60}=\frac{1}{7500} \\ \lambda^3 & =\frac{1}{125} \\ \lambda & =\frac{1}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} n(S) & ={ }^{64} C_2=\frac{64 \times 63}{2} \\ n(E) & =\frac{64 \times 63}{2}-2 \times{ }^8 C_1 \times{ }^7 C_1 \\ & =2016-112=1904 \end{aligned} $

$ \begin{aligned} P(E) & =\frac{1904}{2016} \\ & =\frac{17}{18} \end{aligned} $

$ \begin{aligned} P(A \cap \bar{B}) & =P(A)-P(A \cap B) \\ P(\bar{A} \cap B) & =P(B)-P(A \cap B) \\ P(\bar{A} \cap \bar{B}) & =1-P(A \cup B) \\ P(A \cap B) & =P(B)-P(\bar{A} \cap B) \\ & =0.5-0.2=0.3 \\ P(A) & =P(A \cap \bar{B})+P(A \cap B) \\ & =0.1+0.3=0.4 \\ P(A \cup B) & =0.4+0.5-0.3 \\ & =0.6 \\ \Rightarrow \quad P(\bar{A} \cap \bar{B}) & =0.4 \end{aligned} $

Total number of balls left after second

$ \begin{aligned} \text { draw } & =(7+5+3)-(1+1) \\ & =13 \end{aligned} $

Probability of 3rd ball drawn not be red

$ =\frac{{ }^3 C_1}{{ }^{13} C_1}+\frac{{ }^4 C_1}{{ }^{13} C_1}=\frac{{ }^3 C_1+{ }^4 C_1}{{ }^{13} C_1}=\frac{7}{13} $

$ \text {} \begin{aligned} & P(X=1)+P(X=2)+P(X=3)=1 \\ & \Rightarrow 3 K^3+2 K^2+7-19 K=1 \\ & \Rightarrow 3 K^3+2 K^2-19 K+6=0 \\ & \Rightarrow 3 K^2(K-2)+8 K(K-2)-3(K-2)=0 \\ & \Rightarrow(K-2)\left(3 K^2+9 K-K-3\right)=0 \\ & \Rightarrow(K-2)(3 K-1)(K+3)=0 \\ & K \neq 2 \text { and } K \neq-3, K=\frac{1}{3} \\ & P\left(X=3=7-\frac{19}{3}=\frac{2}{3}\right. \end{aligned} $

Probability of defective unit $=\frac{1}{8}$

Probability of undefective unit

$ =1-\frac{1}{8}=\frac{7}{8} . $

Probability that atmost one unit is defective $=P(1$ unit defective $)+P(0$ unit defective)

$ \begin{aligned} & ={ }^5 C_4 \times \frac{1}{8} \times\left(\frac{7}{8}\right)^4+{ }^5 C_0 \times\left(\frac{7}{8}\right)^5 \\ & =\left(\frac{7}{8}\right)^4\left(\frac{5}{8}+\frac{7}{8}\right)=\left(\frac{7}{8}\right)^4\left(\frac{3}{2}\right) \end{aligned} $

We have 25 numbers in a row, one after another, without skipping any numbers. The smallest number in this set is an odd number.

This means there are 13 odd numbers and 12 even numbers among them. (Odd and even numbers always alternate. If we start with an odd number, we will always have one more odd number than even numbers.)

We need to find the chance (probability) that if we pick 3 of these numbers, their total (sum) is an even number.

For the sum of 3 numbers to be even, two situations can happen:
1. We choose three even numbers. (Even + Even + Even = Even)
2. We choose two odd numbers and one even number. (Odd + Odd + Even = Even)

Counting the Ways

Number of ways to pick 3 even numbers:
There are 12 even numbers. Number of ways = $^{12}C_3$.

Number of ways to pick 2 odd numbers and 1 even number:
There are 13 odd numbers and 12 even numbers.
Ways to pick 2 odd numbers = $^{13}C_2$.
Ways to pick 1 even number = $^{12}C_1$.
Total ways = $^{13}C_2 \times ^{12}C_1$.

Total Possible Ways

The total number of ways to select any 3 numbers from 25 numbers is $^{25}C_3$.

Probability Formula

The probability that the sum is even is:

$ P(A) = \frac{{ }^{13} C_2 \cdot{ }^{12} C_1+{ }^{12} C_3}{{ }^{25} C_3} = \frac{289}{575} $

Let N be the total number of possible outcomes when three dice are thrown.

Each die has 6 faces, so the total number of outcomes is $N = 6 \times 6 \times 6 = 216$.

Let S be the sum of the numbers on the three dice.

The minimum possible sum is $1+1+1=3$.

The maximum possible sum is $6+6+6=18$.

We need to find the probability that the sum S is a prime number. The prime numbers between 3 and 18 (inclusive) are:

3, 5, 7, 11, 13, 17.

Now we need to count the number of ways to obtain each of these prime sums. Let $(d_1, d_2, d_3)$ be the numbers on the three dice. The order matters (e.g., (1,1,2) is different from (1,2,1)).

1. Sum = 3:

   The only combination is (1, 1, 1).

   Number of ways: 1.

2. Sum = 5:

   Possible combinations (listing partitions in non-decreasing order to avoid duplicates, then counting permutations):

   • (1, 1, 3): Permutations: $\frac{3!}{2!1!} = 3$ ways ((1,1,3), (1,3,1), (3,1,1))

   • (1, 2, 2): Permutations: $\frac{3!}{1!2!} = 3$ ways ((1,2,2), (2,1,2), (2,2,1))

   Number of ways: $3 + 3 = 6$.

3. Sum = 7:

   Possible combinations ($d_1 \le d_2 \le d_3$):

   • (1, 1, 5): 3 ways

   • (1, 2, 4): 3! = 6 ways

   • (1, 3, 3): 3 ways

   • (2, 2, 3): 3 ways

   Number of ways: $3 + 6 + 3 + 3 = 15$.

4. Sum = 11:

   Possible combinations ($d_1 \le d_2 \le d_3$):

   • (1, 4, 6): 3! = 6 ways

   • (1, 5, 5): 3 ways

   • (2, 3, 6): 3! = 6 ways

   • (2, 4, 5): 3! = 6 ways

   • (3, 3, 5): 3 ways

   • (3, 4, 4): 3 ways

   Number of ways: $6 + 3 + 6 + 6 + 3 + 3 = 27$.

5. Sum = 13:

   Possible combinations ($d_1 \le d_2 \le d_3$):

   • (1, 6, 6): 3 ways

   • (2, 5, 6): 3! = 6 ways

   • (3, 4, 6): 3! = 6 ways

   • (3, 5, 5): 3 ways

   • (4, 4, 5): 3 ways

   Number of ways: $3 + 6 + 6 + 3 + 3 = 21$.

6. Sum = 17:

   Possible combinations ($d_1 \le d_2 \le d_3$):

   • (5, 6, 6): 3 ways

   (Note: sums like (4,x,y) would give max 4+6+6=16, so 5 must be present)

   Number of ways: 3.

Now, we sum the number of ways for each prime sum to get the total number of favorable outcomes:

Total favorable outcomes = $1 + 6 + 15 + 27 + 21 + 3 = 73$.

The probability of getting a sum that is a prime number is the ratio of favorable outcomes to the total possible outcomes:

Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{73}{216}$.

$E_1=$ company buys tyres from $c_1$

$E_2=$ company buys tyres from $c_2$

$E_3=$ company buys tyres from $c_3$

$E=$ tyre is defective

According to the question,

$ P\left(\frac{E_2}{E}\right)=? $

Using Baye's theorem,

$ P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right)} +P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)$

$ \begin{aligned} & =\frac{\frac{35}{100} \times \frac{3}{100}}{\frac{40}{100} \times \frac{2}{100}+\frac{35}{100}+\frac{3}{100}+\frac{25}{100} \times \frac{4}{100}} \\ & =\frac{35 \times 3}{40 \times 2+35 \times 3+25 \times 4} \\ & =\frac{21}{16+21+20}=\frac{21}{57}=\frac{7}{19} \end{aligned} $

$ \begin{aligned} & \text { Mean }=n p=\frac{4}{3} \\ & \qquad \begin{array}{l} \text { Variance }=n p q=\frac{10}{9} \\ q=\frac{10}{9} \times \frac{3}{4}=\frac{5}{6} \\ p=\frac{1}{6} \Rightarrow n=8 \\ \because p(x \geq 6)=p(x=6)+p(x=7)+p(x=8) \\ ={ }^8 C_6\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^2+{ }^8 C_7\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)+{ }^8 C_8\left(\frac{1}{6}\right)^8 \\ =\frac{1}{6^8}\left({ }^8 C_6 \cdot 25+8(5)+1\right) \\ =\frac{1}{6^8}(28(29)+41) \\ =\frac{1}{6^8}(700+41)=\frac{741}{6^8} \end{array} \end{aligned} $

$ \begin{aligned} & \text { } S=\{1,2, \ldots, 50\} \\ & \because x+\frac{10}{x} \leq 11 \\ & x^2-11 x+10 \leq 0 \\ & (x-10)(x-1) \leq 0 \\ & x \in[1,10] \\ & \therefore \text { Required probability }=\frac{10}{50}=\frac{1}{5} \end{aligned} $

7 Toss - 3 Head 4 Tails

$ \begin{gathered} \therefore \text { Required probability }={ }^5 C_3\left(\frac{1}{2}\right)^7 \\ =10 \times \frac{1}{128}=\frac{5}{64} \end{gathered} $

Step 1: Find Probability of Drawing a Face Card First ($P(A)$)

There are 12 face cards in a deck of 52 cards. So, $P(A) = \frac{12}{52} = \frac{3}{13}$.

Step 2: Find Probability of Drawing a Club Card Second ($P(B)$)

There are 13 clubs in the deck. So, $P(B) = \frac{13}{52} = \frac{1}{4}$.

Step 3: Find $P\left(\frac{\bar{B}}{A}\right)$ (Probability of Not Drawing a Club in Second Draw Given First Card is Face Card)

We use the formula for conditional probability: $P\left(\frac{\bar{B}}{A}\right) = \frac{P(\bar{B} \cap A)}{P(A)}$.

The draws are independent, so $P(\bar{B} \cap A) = P(\bar{B}) \cdot P(A)$.

This means $P\left(\frac{\bar{B}}{A}\right) = \frac{P(\bar{B}) \cdot P(A)}{P(A)} = P(\bar{B})$.

We already have $P(B)$, so $P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}$.

$ \begin{aligned} & \text { } P(X=k)=\frac{2 k+3}{3^k} \cdot C \\ & \Rightarrow \quad \sum_{K=0}^{\infty} P(X=k)=1 \\ & \Rightarrow \quad \sum_{K=0}^{\infty} \frac{2 k+3}{3^k} \cdot c=1 \\ & \quad S=\sum_{K=0}^{\infty} \frac{2 k+3}{3^k} \\ & \Rightarrow \quad S=\frac{3}{3^0}+\frac{5}{3^1}+\frac{7}{3^2}+\frac{9}{3^3}+\ldots \\ & \Rightarrow \quad \frac{1}{3} S=\frac{3}{3^1}+\frac{5}{3^2}+\frac{7}{3^3}+\ldots \\ & \Rightarrow \quad \frac{2 S}{3}=3+2\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots\right) \end{aligned} $

$ \begin{aligned} & =3+2\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)=3+1 \\ & \Rightarrow \quad \frac{2 S}{3}=4 \Rightarrow S=6 \\ & \therefore \quad 6 c=1 \Rightarrow c=\frac{1}{6} \\ & \text { Now, } P(X=3)=\frac{6+3}{3^3} \times \frac{1}{6}=\frac{1}{18} . \end{aligned} $

Given, matrix $P=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{array}\right]$

Odd elements $=\{1,3,5,7,9\}$

Total = 5

Even elements $=\{2,4,6,8\}$

Total =4

Case I 3 odd numbers numbers of ways $={ }^5 C_3=10$

Case II 1 odd and 2 even numbers of ways $={ }^5 C_1 \times{ }^4 C_2$

$ =5 \times 6=30 $

Total favourable cases for $A$

$ =10+30=40 $

And total ways for selecting 3 elements from 9 elements $={ }^9 C_3=84$

$ \therefore \quad P(A)=\frac{40}{84} $

Selecting 3 elements in a row or column.

3 rows $\times 1$ way $=3$ ways

3 column $\times 1$ way $=3$ ways

total $=6$ ways

Now, Row $1: 1+2+3=6$ (even)

Row 2 : $4+5+6=15$ (odd)

Row $3: 7+8+9=24$ (even)

Column I : $1+4+7=12$ (even)

Column II : $2+5+8=15$ (odd)

Column III : $3+6+9=18$ (even)

$ \begin{aligned} \therefore P(A \cap B) & =\frac{2}{84} \\ P(B) & =\frac{6}{84} \end{aligned} $

$ \begin{aligned} & \therefore \quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{2}{84}}{\frac{6}{84}}=\frac{1}{3} \\ & \begin{aligned} \therefore P(A)+P\left(\frac{A}{B}\right) & =\frac{40}{84}+\frac{1}{3} \\ & =\frac{40+28}{84} \\ & =\frac{68}{84}=\frac{17}{21} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, } P\left(B_1\right)=0.25, P\left(B_2\right)=0.30 \text {, }\\ &\begin{aligned} & P\left(B_3\right)=0.45, P\left(\frac{A}{B_1}\right)=0.05 \\ & P\left(\frac{A}{B_2}\right)=0.04, P\left(\frac{A}{B_3}\right)=0.03 \\ & \begin{array}{r} \therefore P\left(\frac{B_2}{A}\right)=\frac{P\left(\frac{A}{B_2}\right) P\left(B_2\right)}{P\left(\frac{A}{B_2}\right) P\left(B_2\right)+P\left(\frac{A}{B_3}\right) P\left(B_3\right)} \\ +P\left(\frac{A}{B_1}\right) P\left(B_1\right) \end{array} \\ & =\frac{0.04 \times 0.30}{0.04 \times 0.30+0.03 \times 0.45+0.05 \times 0.25} \\ & =\frac{0.012}{0.012+0.0135+0.0125} \\ & =\frac{0.012}{0.038}=\frac{6}{19} \end{aligned} \end{aligned} $

Given, $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$,

$ \begin{aligned} P(A \cap B) & =\frac{1}{4} \\ \therefore \quad P\left(\frac{A}{B}\right) & =\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4} \\ P\left(\frac{A^{\prime}}{B^{\prime}}\right) & =\frac{P\left(A^{\prime} \cap B^{\prime}\right)}{P\left(B^{\prime}\right)} \\ & =\frac{P\left((A \cup B)^{\prime}\right)}{P\left(B^{\prime}\right)} \end{aligned} $

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

$ \begin{aligned} & =\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \\ & =\frac{6+4-3}{12}=\frac{7}{12} \end{aligned} $

$ \begin{aligned} & \therefore P\left((A \cup B)^{\prime}=1-\frac{7}{12}=\frac{5}{12}\right. \\ & \text { And } P\left(B^{\prime}\right)=\frac{2}{3} \\ & \therefore P\left(\frac{A^{\prime}}{B^{\prime}}\right)=\frac{\frac{5}{12}}{\frac{2}{3}}=\frac{5}{12} \times \frac{3}{2}=\frac{5}{8} \\ & \therefore P\left(\frac{A^{\prime}}{B^{\prime}}\right)+P\left(\frac{A}{B}\right)=\frac{5}{8}+\frac{3}{4}=\frac{11}{8} \end{aligned} $

Total outcomes when two dice are thrown $=36$

Even sum : happen when sum

$ =2,4,6,8,10,12 $

Odd sum : happens when sum

$ =3,5,7,9,11 $

$ \begin{array}{clc} \hline \text { Sum } & \text { Combination } & \text { Even } \\ \hline 2 & (1,1) & 1 \\ \hline 4 & (1,3),(2,2),(3,1) & 3 \\ \hline 6 & (1,5),(2,4),(3,3),(4,2),(5,1) & 5 \\ \hline 8 & (2,6),(3,5),(4,4),(5,3),(6,2) & 5 \\ \hline 10 & (4,6),(5,5),(6,4) & 3 \\ \hline 12 & (6,6) & 1 \\ \hline \end{array} $

$\therefore$ Even sum $=18$ outcomes

Odd sum $=36-18=18$ outcomes

Sum is even, $A$ gets $=\frac{1}{2}$, point with probability

$ =\frac{18}{36}=\frac{1}{2} $

Sum is odd, $A$ gets $=\frac{1}{2}$, point with probability

$ =\frac{18}{36}=\frac{1}{2} $

Expected value $E(X)=\frac{1}{2} \times \frac{1}{2}+1 \times \frac{1}{2}$

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

Step 1: Finding the average number of errors

The typist makes 1 mistake for every 10 pages. For 40 pages, expected errors = $ \frac{40}{10} = 4 $.

Step 2: Setting up the Poisson distribution

We use the Poisson formula to find the chance of getting up to 2 errors when the average (λ) is 4.

$ P(X=k) = \frac{e^{-4} 4^k}{k!} $

Step 3: Find probability of at most 2 errors

We need to calculate $ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) $:

$ P(X=0) = \frac{e^{-4} \cdot 4^0}{0!} = e^{-4} $
$ P(X=1) = \frac{e^{-4} \cdot 4^1}{1!} = 4e^{-4} $
$ P(X=2) = \frac{e^{-4} \cdot 4^2}{2!} = 8e^{-4} $

Add them up:
$ p = e^{-4} + 4e^{-4} + 8e^{-4} = 13e^{-4} $

Step 4: Multiply by $ e^2 $ as asked in the question

$ e^2p = 13e^{-4} \times e^2 = 13e^{-2} $

Total number of ways to choose any 3 distinct squares $={ }^{64} C_3$

$ \Rightarrow \quad \frac{64!}{3!61!}=\frac{64 \times 63 \times 62}{6}=41664 $

Since, each row has 8 squares.

In each row, the number of ways to choose 3 adjacent squares $=8-3+1=6$

So, total for 8 rows $=8 \times 6=48$

Similarly, total for 8 columns $=8 \times 6=48$

$\therefore$ Total favourable outcomes

$ \Rightarrow \quad 48+48=96 $

Hence, required probability

$ \Rightarrow \quad \frac{96}{41664}=\frac{1}{434} $

Number of spade cards $=13$, number of kings $=4$

Prime numbered cards are $2,3,5$ and 7

So, total primes cards $=4$ suits $\times 4$

numbers

$\Rightarrow 16$ cards

But, a single card can be both a king and a spade or a prime and a spade.

So, the number of favourable outcomes.

Case I Spade that is not a king or prime $=13-1-4=8$

Case II King that is not a spade or prime = 3

Case III Prime card that is not a spade or king $=16-4=12$

So, total favourable cases

$ \Rightarrow \quad 8 \times 3 \times 12=288 $

and total number of ways to choose 3 cards from 52 cards $={ }^{52} C_3$

$ =\frac{52!}{3!49!}=\frac{52 \times 51 \times 50}{6}=22100 $

Thus, required probability

$ =\frac{288}{22100}=\frac{72}{5525} $

Since, there are three Urn.

So, $P(\operatorname{Urn} A)=\frac{1}{3}=P(\operatorname{Urn} B)=P(\operatorname{Urn} C)$

Now, probability of drawing a white ball from each Urn

Urn A 6 white balls, 2 black balls.

Total = 8 balls

$P($ White/Urn A$)=\frac{6}{8}=\frac{3}{4}$

Urn B 5 white balls, 3 black balls

Total = 8 balls

$P($ White/Urn B $)=\frac{5}{8}$

Urn C 4 white balls, 4 black balls

Total $=8$ balls

$P($ White $/$ Urn C$)=\frac{4}{8}=\frac{1}{2}$

So, the total probability of drawing a white ball is

$P($ White $)=P($ White $/$ Urn A$) \cdot P($ Urn A$)$

$+P($ White $/$ Urn B$) \cdot P($ Urn B$)$

$+P($ White $/$ Urn C$) \cdot P($ Urn C$)$

$ \begin{aligned} & =\frac{1}{3} \times \frac{3}{4}+\frac{1}{3} \times \frac{5}{8}+\frac{1}{3} \times \frac{1}{2} \\ & =\frac{1}{4}+\frac{5}{24}+\frac{1}{6} \\ & =\frac{6+5+4}{24}=\frac{15}{24}=\frac{5}{8} \end{aligned} $

We have the Numbers $ 2,3,5,7,11 $ and 13.

3 Number are chosen

$ \therefore \quad n(s)={ }^{6} C_{3}+20 $

$ E= $ sum of number is divisible by 3

$ E=\{(2,37)(2,3,13)(3,5,7)(3,5,13) $

$ (3,11,13)\}(3,7,11)(2,5,11) $

$ n(E)=7 $

$ P(E)=\frac{7}{20} $

To determine the probability of getting a multiple of 3 as the sum of the numbers on the top faces of two dice, given that their sum is an odd number, we define the following:

Let $ A $ be the event that the sum is a multiple of 3.

Let $ B $ be the event that the sum is an odd number.

First, we calculate the total probabilities:

The probability of event $ A $, $ P(A) $, is the number of favorable outcomes divided by the total outcomes, i.e., $ \frac{12}{36} $.

The probability of event $ B $, $ P(B) $, is similarly $ \frac{18}{36} $.

Next, we need to determine the probability of both events occurring simultaneously, $ P(A \cap B) $.

The probability $ P(A \cap B) = \frac{6}{36} $, which represents the outcomes where the sum is both a multiple of 3 and an odd number.

Finally, we calculate the conditional probability of $ A $ given $ B $, denoted as $ P(A|B) $:

$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{6}{36}}{\frac{18}{36}} = \frac{6}{18} = \frac{1}{3} $

Therefore, the probability of getting a multiple of 3 as the sum, given that the sum is an odd number, is $\frac{1}{3}$.

$ \Sigma p_{i}=1 $

$ \begin{array}{l} 4 k^{2}+3 k=1 \\\\ 4 k^{2}+4 k-k-1=0 \\\\ 4 k(k+1)-1(k+1)=0 \\\\ (k+1)(4 k-1)=0 \\\\ k=\frac{1}{4} \end{array} $

$ \operatorname{Var}(X)=\frac{15}{16} $

Mean $ =n p=\frac{16}{5} $

Variance $ =n p q=\frac{48}{25} $

$ \begin{array}{l} q=\frac{48}{25} \times \frac{5}{16}=\frac{3}{5} \\\\ p=\frac{2}{5} \end{array} $

$ \begin{aligned} & n \times \frac{2}{5}=\frac{16}{5} \\\\ n & =8 \\\\ P(x > 1) & =1-P(x=0)-P(x=1) \\\\ = & 1-{ }^{8} C_{0}\left(\frac{3}{5}\right)^{8}-{ }^{8} C_{1}\left(\frac{2}{5}\right)\left(\frac{3}{5}\right)^{7} \\\\ = & 1-\left(\frac{3}{5}\right)^{7}\left\{\frac{3}{5}+\frac{16}{5}\right\} \\\\ = & 1-\left(\frac{3}{5}\right)^{7}\left(\frac{19}{5}\right) \\\\ \because \quad k & =\frac{19}{5} \\\\ \quad 5 k & =19 \end{aligned} $

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} $

Total $3 \times 3$ matrix using

$ \{-1,0,1\}=3^9=19683 $

Total $3 \times 3$ non-zero matrix

$ =3^9-1=19682 $

Now, skew-symmetric matrices

$ a_{11}=a_{22}=a_{33}=0 $

$a_{12}, a_{13}, a_{23}$ can be filled in $3^3=27$ ways

But all cannot be zero, so 26 ways.

$ P=\frac{26}{19682}=\frac{1}{757} $

$ \begin{aligned} &\text { 1st throw 2nd 3rd 4th 5th 6th 7th }\\ &\begin{array}{lll} 1 \rightarrow \frac{1}{6} & 6 \rightarrow \frac{1}{6} & 7 \rightarrow\left(\frac{1}{6}\right)^2 \\ 1 \rightarrow \frac{1}{6} & 1 \rightarrow \frac{1}{6} 5 \rightarrow \frac{1}{6} & 7 \rightarrow\left(\frac{1}{6}\right)^3 \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & 1 \rightarrow \frac{1}{6} 1 \rightarrow \frac{1}{6} 1 \rightarrow \frac{1}{6} 4 \rightarrow \frac{1}{6} \\ & 7 \rightarrow\left(\frac{1}{6}\right)^4 \\ & \text { Total }=\left(\frac{1}{6}\right)^2+\left(\frac{1}{6}\right)^3 \ldots\left(\frac{1}{6}\right)^6 \end{aligned}\\ &\text { Given series is GP, } a=\frac{1}{36}, r=\frac{1}{6} \text { and } n=5\\ &P=\frac{1}{36}\left(\frac{1-\frac{1}{6^5}}{1-\frac{1}{6}}\right)=\frac{1}{30}\left(1-\frac{1}{6^5}\right) \end{aligned} $