Probability

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} $

Total number of possible words in the word 'CURVE' is

$ \begin{aligned} & \text { 2-letter words }={ }^5 P_2=\frac{5!}{(5-2)!}=5 \times 4=20 \\ & \text { 3-letter words }={ }^5 P_3=5 \times 4 \times 3=60 \\ & \text { 4-letter words }={ }^5 P_4=5 \times 4 \times 3 \times 2=120 \\ & \text { 5-letter words } \\ & ={ }^5 P_5=5 \times 4 \times 3 \times 2 \times 1=120 \end{aligned} $

So, total words $=20+60+120+120$

$ =320 $

$\therefore$ P(3-letter words)

$ \begin{aligned} & =\frac{\text { Number of } 3 \text { - letter words }}{\text { Total words }} \\ & =\frac{60}{320}=\frac{3}{16} \end{aligned} $

Total ways of choosing 3 numbers

$ \begin{aligned} & ={ }^{30} C_3=\frac{30 \times 29 \times 28}{3 \times 2 \times 1} \\ & =4060 \end{aligned} $

The number of sets of three consecutive numbers

$ \begin{aligned} & \{(1,2,3),(2,3,4) \ldots(28,29,30)\} \\ & \Rightarrow 28 \text { sets. } \end{aligned} $

So, the number of sets that are not consecutive numbers $=4060-28=4032$

Hence, required probability

$ =\frac{4032}{4060}=\frac{144}{145} $

$P(\bar{A})=0.3, P(B)=0.4, P(A \cap \bar{B})=0.5$

So, $P(A)=1-P(\bar{A})=0.7$,

$ P(\bar{B})=1-0.4=0.6 $

Now, $P(A \cap B)$

$ \because \quad P(A)=P(A \cap B)+P(A \cap \bar{B}) $

$ \begin{aligned} & \Rightarrow \quad 0.7=P(A \cap B)+0.5 \\ & \Rightarrow \quad P(A \cap B)=0.2 \\ & \text { then, } P(B / A \cup \bar{B})=\frac{P(B \cap(A \cup \bar{B}))}{P(A \cup \bar{B})} \\ & =\frac{P((B \cap A) \cup(B \cap \bar{B})}{P(A)+P(\bar{B})-P(A \cap \bar{B})} \\ & =\frac{P(A \cap B)}{0.7+0.6-0.5}=\frac{0.2}{0.8}=0.25 \end{aligned} $

• Probability that A will get the job, $ P(A) = 0.8 $

• Probability that B will get the job, $ P(B) = 0.7 $

We are asked to find the probability that at least one of them will get the job.

Step 1: Formula for "at least one"

$ P(\text{at least one}) = P(A \cup B) = P(A) + P(B) - P(A \cap B) $

Step 2: If events $A$ and $B$ are independent:

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

$ P(A \cap B) = 0.8 \times 0.7 = 0.56 $

Step 3: Substitute values

$ P(A \cup B) = 0.8 + 0.7 - 0.56 = 0.94 $

✅ Final Answer: 0.94

Correct Option: (B) 0.94

$ \begin{aligned} & \text { } \because P(x=k)={ }^n C_k p^{n-k} q^k \\ & \text { Here, } p=q=1 / 2 \\ & \text { and } 2 P(X=5)=P(X=4)+P(X=6) \\ & \Rightarrow 2^n C_5\left(\frac{1}{2}\right)^{n-5}\left(\frac{1}{2}\right)^5={ }^n C_4\left(\frac{1}{2}\right)^{n-4}\left(\frac{1}{2}\right)^4 +{ }^n C_6\left(\frac{1}{2}\right)^{n-6}\left(\frac{1}{2}\right)^6 \end{aligned} $

$ \begin{aligned} & \Rightarrow 2^n C_5\left(\frac{1}{2}\right)^n={ }^n C_4\left(\frac{1}{2}\right)^n+{ }^n C_6\left(\frac{1}{2}\right)^n \\ & \Rightarrow 2\left[\frac{n!}{5!(n-5)!}\right]=\frac{n!}{4!(n-4)!}+\frac{n!}{6!(n-6)!} \\ & \Rightarrow \frac{2}{5!(n-5)}=\frac{1}{4!}+\frac{1}{6!(n-6)(n-5)} \\ & \Rightarrow \frac{2}{5(n-5)}=1+\frac{1}{30(n-6)(n-9)} \\ & \Rightarrow \frac{2}{5(n-5)}=\frac{30\left(n^2-11 n+30\right)+1}{30(n-6)(n-9)} \\ & \Rightarrow 12(n-6)=30\left(n^2-11 n+30\right)+1 \\ & \Rightarrow n^2-21 n+98=0 \Rightarrow(n-14)(n-7)=0 \end{aligned} $

So, $n=7$ or 14

Hence, largest value of $n=14$

$ \begin{array}{clllcc} \hline \boldsymbol{X}=x_i & -2 & -1 & 0 & 1 & 2 \\ \hline \boldsymbol{P}\left(\boldsymbol{X}=x_i\right) & k^2 / 3 & k^2 & 2 k^2 / 3 k / 2 & k / 2 \\ \hline \end{array} $

$ \begin{aligned} & \because \quad \sum_{i=1}^n P\left(X=x_i\right)=1 \\ & \Rightarrow \quad \frac{k^2}{3}+k^2+\frac{2 k^2}{3}+k / 2+k / 2=1 \\ & \Rightarrow \quad 2 k^2+k-1=0 \\ & \Rightarrow \quad 2 k^2+2 k-k-1=0 \\ & \Rightarrow \quad(2 k-1)(k+1)=0 \\ & \text { so, } k=-1,1 / 2 \\ & \text { for } k=1 / 2 \\ & k^2 / 3=1 / 12>0, k^2=1 / 4>0 \\ & \frac{2 k^2}{3}=1 / 6>0 \end{aligned} $

$ k / 2=1 / 4>0 $

and sum of

$1 / 12+1 / 4+1 / 6+1 / 4+1 / 4=1$

thus, $k=1 / 2$ is a valid solution

Now, $E(X)=\sum x_i P\left(X=x_i\right)$

$ \begin{aligned} &=-2 \times 1 / 12+(-1) \times 1 / 4+0 \times 1 / 6 +1 \times 1 / 4+2 \times 1 / 4 \\ &=-1 / 6-1 / 4+1 / 4+1 / 2 \end{aligned} $

$ =1 / 2-1 / 6=\frac{3-1}{6}=1 / 3 $

Let $A=$ first student get qualified in the exam and $B=$ second students get qualified in the exam $\because P(A)=\frac{1}{4}$ and $P(B)=\frac{2}{5}$ and $A, B$ are independent events.

$ \begin{aligned} P(A \cup B) & =1-P(\overline{A \cup B}) \\ & =1-P(\bar{A} \cap \bar{B}) \\ & =1-P(\bar{A}) \cdot P(\bar{B}) \\ & =1-(1-P(A))(1-P(B)) \\ & =1-\left(1-\frac{1}{4}\right)\left(1-\frac{2}{5}\right) \\ & =1-\frac{3}{4} \times \frac{3}{5}=\frac{11}{20} \end{aligned} $

$P$ (exactly one or $A$ or $B$ occurs) $=P$ (exactly one $B$ or $C$ occurs) $=P$ (exactly one of $C$ or $A$ occurs) $=\frac{1}{4}$

$\therefore \quad P(A \cap B \cap C)=\frac{1}{16}$

$ \begin{aligned} & P(A)+P(B)-2 P(A \cap B)=\frac{1}{4} \\ & P(B)+P(C)-2 P(B \cap C)=\frac{1}{4} \\ & P(C)+P(A)-2 P(C \cap A)=\frac{1}{4} \end{aligned} $

On adding Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & 2[P(A)+P(B)+P(C)-P(A \cap B) \\ & -P(B \cap C)-P(C \cap A)]=\frac{3}{4} \end{aligned} $

$ \begin{aligned} &\text {}\\ &\begin{aligned} Now, \,\, & P(A \cup B \cup C)=P(A)+P(B)+P(C) \\ & -P(A \cap B)-P(B \cap C) \\ & -P(C \cap A)+P(A \cap B \cap C) \\ & =\frac{3}{8}+\frac{1}{16}=\frac{6+1}{16}=\frac{7}{16} \end{aligned} \end{aligned} $

Probability out of 3 balls 2 are black and one red.

$ \begin{aligned} & =\frac{(1 R)_P(2 B)_Q+(1 B)_P(1 R \cdot 1 B)_Q}{(1 \text { ball })_P \cdot(2 \text { balls })_Q} \\ & =\frac{{ }^4 C_1 \cdot{ }^6 C_2+{ }^5 C_1 \cdot{ }^3 C_1 \cdot{ }^6 C_1}{{ }^9 C_1 \cdot{ }^9 C_2} \\ & =\frac{4 \times \frac{6 \times 5}{2 \times 1}+5 \times 3 \times 6}{9 \times \frac{9 \times 8}{2 \times 1}} \\ & =\frac{60+90}{36 \times 9}=\frac{150}{36 \times 9}=\frac{25}{54} \end{aligned} $

Let, $A=$ Watching TV

$B=$ Read book and $E=$ Student sleep

$ \begin{aligned}Now,\,\, & v, P(A)=\frac{4}{5} \text { and } P(B)=1-\frac{4}{5}=\frac{1}{5} \\ & P\left(\frac{E}{A}\right)=\frac{3}{4} \Rightarrow P\left(\frac{E}{B}\right)=\frac{1}{4} \end{aligned} $

$ \begin{aligned}Now,\,\, P\left(\frac{A}{E}\right) & =\frac{P(A) \cdot P\left(\frac{E}{A}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)} \\ & =\frac{\frac{4}{5} \times \frac{3}{4}}{\frac{4}{5} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{5}}=\frac{12}{13} \end{aligned} $

We have, $P(X=k)=\frac{1}{n}$

$ P(X=1)=P(X=2) \ldots, P(X=n)=\frac{1}{n} $

Now, variance

$ \begin{aligned} & \Sigma P_i x_i^2-\left(\Sigma P_i x_i\right)^2 \\ & \frac{1}{n}\left(1^2+2^2+\ldots+n^2\right)-\left(\frac{1}{n}(1+2+\ldots+n)\right)^2 \\ & \frac{(n+1)}{6}(2 n+1)-\left(\frac{n+1}{2}\right)^2 \\ & (n+1)\left(\frac{2 n+1}{6}-\frac{n+1}{4}\right) \\ & (n+1)\left(\frac{4 n+2-3 n-3}{12}\right) \\ & \quad=\frac{n^2-1}{12} \end{aligned} $

$p=$ Probability of detect an enemy plane $=\frac{1}{10}$

and $q=$ Probability of not detect an enemy plane $=1-\frac{1}{10}=\frac{9}{10}$.

$ \begin{aligned} & n=4 \\ P(X \geq 2)= & 1-P(X=0)-P(X=1) \\ = & 1-{ }^4 C_0 P^0 q^4-{ }^4 C_1 p^1 q^3 \\ = & 1-\left(\frac{9}{10}\right)^4-4\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^3 \\ = & \frac{10000-6561-2916}{104} \\ = & \frac{523}{10000}=0.0523 \end{aligned} $

Total ways of distributing 5 identical samples among 12 house, with atmost one sample $={ }^{12} \mathrm{C}_5$

Number of ways to select 5 house with no two consecutive

$ \begin{gathered} ={ }^{n-r+1} C_r={ }^{12-5+1} C_5={ }^8 C_5 \\ \therefore \text { Required probability }=\frac{{ }^8 C_5}{{ }^{12} C_6}=\frac{\frac{8!}{3!5}}{\frac{12!}{5!7}} \\ \quad=\frac{8!7!}{12!3!}=\frac{7 \times 6 \times 5 \times 4}{12 \times 11 \times 10 \times 9}=\frac{7}{99} \end{gathered} $

Step 1: Show what we know.

$A$ and $B$ are independent, so $P(A \cap B) = P(A) \cdot P(B) = \frac{1}{6}$.

The chance that neither happens is $P(\bar{A} \cap \bar{B}) = (1 - P(A)) \cdot (1 - P(B)) = \frac{1}{3}$.

Step 2: Set up equations.

From above, we have two equations:

• $P(A) \cdot P(B) = \frac{1}{6}$
• $(1 - P(A)) \cdot (1 - P(B)) = \frac{1}{3}$

Step 3: Expand the second equation.

Expand $(1 - P(A)) \cdot (1 - P(B)) = 1 - P(A) - P(B) + P(A) \cdot P(B) = \frac{1}{3}$.

Step 4: Replace $P(A) \cdot P(B)$.

Since $P(A) \cdot P(B) = \frac{1}{6}$, substitute:

• $1 - P(A) - P(B) + \frac{1}{6} = \frac{1}{3}$

• $1 + \frac{1}{6} - \frac{1}{3} = P(A) + P(B)$
• $1 + \frac{1}{6} - \frac{2}{6} = P(A) + P(B)$
• $1 - \frac{1}{6} = P(A) + P(B)$
• $\frac{5}{6} = P(A) + P(B)$

Step 5: Build a quadratic equation.

Let $P(A) = x$. Then $P(B) = \frac{5}{6} - x$.

Also, $x \cdot P(B) = \frac{1}{6}$.

So, $x \left( \frac{5}{6} - x \right) = \frac{1}{6}$

$5x/6 - x^2 = 1/6$

Multiply both sides by 6: $5x - 6x^2 = 1$

$6x^2 - 5x + 1 = 0$

Step 6: Solve the quadratic equation.

$x = \frac{5 \pm \sqrt{25-24}}{12}$

$x = \frac{5 \pm 1}{12}$, so $x = \frac{6}{12} = \frac{1}{2}$ and $x = \frac{4}{12} = \frac{1}{3}$

Step 7: Choose the correct values using $P(A) > P(B)$.

Since $P(A) > P(B)$, $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$.

Step 1: Understand events A and B

Event $A$ means the sum that comes from the two dice is a prime number.

Event $B$ means the sum is more than 8.

$\bar{B}$ means the sum is 8 or less.

Step 2: List all possible sums when rolling two dice

When rolling two dice, the possible sums go from 2 to 12.

Step 3: Find all ways to get a prime sum ($A$)

The prime numbers between 2 and 12 are: 2, 3, 5, 7, and 11.

We need to count all pairs of dice rolls that add up to one of these numbers.

Step 4: Find the pairs for $A \cap \bar{B}$

We only want the pairs where the sum is a prime number and also less than or equal to 8.

The prime numbers that are 8 or less are: 2, 3, 5, and 7.

Let’s list all pairs for each:

Sum = 2: (1,1)
Sum = 3: (1,2), (2,1)
Sum = 5: (1,4), (2,3), (3,2), (4,1)
Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

Counting all these, there are 13 pairs.

Step 5: Total number of possible outcomes

There are 36 possible combinations when rolling two dice (6 sides on the first die times 6 sides on the second die).

Step 6: Write the probability

The probability of both $A$ happening and $\bar{B}$ (sum is a prime less than or equal to 8) happening is:

$P(A \cap \bar{B})=\frac{13}{36}$

$\Sigma_1=$ In the family of 8 , men and women are in equal number $\Sigma_i=$ In the family of 8 , men and women are not in equal number.

Total cases $=9$ for each case $P\left(\Sigma_1\right)=1 / 9 A=4$ persons are selected in which ${ }^2$ are men and 2 are women.

$ \begin{aligned} P\left(\Sigma_1 / A\right) & =\frac{P(A) \cdot P\left(A / \Sigma_1\right)}{P(A) \cdot P\left(A / \Sigma_1\right)+P\left(\Sigma_i\right) \cdot P\left(A / \Sigma_i\right)} \\ & =\frac{\frac{1}{9} \cdot \frac{{ }^4 C_2 \cdot{ }^4 C_2}{{ }^8 C_4}}{\frac{1}{9} \cdot \frac{{ }^4 C_2 \cdot{ }^4 C_2}{{ }^8 C_4}+\frac{1}{9} \frac{\left({ }^2 C_2 \cdot{ }^6 C_2+{ }^3 C_2 \cdot{ }^5 C_2\right)_2}{{ }^8 C_4}} \\ & =\frac{1}{1+\frac{2(15+30)}{36}}=\frac{1}{1+\frac{45}{18}}=\frac{1}{1+\frac{5}{2}}=\frac{2}{7} \end{aligned} $

$ \begin{aligned} &\text { We have, } n=6\\ &\begin{aligned} & \text { Mean - Variance }=27 / 8 \\ & \quad\Rightarrow n p-n p q=\frac{27}{8} \Rightarrow n p(1-q)=\frac{27}{8} \\ & \quad\Rightarrow n p^2=\frac{27}{8} \Rightarrow p^2=\frac{27}{8 \times 6} \Rightarrow p^2=\frac{9}{16} \Rightarrow p=\frac{3}{4} \\ & \begin{aligned} \therefore P(X \leq 2) & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^6 C_0\left(\frac{3}{4}\right)^0\left(\frac{1}{4}\right)^6+{ }^6 C_1\left(\frac{3}{4}\right)^1\left(\frac{1}{4}\right)^5+{ }^6 C_2\left(\frac{3}{4}\right)^2 \cdot\left(\frac{1}{4}\right)^2 \\ & =\frac{1}{4^6}(1+18+135)=\frac{154}{4^6} \end{aligned} \end{aligned} \end{aligned} $

The mean of a binomial distribution is $\mu = n p$.

The variance is $\sigma^2 = n p q$ where $q = 1 - p$.

Given $\mu = 2\sigma^2$, so $n p = 2(n p q)$. This means $n p = 2 n p q$.

Dividing both sides by $n p$ (assuming $n p \ne 0$), we get $1 = 2q$, so $q = \frac{1}{2}$.

Since $q = \frac{1}{2}$, then $p = 1 - q = \frac{1}{2}$.

Also, we are told that $\mu + \sigma^2 = 3$. Substitute the formulas to get: $n p + n p q = 3$.

Factor $n p$ to get $n p (1+q) = 3$.

Now, substitute $p = \frac{1}{2}$ and $q = \frac{1}{2}$ into the equation: $n \left(\frac{1}{2}\right)\left(1+\frac{1}{2}\right) = 3$.

Rewrite $1+\frac{1}{2} = \frac{3}{2}$, so $n \left(\frac{1}{2}\right)\left(\frac{3}{2}\right) = 3$.

This becomes $n \cdot \frac{3}{4} = 3$. Divide both sides by $\frac{3}{4}$: $n = 3 \div \frac{3}{4} = 3 \times \frac{4}{3} = 4$.

The distribution is $B(4, \frac{1}{2})$.

We need to find $P(X \leq 3)$.

For a binomial distribution, $P(X \leq 3) = 1 - P(X = 4)$, because the only other possible value for $X$ is $4$ when $n = 4$.

$P(X=4) = {4 \choose 4}\left(\frac{1}{2}\right)^4 = 1 \times \left(\frac{1}{16}\right) = \frac{1}{16}$

Therefore, $P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}$.

Total number of fruits in one basket $=5$ apples +7 oranges $=12$

Probability of picking an apple from Ist basket $=\frac{5}{12}=P\left(A_1\right)$ Probability of picking an orange from Ist basket $=\frac{7}{12}=P\left(B_1\right)$ Total number of fruits in second baskd $=4$ apples +8 oranges $=12$ Probability of picking an apples from IInd basket, $P\left(A_2\right)=\frac{4}{12}$

Probability of picking an oranges from IInd basket $P\left(B_2\right)=\frac{8}{12}$

Probability of picking an apple from Ist and an orange from IInd basket

$ =\frac{5}{12} \times \frac{8}{12}=\frac{40}{144}=\frac{10}{36}=\frac{5}{18} $

Probability of picking an apple from IInd basket and an orange from Ist basket $=\frac{4}{12} \times \frac{7}{12}=\frac{28}{144}$

So, total probability

$ =\frac{40}{144}+\frac{28}{144}=\frac{68}{144}=\frac{17}{36} $

Probability of drawing a black queen first $=\frac{2}{52}$

Remaining black face card $=5$

Probability of drawing a black face card after a black queen $=\frac{5}{51}$

Now, probability of red queen first $=\frac{2}{52}$

Probability of a black face card after a red queen $=\frac{6}{51}$

So, total probability $=\frac{2}{52} \times \frac{5}{51}+\frac{2}{52} \times \frac{6}{51}$

$ =\frac{10}{2652}+\frac{12}{2652}=\frac{22}{2652}=\frac{11}{1326} $

Let $D$ be the event that an item is defective, and $N$ be the event that an item is not defective.

So, $P(D)=0.3$ and $P(N)=1-0.3=0.7$

Probability of an accurate result

$ =\frac{8}{10}=0.8=P\left(R_D / D\right) $

Probability of an inaccurate result

$ \begin{aligned} & =1-0.8=0.2=P\left(R_N / D\right) \\ & \text { So, } P\left(D \cap R_N\right)=P\left(R_N / D\right) \times P(D) \\ & =0.2 \times 0.3=0.06 \\ & \begin{aligned} P\left(N \cap R_N\right) & =P\left(R_N / N\right) \times P(N) \\ & =0.8 \times 0.7=0.56 \end{aligned} \end{aligned} $

$ \begin{aligned} 0, P\left(R_N\right) & =P\left(D \cap R_N\right)+P\left(N \cap R_N\right) \\ & =0.06+0.56=0.62 \\ P\left(D / R_N\right) & =\frac{P\left(D \cap R_N\right)}{P\left(R_N\right)}=\frac{0.06}{0.62}=\frac{6}{62}=\frac{3}{31} \end{aligned} $

$\therefore$ The probability of an item is actually defective given device reports it as not defective is $\frac{3}{31}$.

Let $A, B$ and $C$ be the event that a student is from section $A, B$ and $C$.

So, $P(A)=0.2, P(B)=0.3, P(C)=0.5$

Now, section A : 20 girls, 30 boys, total 50 students

Section B : 40 girls, 20 boys, total 60 students

Section C : 10 girls, 30 boys, total 40 students

$ \begin{aligned} P(G / A) & =\frac{20}{50}=0.4, P(G / B)=\frac{40}{60}=\frac{2}{3} \\ P(G / C) & =\frac{10}{40}=0.25 \\ P(G) & =P(G / A) \cdot P(A)+P(G / B) \cdot P(B)+P(G / C) \cdot P(C) \\ & =0.4 \times 0.2+\frac{2}{3} \times 0.3+0.25 \times 0.5 \\ & =0.08+0.2+0.125 \\ & =0.405 \end{aligned} $

Now, $P(G \cap A)=P(G / A) \cdot P(A)$

$ =0.4 \times 0.2=0.08 $

Using Bayes' theorem,

$ \begin{aligned} & P(A / G)=\frac{P(A \cap G)}{P(G)} \\ & =\frac{0.08}{0.405}=\frac{80}{405}=\frac{16}{81} \end{aligned} $

So, the probability that a randomly selected girl belongs to section $A$ is $\frac{16}{81}$.

$ \begin{array}{|l|l|l|l|l|} \hline X=x_i & -1 & 0 & 1 & 2 \\ \hline P\left(X=x_i\right) & k^3 & 2 k^3+k & 4 k-10 k^2 & 4 k-1 \\ \hline \end{array} $

$ \begin{aligned} & \sum_{i=-1}^2 P\left(X=x_i\right)=1 \\ \Rightarrow & k^3+2 k^3+k+4 k-10 k^2+4 k-1=1 \\ \Rightarrow & 3 k^3-10 k^2+9 k-2=0 \\ \Rightarrow & (k-1)\left(3 k^2-7 k+2\right)=0 \\ \Rightarrow & 3(k-1)(k-2)\left(k-\frac{1}{3}\right)=0 \end{aligned} $

So, $k=1,2$ or $\frac{1}{3}$

$ k \neq 1,2 $

For $k=\frac{1}{3^{\prime}}$,

$ \begin{aligned} \mu & =-1 \cdot\left(\frac{1}{3}\right)^3+1 \cdot\left(\frac{4}{3}-\frac{10}{9}\right)+2 \cdot\left(\frac{4}{3}-1\right) \\ & =\frac{-1}{27}+0+\frac{12-10}{9}+2 \cdot \frac{1}{3} \\ & =\frac{-1}{27}+\frac{2}{9}+\frac{2}{3}=\frac{-1+6+18}{27}=\frac{23}{27} \end{aligned} $

So, $\mu=\frac{23}{27}$

Mean, $\mu=\frac{16}{5}$, variance, $\sigma^2=\frac{48}{25}$

We know that, $\mu=n p$ and $\sigma^2=n p(1-p)$

So, $\frac{16}{5}=n p$ and $\frac{48}{25}=n p(1-p)$

$ \begin{aligned} & \Rightarrow \quad \frac{16}{5}(1-p)=\frac{48}{25} \\ & \Rightarrow \quad 1-p=\frac{48}{25} \times \frac{5}{16}=\frac{3}{5} \end{aligned} $

$ \Rightarrow \quad p=1-\frac{3}{5}=\frac{2}{5}, 1-p=\frac{3}{5} $

Now, $n p=\frac{16}{5}$

$ \begin{aligned} & \Rightarrow \quad n \cdot \frac{2}{5}=\frac{16}{5} \\ & \Rightarrow \quad n=8 \end{aligned} $

$ \begin{aligned} & \text { Now, } \\ & \begin{array}{r} P(X \leq 2)=P(X=0)+P(X=1)+P(X=2) \\ =\binom{8}{0} P^0(1-p)^8+\binom{8}{1} p^1(1-p)^7 +\binom{8}{2} p^2(1-p)^6 \end{array} \\ & =1 \cdot\left(\frac{2}{5}\right)^0 \cdot\left(\frac{3}{5}\right)^8+8 \cdot \frac{2}{5} \cdot\left(\frac{3}{5}\right)^7 +28 \cdot\left(\frac{2}{5}\right)^2 \cdot\left(\frac{3}{5}\right)^6 \end{aligned} \text { } \text { } $

$ \begin{aligned} & =\left(\frac{3}{5}\right)^8+\frac{16 \cdot 3^7}{5^8}+\frac{28 \cdot 4 \cdot 3^6}{5^8} \\ & =\frac{3^8+16 \cdot 3^7+28 \cdot 4 \cdot 3^6}{5^8} \\ & =\frac{3^6\left(3^2+16 \cdot 3+112\right)}{5^8}=\frac{3^6(169)}{5^8} \end{aligned} $

Total childrens $=15(8 B+7 G)$

Total ways of choosing 3 childrens out of 15 is

$ ={ }^{15} C_3=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455 $

Case I Choosing 1 boy and 2 girls

$ ={ }^8 C_1 \cdot{ }^7 C_2=\frac{8 \times 7 \times 6}{2}=168 $

Case II Choosing 2 boys and 1 girl

$ ={ }^8 C_2 \cdot{ }^7 C_1=\frac{8 \times 7}{2} \times 7=196 $

Thus, required probability

$ =\frac{168+196}{455}=\frac{364}{455}=\frac{4}{5} $

Total people $=14(9 \mathrm{M}+5 \mathrm{~W})$

Total ways to choose 4 peoples from 14 is

$ ={ }^{14} C_4 $

Total ways to choose all men is $={ }^9 C_4$

So, total ways to choose atleast one women

$ \begin{aligned} & ={ }^{14} C_4-{ }^9 C_4=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} -\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \\ & =1001-126=875 \end{aligned} $

Hence, required probability

$ =\frac{875}{1001}=\frac{125}{143} $

Total outcomes $=\{1,2,3,4,5,6\}=6$

Prime number $=\{2,3,5\}=3$

So, $P(A)=3 / 6=1 / 2$

and even numbers $=\{2,4,6\}=3$

$\therefore P(B)=3 / 6=1 / 2$

Thus, $P(\bar{B})=1-P(B)=1-1 / 2=1 / 2$

Total sample for 2 throws $=6 \times 6=36$

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

First throw is prime $=\{2,3,5\}$

Second throw is not even $=\{1,3,5\}$

Pairs where first is in $\{2,3,5\}$ and second is in $\{1,3,5\}$

$ \begin{aligned} &=3 \times 3=9 \text { outcomes }\\ &\text { So, } P(A \cap \bar{B})=9 / 36=1 / 4\\ &\text { Therefore, } P(A / \bar{B})=\frac{(1 / 4)}{(1 / 2)}=1 / 2 \end{aligned} $

Let $R_k$ be the event that are exactly $k$ red balls is in bag, where

$ k=1,2,3,4,5 $

So, $P\left(R_k\right)=\frac{1}{5}$ for $k=1,2,3,4,5$

Let $A$ be the event that a red ball is drawn.

$ \begin{aligned} & \begin{aligned} P\left(R_1 / A\right) & =\frac{P\left(R_1 \cap A\right)}{P(A)} \\ \because \quad P\left(A / R_1\right) & =\frac{1}{5} \\ \text { and } P\left(R_1 \cap A\right) & =P\left(R_1\right) \cdot P\left(A / R_1\right) \\ & =\frac{1}{5} \cdot \frac{1}{5}=\frac{1}{25} \end{aligned} \\ & \text { } \begin{aligned}and\,\, & P(A)=\sum_{k=0}^5 P\left(R_k\right) \cdot P\left(A / R_k\right) \\ &=\frac{1}{5} \sum_{k=0}^5 \frac{k}{5}=\frac{1}{25}(1+2+3+4+5) \\ &=\frac{1}{25} \times 15=\frac{15}{25} \\ & \text { Hence, } P\left(R_1 / A\right)=\frac{P\left(R_1 \cap A\right)}{P(A)}=\frac{\frac{1}{25}}{\frac{15}{25}}=\frac{1}{15} \end{aligned} \end{aligned} $

$\because P(X=r)={ }^n C_r p^r(1-p)^{n-r}$

$ \begin{array}{ll} \text { and } & P(X=3)=P(X=6) \\ \Rightarrow & { }^9 C_3 p^3(1-p)^6={ }^9 C_6 p^6(1-p)^3 \\ \Rightarrow & (1-p)^3=p^3 \\ \Rightarrow & \frac{1-p}{p}=1 \\ \Rightarrow & p=1 / 2 \end{array} $

So, $P(X<3)=P(X=0)+P(X=1)+P(X=2)$

$ \begin{aligned} & ={ }^9 C_0\left(\frac{1}{2}\right)^9+{ }^9 C_1\left(\frac{1}{2}\right)^9+{ }^9 C_2\left(\frac{1}{2}\right)^9 \\ & =\frac{1}{512}+\frac{9}{512}+\frac{36}{512}=\frac{46}{512}=23 / 256 \end{aligned} $

We can choose three squares in a diagonal line parallel to $B D$ in the $\triangle A B D$. Clearly, three squares in $\triangle A B D$ and in a diagonal line parallel to $B D$ can be chosen in

$ { }^3 C_3+{ }^4 C_3+{ }^5 C_3+{ }^6 C_3+{ }^7 C_3+{ }^8 C_3 \text { ways } $

Similarly, in $\triangle B C D$ the squares can be chosen parallel to $B D$ in an equal number of ways.

Hence, the total number of ways in which three squares can be chosen in a diagonal line parallel to $B D$ is

$ =2\left({ }^3 C_3+{ }^4 C_3+{ }^5 C_3+{ }^6 C_3+{ }^7 C_3\right)+{ }^8 C_3 $

( $\because B D$ is common to both the triangles) Similarly, squares can be chosen in a diagonal line parallel to $A C$ and hence the total number of favourable ways

$ \begin{aligned} =4\left({ }^3 C_3+{ }^4 C_3+{ }^5 C_3\right. & \left.+{ }^6 C_3+{ }^7 C_3\right) +2 \cdot{ }^8 C_3=392 \end{aligned} $

Hence, the required probability

$ =\frac{392}{{ }^{64} C_3}=\frac{392 \times 6}{64 \times 63 \times 62}=\frac{7}{744} $

Total number of ways to choose 4 shoes from $8={ }^8 C_4=70$

We want to choose 4 shoes that none of them form a pair.

Firstly, choose 4 different pairs from the 4 available pairs $={ }^4 C_4=1$

For each of the 4 pairs, 2 choices (left or right) $=2^4=16$.

Hence, total number of ways to choose 4 shoes with no pair $=1 \times 2^4=16$.

Thus, number of ways with atleast one pair $=70-16=54$.

$\therefore$ Required probability $=\frac{54}{70}=\frac{27}{35}$