Probability

$ \begin{aligned} &\text { Given, there are } 10 \text { coins in a box. }\\ &\begin{aligned} & 8 \text { normal }+2 \text { biased } \\ & P(\text { biased coin/appear head }) \\ & \quad=\frac{P(\text { biased coin } \cap \text { appear head })}{P(\text { appear head })} \\ & \quad=\frac{\frac{2}{10} \times\left(\frac{2}{2}\right)^6}{\frac{2}{10} \times\left(\frac{2}{2}\right)^6+\frac{8}{10} \times\left(\frac{1}{2}\right)^6} \\ & \quad=\frac{2}{2+8 \times \frac{1}{64}}=\frac{2 \times 8}{16+1}=\frac{16}{17} \end{aligned} \end{aligned} $

$ \begin{array}{ccccccc} \hline \boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}} & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{P}\left(\boldsymbol{X}=\boldsymbol{x}_{\boldsymbol{i}}\right) & 0.1 & k & 0.2 & 2 k & 3 k & k \\ \hline \end{array} $

$ \begin{aligned} &\text { Total probability }\\ &\begin{aligned} & 0.1+k+0.2+2 k+3 k+k=1 \\ & \Rightarrow \quad 7 k+0.3=1 \\ & \Rightarrow \quad k=0.1 \\ & \therefore \sigma^2=\sum x^2 p(x)-\left(\sum x p(x)\right)^2 \\ & \text { Now, } \sum x^2 p(x)=4 \times(0.1)+1(k)+0(0.2) +1(2 k)+4(3 k)+9(k) \\ & =2.8 \\ & \text { and } \sum x p(x)=-0.2-0.1+0+0.2 +0.6+0.3=0.8 \\ & \Rightarrow \quad \sigma^2=2.8-(0.8)^2=2.16 \end{aligned} \end{aligned} $

Total balls $=4$

Out of 4 balls, 2 balls are white.

$E_1=$ Event that all balls white.

$E_2=$ Event that 3 balls are white.

$E_3=$ Event that 2 balls are white.

$ \because P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3} $

So, $P\left(E_1 / A\right)$

$ \begin{gathered} =\frac{P\left(A / E_1\right) \cdot P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)} \\ +P\left(A / E_3\right) P\left(E_3\right) \\ \because P\left(A / E_1\right)={ }^4 C_2 /{ }^4 C_2=1, \\ P\left(A / E_2\right)=\frac{{ }^3 C_2}{{ }^4 C_2}=1 / 2 \\ P\left(A / E_3\right)={ }^2 C_2 /{ }^4 C_2=1 / 6 \\ \therefore P\left(E_1 / A\right)=\frac{1 \times 1 / 3}{1 / 3+1 / 2 \cdot 1 / 3+1 / 6 \cdot 1 / 3} \\ =3 / 5 \end{gathered} $

If the equation $x^2+a x+b=0$ has real roots, then

$ a^2-4 b \geq 0 \Rightarrow a^2 \geq 4 b $

$\because a \in A$ and $A=\{3,4,5\}$

and $b \in B$ and $B=\{1,2,3,4\}$

When, $a=3 \Rightarrow a^2=9$

Then, possible values of $b=1,2$

When, $a=4 \Rightarrow a^2=16$

Then, possible values of $b=1,2,3,4$ and when $a=5 \Rightarrow a^2=25$

Then, possible values of $b=1,2,3,4$

Therefore, total possible cases

$ =2+4+4=10 $

and total combinations of $a$ and

$ b=3 \times 4=12 $

Hence, probability that the equation

$ x^2+a x+b=0 $

has real roots $=\frac{10}{12}=\frac{5}{6}$

$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{X}=\boldsymbol{x} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \boldsymbol{P} & 0.15 & 0.23 & k & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\ (\boldsymbol{X}=\boldsymbol{x}) & & & & & & & & \\ \hline \end{array} $

$ \begin{aligned} & \because \sum P(X=x)=1 \\ & \Rightarrow 0 \cdot 15+0 \cdot 23+k+0 \cdot 10+ \\ & 0 \cdot 20+0 \cdot 08+0 \cdot 07+0 \cdot 05=1 \\ & \Rightarrow \quad k+0.88=1 \\ & \Rightarrow \quad k=0.12 \\ & \because E=\{x / x, \text { is a prime number }\} \\ & =\{2,3,5,7\} \\ & F=\{x \mid x<4\}=\{1,2,3\} \\ & \text { Then, } E \cup F=\{1,2,3,5,7\} \\ & \text { So, } P(E \cup F)=P(X=1) \\ & +P(X=2)+P(X=3)+P(X=5)+P(X=7) \\ & =0.15+0.23+0.12+0.20+0.07 \\ & =0.77 \end{aligned} $

$\because$ Total floor $=7$,

at first floor, total persons $=5$

Then, remaining floor $=6$

Now, total number of ways in which each of the 5 persons can leave cabin at any of the 6 floors $=6^5$ (total)

And favourable number of ways, that is number of ways in which 5 persons leave at different floors $={ }^6 P_5=6$

Therefore, required probability $=\frac{{ }^6 P_5}{6^5}$

$ =\frac{\left(\frac{6!}{1!}\right)}{6^5}=\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 6 \cdot 6 \cdot 6 \cdot 6}=\frac{5}{54} $

∵ A bag contains 3 red balls, 5 black balls and 7 blue balls

$ \text { Total outcomes }={ }^{15} \mathrm{C}_3 $

∴ Probability of getting atleast two blue balls

$ \begin{aligned} & =\frac{\text { Favourable outcomes }}{\text { Total outcomes }} \\ & =\frac{{ }^7 C_2 \cdot{ }^3 C_1+{ }^7 C_2 \cdot{ }^5 C_1+{ }^7 C_3}{{ }^{15} C_3} \\ & =\frac{\frac{7 \times 6}{2} \times 3+\frac{7 \times 6}{2} \times 5+\frac{7 \times 6 \times 5}{3 \times 2}}{\frac{15 \times 14 \times 13}{3 \times 2}} \\ & =\frac{203}{455}=\frac{29}{65} \end{aligned} $

We have, two independent events in which two dice are thrown simultaneously by person A and two cards are drown at random simultaneously from a pack of 52 playing cards by a person $B$.

Consider the first event when $A$ thrown two dice.

Total possible outcomes, $n(s)=36$

Favourable outcomes when sum is prime

$ [(1,1)]\left[\begin{array}{l} (1,2) \\ (2,1) \end{array}\right] \ldots \ldots \ldots $

So, number of favourable outcome $=15 \Rightarrow P(A$ win the game $)=15 / 36$

And, now second event when $B$ drawns two cards. Total possible outcomes $={ }^{52} C_2$

$B$ win the game when $B$ gets a face card, a card having prime number.

Total face card = 12 and total card having prime number $=16$

$ \begin{aligned} \Rightarrow \text { Favourable outcomes } & =\frac{{ }^{12} C_1 \times{ }^{16} C_1}{{ }^{52} C_2} \\ & =\frac{12 \times 16}{\frac{52 \times 51}{2}}=\frac{192}{1326} \end{aligned} $

Hence, $P$ (both $A$ and $B$ win)

$ =\frac{15}{36} \times \frac{192}{1326}=\frac{40}{663} $

[ ∵ both events are independent]

Sample space when three coin tossed simultaneously {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Winning probability in each toss

$ (A \text { or } B)=\frac{5}{8} $

And, loosing probability in each toss

$ (A \text { or } B)=\frac{3}{8} $

∴ Required probability

$ \begin{aligned} & =\frac{5}{8} \times \frac{3}{8}+\frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{3}{8} \\ & \quad+\frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{5}{8} \times \frac{3}{8}+\ldots \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{5}{8} \times \frac{3}{8}+\left(\frac{5}{8}\right)^3 \times \frac{3}{8}+\left(\frac{5}{8}\right)^5 \times \frac{3}{8}+\ldots \\ & =\frac{5}{8} \times \frac{3}{8}\left[1+\left(\frac{5}{8}\right)^2+\left(\frac{5}{8}\right)^4+\ldots\right] \\ & =\frac{15}{64} \times \frac{1}{1-\frac{25}{64}} \end{aligned}\\ &\text { [ ∵ Sum of infinite, GP }=\frac{a}{1-r} \text { ] }\\ &=\frac{15}{64} \times \frac{64}{39}=\frac{15}{39} \end{aligned} $

Number of ways to select two cards

$ \begin{aligned} & ={ }^{52} C_2 \\ & =\frac{52 \times 51}{2}=1326 \end{aligned} $

Number of ways of getting a face card and a spade card other than the face card

$ \begin{aligned} & ={ }^{12} C_1 \times{ }^{10} C_1 \\ & =12 \times 10=120 \end{aligned} $

$ \begin{aligned} &\text { Required probability }\\ &=\frac{120}{1326}=\frac{20}{221} \end{aligned} $

$ \begin{aligned} & \text { Total outcomes }=6 \times 6 \times 6=216 \\ & \text { Favourable outcomes }=6 \times 5 \times 4=120 \\ & \text { Required probability }=\frac{120}{216}=\frac{5}{9} \end{aligned} $

Chance of $A, B, C$ in getting through the test $=1: 2: 3$

Chance of $A: P(A)=\frac{1}{6}$

Chance of $B: P(B)=\frac{2}{6}$

Chance of $C: P(C)=\frac{3}{6}$

If $I$ denotes the facing interview successfully, then

$ \begin{aligned} P(I \mid A) & =0.8 \\ P(I \mid B) & =0.7 \\ P(I \mid C) & =0.6 \\ P(I)= & P(A) P(I \mid A)+P(B) P(I \mid B)+P(C) P(I \mid C) \\ = & \frac{1}{6} \times 0.8+\frac{2}{6} \times 0.7+\frac{3}{6} \times 0.6=\frac{4}{6}=\frac{2}{3} \\ \text { Now, } P(A \mid I)= & \frac{P(A) P(I \mid A)}{P(I)}=\frac{0.8 \times \frac{1}{6}}{\frac{4}{6}}=\frac{1}{5} \end{aligned} $

Let $X$ be the number of spades, then $X=0,1,2$

$ \begin{gathered} P(X=0)=\frac{39}{52} \times \frac{39}{52}=\left(\frac{39}{52}\right)^2 \\ P(X=1)=\frac{13}{52} \times \frac{39}{52} \times 2=\frac{1014}{(52)^2} \\ P(X=2)=\frac{13}{(52)} \times \frac{13}{(52)}=\frac{169}{(52)^2} \\ \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ =\left(0^2 \times \frac{39^2}{(52)^2}+1^2 \times \frac{1014}{(52)^2}+2^2 \times \frac{169}{(52)^2}\right) \\ \quad-\left(0 \times \frac{39^2}{(52)^2}+1 \times \frac{1014}{(52)^2}+2 \times \frac{169}{(52)^2}\right) \\ =\frac{1690}{52^2}-\frac{(1352)^2}{(52)^4}=\frac{1014}{52^2}=\frac{3}{8} \end{gathered} $

Given, $P(A \cup B)=P(A \cap B)$

We know,

$ \begin{aligned} & P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(A)+P(B)=2 P(A \cap B) \neq 1 \end{aligned} $

Hence, option (d) is not true.

Let us consider, $A, B, C$, where $C$ is between $A$ and $B$. Now choice for $C$ is ${ }^4 C_1$ and $A, B$ interchange their placer in 2 ways.

Now, $A, B, C$ and other 3 students stand in a row in 4! ways.

So, total number of ways to stand in a row such that two $A, B$ and separated by one students in between them is

$ 4 \times 2 \times 4! $

Total number of ways to stand without restriction $=6!$

$ \begin{aligned} \text { Probability } & =\frac{4 \times 2 \times 4!}{6!}=\frac{4 \times 2 \times 4!}{6 \times 5 \times 4!} \\ & =\frac{8}{30}=\frac{4}{15} \end{aligned} $

Let ' $E$ ' be the event of any one cutting a spade in one out first, and ' $S$ ' be the sample space.

Then, $n(E)={ }^{13} C_1$ and $n(S)={ }^{52} C_1$

Now, $p(E)=\frac{n(E)}{n(S)}=\frac{{ }^{13} C_1}{{ }^{52} C_1}=\frac{13}{52}=\frac{1}{4}$

$ \therefore \quad q=1-p=1-\frac{1}{4}=\frac{3}{4} $

Now, probability of $A$ wins the game is

$ \begin{aligned} & =P+q^4 P+q^8 P+\ldots \infty \\ & =\frac{P}{1-q^4}=\frac{1 / 4}{1-\left(\frac{3}{4}\right)^4}=\frac{64}{256-81}=\frac{64}{175} \end{aligned} $

Since, Their are total two bad eggs. So, while drawing 3 eggs we can draw 0 bad eggs, or 1 bad eggs or 2 bad eggs.

Let ' $x$ ' be the random variable denoting number of bad eggs that can be drawn in each draw.

Thus, $P(x=0)$

$ =\frac{{ }^2 C_0 \times{ }^{10} C_3}{{ }^{12} C_3}=\frac{120}{220}=\frac{6}{11} $

Similarly, $P(x=1)=\frac{{ }^2 C_1 \times{ }^{10} C_2}{{ }^{12} C_3}=\frac{9}{22}$

Now,

$ \begin{array}{cccc} \hline \boldsymbol{x} & \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{x} \boldsymbol{P}_{\boldsymbol{i}} & \boldsymbol{x}^2 \boldsymbol{P}_{\boldsymbol{i}} \\ \hline 0 & 6 / 11 & 0 & 0 \\ \hline 1 & 9 / 22 & 9 / 22 & 9 / 22 \\ \hline 2 & 1 / 22 & 1 / 11 & 2 / 11 \\ \hline \end{array} $

$ \begin{aligned} &\begin{aligned} \therefore \mu & =\sum_{i=0,1,2} x P_i=0+\frac{9}{22}+\frac{1}{11}=\frac{11}{22}=\frac{1}{2} \\ \sigma^2 & =\sum x^2 P_i-\mu^2 \\ & =\frac{9}{22}+\frac{2}{11}-\frac{1}{4}=\frac{18+8-11}{44}=\frac{15}{44} \end{aligned}\\ &\text { Thus, variance is } \frac{15}{44} \text {. } \end{aligned} $

$n=6$

$P=$ Probability of a question marked with correct answer $=1 / 2$

$ \begin{aligned} q & =\frac{1}{2} \\ P(X \geq 4) & ={ }^6 C_4\left(\frac{1}{2}\right)^6+{ }^6 C_5\left(\frac{1}{2}\right)^6+{ }^6 C_6\left(\frac{1}{2}\right)^6 \\ & =\frac{1}{64}(15+6+1)=\frac{22}{64}=\frac{11}{32} \end{aligned} $

To find the value of the constant $c$, we need to ensure that the probability distribution function sums to 1. This means:

Write the sum of the probability mass function:

$\sum\limits_{x=1}^{\infty} c\left(\frac{2}{3}\right)^x = 1.$

Factor out the constant $c$:

$c \sum\limits_{x=1}^{\infty} \left(\frac{2}{3}\right)^x = 1.$

Recognize that the series

$\sum\limits_{x=1}^{\infty} \left(\frac{2}{3}\right)^x$

is a geometric series with first term

$a = \left(\frac{2}{3}\right)$

and common ratio

$r = \frac{2}{3}.$

The sum of a geometric series starting at $x=1$ is given by:

$\sum\limits_{x=1}^{\infty} a\,r^{x-1} = \frac{a}{1-r}.$

However, since our first term is already $\left(\frac{2}{3}\right)^1$, we have:

$\sum\limits_{x=1}^{\infty} \left(\frac{2}{3}\right)^x = \frac{\frac{2}{3}}{1 - \frac{2}{3}}.$

Simplify the series sum:

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

Substitute back into the equation:

$c \cdot 2 = 1.$

Solve for $c$:

$c = \frac{1}{2}.$

Thus, the value of $c$ is $\frac{1}{2}$, which corresponds to Option C.

In a non-leap year, we need to determine the probability of having either 53 Sundays, 53 Tuesdays, or 53 Thursdays.

Calculation

A non-leap year consists of 365 days.

Since each week contains 7 days, a year typically has $\frac{365}{7} = 52$ full weeks with 1 extra day.

The extra day determines whether one of the days of the week occurs 53 times. The extra day can be any day of the week:

Sample Space (S): {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Thus, the probability sample space $n(S)$ is 7.

Event A: {Sunday, Tuesday, Thursday}

Since Sunday, Tuesday, and Thursday can each be this extra day, we have:

Number of favorable outcomes $n(A) = 3$ (Sunday, Tuesday, Thursday)

Using the probability formula, the probability of Event A is:

$ P(A) = \frac{n(A)}{n(S)} = \frac{3}{7} $

Thus, the probability of having either 53 Sundays, 53 Tuesdays, or 53 Thursdays in a non-leap year is $\frac{3}{7}$.

Wrong question

Given the problem statement that $P(A) + P(B) = 2 P(A \cap B)$, we can deduce several points based on probability theory.

First, recall the formula for the probability of the union of two events:

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

From the given condition, substitute the information provided:

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

This simplifies to:

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

This equation implies that the probability of the intersection of events $A$ and $B$ is equal to the probability of their union. For this statement to hold true, there are specific conditions that must be satisfied:

The events $A$ and $B$ are such that the probability of their overlap (intersection) is the entire probability of the union, meaning that the events are not just overlapping — they are covering the entire probability space available to them for their union. This can only happen under specific circumstances, which include:

Both events might be equivalent, which implies $P(A) = P(B)$. This means both events have the same probability and overlap completely.

Thus, from the given condition, $P(A) = P(B)$ is a valid conclusion.

Two cases

(i) A spade king + a spade card Total ways $=12$

(ii) A non spade king + a spade card Total ways $=3 \times 13$

$ \begin{aligned} \therefore \text { Required probability } & =\frac{3 \times 13+12}{{ }^{52} C_2} \\ & =\frac{51}{52 \times 51} \times 2=\frac{2}{52}=\frac{1}{26} \end{aligned} $

$P_1=P$ (first card is queen) $\cdot P$ (second card is diamond)

$ =\frac{4}{52} \times \frac{13}{52}=\frac{1}{52} $

$P_2$ : Case $I$ queen is not diamond

$ \rightarrow \frac{3}{52} \times \frac{13}{51} $

Case II queen is diamond $\rightarrow \frac{1}{52} \times \frac{12}{51}$

$ \begin{aligned} \therefore \quad P_2 & =\frac{3}{52} \times \frac{13}{51}+\frac{1}{52} \times \frac{12}{51} \\ & =\frac{39+12}{52 \times 51}=\frac{51}{52 \times 51}=\frac{1}{52} \end{aligned} $

Therefore, $\frac{P_1}{P_2}=\frac{1 / 52}{1 / 52}=1$

Let $E_1=$ Event that the ball is drawn from bag $A$.

$E_2=$ Event that the ball is drawn from bag B.

$E_3=$ Event that the ball is drawn from bag $C$. .

Let $E$ be the event that the ball drawn is black.

$ \begin{aligned} P\left(E_1\right)= & P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3} \\ P\left(\frac{E}{E_1}\right)= & \frac{2}{6}, P\left(\frac{E}{E_2}\right)=\frac{3}{6}, P\left(\frac{E}{E_3}\right)=\frac{4}{6} \\ P(E)= & P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right) \\ & +P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right) \\ = & \frac{1}{3} \times \frac{2}{6}+\frac{1}{3} \times \frac{3}{6}+\frac{1}{3} \times \frac{4}{6} \\ = & \frac{2+3+4}{18}=\frac{9}{18}=\frac{1}{2} \end{aligned} $

$ \begin{array}{lccccccccc} \hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline P\left(x=x_i\right) & 10 k & 9 k & 8 k & 8 k & 6 k & 5 k & 4 k & 3 k & k \\ \hline \end{array} $

$ \begin{aligned} &\text { We know, } \Sigma P=1\\ &\begin{aligned} \Rightarrow 10 k+9 k+8 k+8 k & +6 k+5 k +4 k+3 k+k=1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 54 k=1 \Rightarrow k=\frac{1}{54} \\ & A=\left\{x_i \mid x_i\right. \text { is a prime number } \\ & B=\left\{x_i \mid x_i>5\right\} \\ & P(A)=9 k+8 k+6 k+4 k \\ &(\because 2,3,5,7 \text { are prime numbers }) \\ &=27 \times \frac{1}{54}=\frac{27}{54} \\ & P(B)=5 k+4 k+3 k+k \\ &(\because 6,7,8,9 \text { are more than } 5) \\ &=13 \times \frac{1}{54}=\frac{13}{54} \end{aligned} $

$ P(A \cap B)=4 k $

( $\because 7$ is only prime which is more than 5 in the, given problem)

$ =4 \times \frac{1}{54}=\frac{4}{54} $

$ \begin{aligned} P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{27+13-4}{54} \\ & =\frac{36}{54}=\frac{2}{3} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{5}{3} k=P(X=2)=P(X=3) \\ & P(X=x)=\frac{\lambda^x e^{-\lambda}}{x!} \\ & P(X=2)=P(X=3) \\ & \Rightarrow \quad \frac{\lambda^2 e^{-\lambda}}{2!}=\frac{\lambda^3 e^{-\lambda}}{3!} \Rightarrow \frac{1}{2}=\frac{\lambda}{6} \\ &\Rightarrow \lambda=3 \\ & \text { Now, } P\left(X=9=\frac{\lambda^5 e^{-\lambda}}{5!}\right. \\ & =\lambda^3\left(\frac{\lambda^2 e^{-\lambda}}{2!}\right) \cdot \frac{2!}{5!} \\ & =(3)^3\left(\frac{5}{3} k\right) \cdot \frac{2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} \\ & =\frac{3}{4} k \end{aligned} \end{aligned} $

9 identical black balls numbered from 1 to 9

4 identical white balls numbered from 1 to 4

3 balls are drawn

$P$ (at least 1 white ball)

$ \begin{aligned} & =P(1 W, 2 R)+P(2 W, 1 R)+P(3 W, 0 R) \\ & =\frac{\left(4{ }^4 C_1 \times{ }^9 C_2\right)+\left({ }^4 C_2 \times{ }^9 C_1\right)+\left({ }^4 C_3 \times{ }^9 C_0\right)}{{ }^{13} C_3} \\ & =\frac{(4 \times 36)+(6 \times 9)+(4 \times 1)}{\frac{13 \times 12 \times 11}{3 \times 2}}=\frac{101}{143} \end{aligned} $

Let $P(A)=\frac{1}{4}, P(B)=\frac{1}{5}$ probability of target hit

$ \begin{aligned} & =P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(A) \cdot P(B) \\ & =\frac{1}{4} \times \frac{4}{5}+\frac{3}{4} \times \frac{1}{5}+\frac{1}{4} \times \frac{1}{5}=\frac{8}{20}=\frac{2}{5} \end{aligned} $

3 dice are thrown

$A=$ Sum of the number appeared to be 15

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

$B=$ The number 5 does not appear on any of the dice

$ \begin{aligned} & =\{(6,6,3),(6,3,6),(3,6,6)\} \\ \therefore \quad P\left(\frac{B}{A}\right) & =\frac{n(A \cap B)}{n(A)}=\frac{3}{10} \end{aligned} $

We know that sum of total probability is 1 .

$ \begin{array}{ll} \Rightarrow & 0+k+2 k+5 k^2+2 k^2+3 k=1 \\ \Rightarrow & 7 k^2+6 k-1=0 \\ \Rightarrow & 7 k^2+7 k-k-1=0 \\ \Rightarrow & (7 k-1)(k+1)=0 \Rightarrow k=\frac{1}{7}, k \neq-1 \end{array} $

Mean of

$ \begin{aligned} X & =0+2 k+8 k+30 k^2+16 k^2+30 k \\ & =40 k+46 k^2=k(40+46 k) \\ & =\frac{1}{7}\left(40+\frac{46}{7}\right)=\frac{1}{7} \times \frac{326}{7}=\frac{326}{49} \end{aligned} $

Let probability of success is $p$ and that of failure is $q$.

Given, $p=5 q$

We know that, $p+q=1 \Rightarrow 5 q+q=1$

$ \Rightarrow \quad q=\frac{1}{6}, p=\frac{5}{6} $

p(atmost one success)

$ \begin{aligned} & =p(0 \text { success })+p(1 \text { success }) \\ & ={ }^5 C_0 p^0 q^5+{ }^5 C_1 p^1 q^4 \\ & =1 \times\left(\frac{5}{6}\right)^0 \times\left(\frac{1}{6}\right)^5+5 \times \frac{5}{6} \times\left(\frac{1}{6}\right)^4=\frac{26}{6^5} \end{aligned} $

There are total 9 balls in which 3 balls are white and 6 balls are red.

Total number of observation $={ }^9 C_4$

Number of favourable outcomes $=(2$

Red 2 Black

$ \begin{gathered} +3 \text { Red } \times 1 \text { Black }+4 \text { Red) balls } \\ ={ }^6 C_2 \times{ }^3 C_2+{ }^6 C_3 \times{ }^3 C_1+{ }^6 C_4 \end{gathered} $

$P$ (atleast two red balls)

$ \begin{aligned} & =\frac{{ }^6 C_2 \times{ }^3 C_2+{ }^6 C_3 \times{ }^3 C_1+{ }^6 C_4}{{ }^9 C_4} \\ & =\frac{15 \times 3+20 \times 3+15}{126} \\ & =\frac{45+60+15}{126}=\frac{120}{126}=\frac{60}{63} \end{aligned} $

$P(A)=\frac{2}{5}$ and $P(B)=\frac{1}{3}$

$A$ and $B$ are independent events,

$ \begin{aligned} P(A \cap B) & =P(A) \times P(B) \\ & =\frac{2}{5} \times \frac{1}{3}=\frac{2}{15} \\ P(A-B) & =P(A)-P(A \cap B) \\ & =\frac{2}{5}-\frac{2}{15}=\frac{4}{15} \end{aligned} $

$ \begin{aligned} P(\bar{A} \cup B) & =1-P(A-B)=1-\frac{4}{15}=\frac{11}{15} \\ P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{2}{5}+\frac{1}{3}-\frac{2}{15} \\ & =\frac{6+5-2}{15}=\frac{3}{5} \end{aligned} $

$ \begin{aligned} P\left(\frac{A}{\bar{B}}\right) & =\frac{P(A \cap \bar{B})}{P(\bar{B})}=\frac{P(A-B)}{1-P(B)} \\ & =\frac{4}{15}+\left(1-\frac{1}{3}\right)=\frac{4}{15} \times \frac{3}{2}=\frac{2}{5} \\ P\left(\frac{\bar{B}}{A}\right) & =\frac{P(\bar{B} \cap A)}{P(A)}=\frac{P(A-B)}{P(A)} \\ & =\frac{4}{15}+\frac{2}{5}=\frac{2}{3} \end{aligned} $

The total possible outcomes are (H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6).

∴ Total number of outcomes $=12$

$ \begin{aligned} & \text { Probability of winning }=\frac{1}{12} \\ & \text { and Probability of losing }=\frac{11}{12} \end{aligned} $

$P$ (B wins)

$ \begin{aligned} = & P(\bar{A} B+\overline{A B A} B+\overline{A B A B A} B+\ldots) \\ = & \frac{11}{12} \times \frac{1}{12}+\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12} \\ & +\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}+\ldots \infty \\ = & \frac{11}{144}\left[1+\left(\frac{11}{12}\right)^2+\left(\frac{11}{12}\right)^4+\ldots \infty\right] \end{aligned} $

which form a GP.

$ =\frac{11}{144}\left[\frac{1}{1-\frac{121}{144}}\right]=\frac{11}{144} \times \frac{144}{23}=\frac{11}{23} $

When two dices are thrown there are total 36 outcomes.

Here, $X$ denotes the sum of the number that shows the faces of dice.

The sum of two dice can be

$ 2,3,4,5,6,7,8,9,10,11,12 $

$ \begin{aligned} & P(X=2)=\frac{1}{36} \\ & P(X=3)=\frac{2}{36} \\ & P(X=4)=\frac{3}{36} \\ & P(X=5)=\frac{4}{36} \\ & P(X=6)=\frac{5}{36} \\ & P(X=7)=\frac{6}{36} \\ & P(X=8)=\frac{5}{36} \\ & P(X=9)=\frac{4}{36} \\ & P(X=10)=\frac{3}{36} \\ & P(X=11)=\frac{2}{36} \\ & P(X=12)=\frac{1}{36} \end{aligned} $

$ \begin{aligned} & \text { Mean }=\Sigma x_i p_i \\ & \begin{aligned} =\frac{2 \times 1}{36}+ & \frac{3 \times 2}{36}+\frac{4 \times 3}{36}+\frac{5 \times 4}{36}+\frac{6 \times 5}{36} \\ & +\frac{7 \times 6}{36}+\frac{8 \times 5}{36}+\frac{9 \times 4}{36}+\frac{10 \times 3}{36} \\ & +\frac{11 \times 2}{36}+\frac{12 \times 1}{36} \\ = & \frac{2+6+12+20+30+42+40+36+30+22+12}{36} \\ = & \frac{252}{36}=7 \end{aligned} \end{aligned} $

Let $p=P$ (chosen an engineering student) $=\frac{1}{5}$

$q=p$ (not chosen an engineering student)

$ \begin{aligned} & =1-\frac{1}{5}=\frac{4}{5} \\ P(X \leq 2) & =P(X=0)+P(X=1)+P(X=2) \\ & ={ }^8 C_0\left(\frac{1}{5}\right)^0\left(\frac{4}{5}\right)^8+{ }^8 C_1\left(\frac{1}{5}\right)^1\left(\frac{4}{5}\right)^7 +{ }^8 C_2\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^6 \\ & =\frac{4^6}{5^8}[16+32+28] \\ & =\frac{4^6}{5^8} \times 76=19 \times \frac{4^7}{5^8} \end{aligned} $

The total number of outcomes is 36.

Favourable outcomes $=(1,1)(1,2)$, $(1,3)(1,4),(2,1),(2,2),(2,3),(2,4)$, $(3,1),(3,2),(3,3),(3,4),(4,1)(4,2) (4,3)$.

Number of favourable outcomes $=15 P\left(\right.$ ordered pair such that $\left.x^2+y^2 \leq 25\right) =\frac{15}{36}=\frac{5}{12}$

There are total 8 cards, which are multiples of 5 i.e. 4 cards of 5 and 4 cards of 10 .

There are total 16 cards of prime number i.e. 4 cards each of $2,3,5$ and 7.

Case I When 5 number cards are selected from multiple of 5 .

Number of ways of selecting the card

$ ={ }^4 C_1 \times{ }^{15} C_1=4 \times 15=60 $

Case II When 10 number cards are selected from multiple of 5 , number of ways of selecting the card $={ }^4 C_1 \times{ }^{16} C_1$

$ =4 \times 16=64 $

$\therefore P$ (card having prime number and a card having a number which is multiple of 5)

$ \begin{aligned} & =\frac{60+64}{{ }^{52} C_2}=\frac{124 \times 2}{52 \times 51} \\ & =\frac{62}{663} \end{aligned} $

$ \begin{aligned} & \text {Given, } P(\bar{A})=\frac{2}{3} \\ & \Rightarrow \quad P(A)=\frac{1}{3} \Rightarrow P(B)=\frac{4}{15} \\ & \Rightarrow \quad P(\bar{B})=\frac{11}{15} \end{aligned} $

$ \begin{aligned} & \text { Given, } P(A \cap \bar{B})=\frac{1}{5} \\ & \therefore P(A)-P(A \cap B)=\frac{1}{5} \\ & P(A \cap B)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15} \end{aligned} $

$ \begin{aligned} &\begin{aligned} P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ & =\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15} \end{aligned}\\ &\begin{aligned} P(A \cup \bar{B}) & =1-P(B-A) \\ & =1-P(B)+P(A \cap B) \\ & =1-\frac{4}{15}+\frac{2}{15}=\frac{13}{15} \end{aligned}\\ &\begin{aligned} P(B \mid 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 \cup \bar{B})} \\ & =\frac{P(B \cap A)}{P(A \cup \bar{B})}=\frac{2}{13} \end{aligned}\\ &\text { So, } \sqrt{195[P(B \mid A \cup \bar{B})+P(A \cup B)]}\\ &\begin{aligned} & =\sqrt{195\left[\frac{2}{13}+\frac{7}{15}\right]}=\sqrt{195\left[\frac{30+91}{195}\right]} \\ & =\sqrt{121}=11 \end{aligned} \end{aligned} $

Given, $P(X=r)=k(1+r) \times 3^{-r}$

$ \begin{gathered} P(X=0)+P(X=1)+P(X=2)+\ldots=1 \\ \frac{k(1+0)}{3^0}+\frac{k(1+1)}{3}+\frac{k(1+2)}{3^2}+\ldots=1 \\ k+\frac{2 k}{3}+\frac{3 k}{3^2}+\ldots=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

Multiply by 3, we get

$ 3 k+2 k+\frac{3 k}{3}+\frac{4 k}{3^2}+\frac{5 k}{3^3}+\ldots=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} &\text { Subtracting Eq. (i) from Eq. (ii), }\\ &\begin{gathered} 3 k+k+\frac{k}{3}+\frac{k}{3^2}+\ldots=2 \\ 3 k+\frac{k}{1-\frac{1}{3}}=2 \\ {\left[\because k+\frac{k}{3}+\frac{k}{3^2}+\ldots \text { form a GP }\right]} \\ 3 k+\frac{3 k}{2}=2 \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{9 k}{2}=2 \Rightarrow k=\frac{4}{9} \\ \quad P(X=0)+P(X=1)+P(X=2) \\ k+\frac{2 k}{3}+\frac{3 k}{3^2} \\ k\left[1+\frac{2}{3}+\frac{1}{3}\right] \\ \frac{4}{9}[1+1]=\frac{8}{9} \end{gathered} \end{aligned} $

Let $p=\frac{\alpha}{\beta}, q=1-\frac{\alpha}{\beta}=\frac{\beta-\alpha}{\beta}$

$P$ (he fails at least $n-1$ times)

$ \begin{aligned} = & P(X=0)+P(X=1) \\ = & { }^n C_0\left(\frac{\alpha}{\beta}\right)^0\left(\frac{\beta-\alpha}{\beta}\right)^n +{ }^n C_1\left(\frac{\alpha}{\beta}\right)\left(\frac{\beta-\alpha}{\beta}\right)^{n-1} \\ = & \frac{(\beta-\alpha)^{n-1}}{\beta^n} \quad[\beta-\alpha+n \alpha] \end{aligned} $

Getting prime on one die $=1 / 2$ Getting composite on one die =1/3 ∴ Probability of getting prime on one die and composite on the other

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

As, $A, B$ and $C$ are pairwise independent, $\bar{B} \cap \bar{C}$ and $A$ are independent.

$ \begin{aligned} P\left(\frac{\bar{B} \cap \bar{C}}{A}\right) & =\frac{P(\bar{B} \cap \bar{C}) \cap A}{P(A)} \\ & =\frac{P(\bar{B} \cap \bar{C}) P(A)}{P(A)} \\ & =P(\bar{B} \cap \bar{C}) \end{aligned} $

$ \begin{aligned} & \text { Now, } P(\overline{B \cup \bar{C}})=1 / 2 \\ & \Rightarrow \quad P(\overline{B \cap C})=1 / 2 \\ & \Rightarrow \quad P(B \cap C)=1-\frac{1}{2}=\frac{1}{2} \\ & \therefore \quad P(B \cup C)=P(B)+P(C)-P(B \cap C) \\ & \begin{array}{l} \Rightarrow \quad c-\frac{1}{2} \\ \therefore \quad P(\bar{B} \cup \bar{C})=1-P(B \cup C) \\ =1-b-c+\frac{1}{2}=\frac{3}{2}-b-c \end{array} \end{aligned} $

$n(s)=36$

$ \begin{aligned} & E=\text { Sum of numbers is a multiple of } 4 \\ & F=4 \text { has appeared at least once } \\ & E=\{(1,3),(2,2),(3,1),(2,6),(3,5), \\ & (4,4),(5,3),(6,2),(6,6)\} \\ & \qquad P(E)=\frac{9}{36} \\ & F=\{(1,4),(2,4),(3,4),(4,1),(4,4),(4,3), \\ & (4,4),(4,5),(4,6),(5,4),(6,4)\} \\ & \qquad P(F)=\frac{11}{36} \end{aligned} $

$ \begin{array}{ll} \therefore & (E \cap F)=\{(4,4)\}, P(E \cap F)=\frac{1}{36} \\ \therefore & p=P\left(\frac{F}{E}\right)=\frac{P(F \cap E)}{P(E)}=\frac{1 / 36}{9 / 36}=\frac{1}{9} \\ \therefore & 3 p+2=3 \times \frac{1}{9}+2=\frac{7}{3} \end{array} $

Here, $n=5$,

$p=$ probability of getting head in single throw of a coin $=\frac{1}{2}$

∴ Mean $=n p=5 / 2$

$ \begin{aligned} &\text { Variance }=2\\ &\begin{aligned} \therefore \quad P(X & =k)=\frac{2^k \cdot e^{-2}}{k!} \\ P(X \geq 3) & =1-[P(X=0)+P(X \\ & \quad=1)+P(X=2)] \\ & =1-e^{-2}[1+2+2] \\ & =\frac{e^2-5}{e^2} \end{aligned} \end{aligned} $

So, volume of cube having edge of length 5 cm and all faces are painted

$ =5^3=125 \mathrm{~cm}^3 $

Let cube be divided into ' $n$ ' small cubes of unit volume.

Volume of big cube $=n \times$ volume of unit volume cube

$ 125=n \times 1 \Rightarrow n=125 $

Now,

Number of cubes whose 3 faces are painted $=8$

Number of cubes whose 2 faces are painted = 36

Number of cubes whose 1 face is painted = 54

Number of cubes whose 0 face is painted

$ =125-(8+36+54)=27 $

∵ Required probability

$=\frac{3 \text { faces painted cubes }}{\text { Cubes whose atleast one face is painted }}$

$ =\frac{8}{8+36+54}=\frac{8}{98}=\frac{4}{49} $

We know that, Prime numbers $=2,3,5$

Composite numbers $=4,6$

Pairs of prime numbers on both the die in first throw

$ \begin{array}{r} =\{(2,2),(2,3),(2,5),(3,2),(3,3),(3,5) \\ (5,2),(5,3),(5,5)\} \end{array} $

⇒ Total pairs $=9$

Pairs of composite numbers on both the die in second throw

$ =\{(4,6),(4,4),(6,4),(6,6)\} $

∴ Required probability $=\frac{9}{36} \times \frac{4}{36}=\frac{1}{36}$

We have, total balls $=4+5+6=15$

Number of ways of selecting 3 balls $={ }^{15} C_3$

Number of ways of selecting 3 different coloured balls

$ ={ }^4 C_1 \times{ }^5 C_1 \times{ }^6 C_1 $

∴ Required probability

$ =\frac{{ }^4 C_1 \times{ }^5 C_1 \times{ }^6 C_1}{{ }^{15} C_3}=\frac{24}{91} $