Probability

$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$

$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$

$P({E_1} \cap {E_2}) = {1 \over 8}$

$P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}$

(A) $P({E_1} \cap {E_2}) = {1 \over 8}$ and $P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$ \Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

(B) $P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})$

$ = 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}$

$P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}$

$ \Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})$

(C) $P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})$

$ = {1 \over 6} - {1 \over 8} = {1 \over {24}}$

$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$

$ \Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$

(D) $P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})$

$ = {1 \over 4} - {1 \over 8} = {1 \over 8}$

$P({E_1})P({E_2}) = {1 \over {24}}$

$ \Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$

$\because$ x is a random variable

$\therefore$ $k + 2k + 4k + 6k + 8k = 1$

$\therefore$ $k = {1 \over {21}}$

Now, $P(1 < x < 4\,|\,x \le 2) = {{4k} \over {7k}} = {4 \over 7}$

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} $

Given in bag containing 5 black balls and 3 white balls.

Random variable $X$ denotes the number of white balls drawn $X(0)=$ No white ball is drawn

$ \begin{aligned} & P(0)=\frac{{ }^3 C_0 \times{ }^5 C_2}{{ }^8 C_2}=\frac{10}{28} \\ & X(1)=1 \text { white ball is drawn } \\ & P(X)=\frac{{ }^5 C_1 \times{ }^3 C_1}{{ }^8 C_2}=\frac{15}{28} \\ & X(2)=2 \text { white balls are drawn } \\ & P(X)=\frac{{ }^3 C_2 \times{ }^5 C_0}{{ }^8 C_2}=\frac{3}{28} \end{aligned} $

Mean of

$ \begin{aligned} X=\Sigma X_i P\left(X_i\right) & =0 \times \frac{10}{28}+1 \times \frac{15}{28}+2 \times \frac{3}{28} \\ & =\frac{21}{28}=\frac{3}{4} \end{aligned} $

We know that in binomial distribution

Mean $=n p$

Variance $=n p q$

$ \begin{aligned} \therefore \quad n p & =4 \text { and } n p q=\frac{4}{3} \\ \Rightarrow \quad q & =\frac{1}{3}, p=\frac{2}{3} \text { and } n=6 \\ P(X & =2)={ }^6 C_2 p^2 q^4 \\ & =\frac{6 \times 5}{2} \times\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^4=\frac{20}{243} \end{aligned} $

When two dice are rolled together, then total outcomes are 36 and sample space is $[(1,1),(1,2),(1,3),(1,4),(1,5)(1,6), (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)]$

So, pairs with sum 9 are $(3,6)(4,5),(5,4),(6,3)$ i.e total 4 pairs.

Total outcomes $=36$

Favourable outcomes $=4$

Probability of getting the sum of 9

$=\frac{4}{36}=\frac{1}{9}$

To determine $P(\bar{A} \mid \bar{B})$, we need to use the definition of conditional probability and the properties of complements.

By definition, the conditional probability of $\bar{A}$ given $\bar{B}$ is:

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

Now, using De Morgan's law, we know that:

$\bar{A} \cap \bar{B} = \overline{A \cup B}$

So, we can rewrite the probability as:

$P(\bar{A} \mid \bar{B}) = \frac{P(\overline{A \cup B})}{P(\bar{B})}$

From the properties of complements, we know:

$P(\overline{A \cup B}) = 1 - P(A \cup B)$

Thus, the expression for $P(\bar{A} \mid \bar{B})$ becomes:

$P(\bar{A} \mid \bar{B}) = \frac{1 - P(A \cup B)}{P(\bar{B})}$

This matches Option C.

Therefore, the correct answer is:

Option C

$\frac{1-P(A \cup B)}{P(\bar{B})}$

To find the probability that both brothers $X$ and $Y$ pass the exam, we need to consider the probability of the intersection of events $A$ and $B$. Assuming the events are independent, the probability of both events occurring is given by the product of their individual probabilities.

Let:

$P(A) = \frac{1}{7}$

$P(B) = \frac{2}{9}$

Since events $A$ and $B$ are independent, the probability of both $A$ and $B$ happening is:

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

Substituting the given probabilities:

$P(A \cap B) = \frac{1}{7} \cdot \frac{2}{9} = \frac{1 \cdot 2}{7 \cdot 9} = \frac{2}{63}$

Therefore, the probability that both brothers $X$ and $Y$ pass the exam is $\frac{2}{63}$.

The correct answer is:

Option C

A red ball can be drawn in two mutually exclusive ways.

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it

Let $E_1, E_2$ and $A$ denote the events defined as follows

$\begin{aligned} & E_1=\text { Selecting bag I } \\ & E_2=\text { Selecting bag II } \end{aligned}$

Since, one of the two bags is selected randomly.

$\therefore P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$

Now, $P\left(\frac{A}{E_1}\right)=$ Probability of drawing a red ball when the first bag has been selected $=4 / 7$

$P\left(\frac{A}{E_2}\right)=$ Probability of drawing a red ball when the second bag has been selected $=2 / 5$

$\begin{aligned} P(\text { red ball }) & =P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right) \\ & =\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{2}{5}=\frac{2}{7}+\frac{1}{5}=\frac{17}{35} \end{aligned}$

Let $X$ be the binomial variate for which mean $=4$ and variance $=3$, then $n p=4$ and $n p q=3$

$\Rightarrow q=\frac{3}{4}$

$\therefore P=(1-q)=\left(1-\frac{3}{4}\right)=\frac{1}{4}$ and $n p=4$

$\Rightarrow n=\frac{4}{1} \times 4=16$

Thus, $n=16, p=\frac{1}{4}$ and $q=\frac{3}{4}$

Hence, the binomial distribution

$ \begin{aligned} & 2^{32} P\left(X=\frac{n}{2}\right)=2^{32} \cdot{ }^{16} C_{\frac{16}{2}} \cdot\left(\frac{1}{4}\right)^{\frac{16}{2}}\left(\frac{3}{4}\right)^{16-\frac{16}{2}} . \\ & =2^{32} \cdot{ }^{16} C_8\left(\frac{1}{4}\right)^8 \times\left(\frac{3}{4}\right)^8=2^{32} \cdot{ }^{16} C_8 \frac{(3)^8}{(4)^{16}} \\ & ={ }^{16} C_8(3)^8 \quad {\left[\because(4)^{16}=(2)^{32}]\right.} \end{aligned}$

For Poisson distribution, mean = variance

$l=m=4$

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

Total number of balls $=(8+7+6)=21$

Let $E=$ Event that the ball drawn is neither red nor green

$\begin{array}{l} =\text { Event that the ball drawn is blue } \\ \therefore n(E) =7 \\ \therefore p(E) =\frac{n(E)}{n(S)}=\frac{7}{21}=\frac{1}{3} \end{array}$