Probability

$n(s)=36$

$ \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

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

We have,

$\begin{aligned} & P(\bar{A} \cup \bar{B})=P(\bar{A} \cap \bar{B}) \\ & \quad=1-P(A \cap B)=1-P(A) P\left(\frac{B}{A}\right) \end{aligned}$

Thus, $P(\bar{A} \cup \bar{B}) \neq 1-P(A \cup B)$

and $P(\bar{A} \cup \bar{B}) \neq P(A \cup B) \neq P(\bar{A})+P(\bar{B})$

A number divisible by 4 formed by the digits $1,2,3,4$ and 5 should have the last two digits 12 or 24 or 32 or 52.

In each cases, the five digit number can be formed using the remaining 3 digits in $3!=6$ ways A number divisible by 4 can be formed in $6 \times 4=24$ ways Total number of 5 -digit numbers that can be formed using the digits $1,2,3,4$ and 5 without repetition $5!=120$

Thus, required probability $=\frac{24}{120}=\frac{1}{5}$

The bigger part could be any number from 10000 to 20000.

Now, if the bigger part is to be at least twice as much as the smaller part, we have

$\begin{aligned} X & \geq 2 y \text { or } X \ge 2(20000-X)\\ \text { or } \quad X & \geq \frac{40000}{3}\end{aligned}$

We know that $X$ lies between 10000 and 20000, the probability that $X$ lies between $\frac{40000}{3}$ and 20000 is $\frac{20000-\frac{40000}{3}}{20000-10000}=\frac{2}{3}$

"The probability of two outcomes can change from one trial to the next" which is not a property of Binomial Distribution.

Given the following information for a Binomial distribution $ B(n, p) $:

• The mean is 15.


• The variance is 10.

We need to find the value of the parameter $ n $.

From the mean formula of a Binomial distribution, we have:

$ n \cdot p = 15 $

From the variance formula of a Binomial distribution, we have:

$ n \cdot p \cdot (1 - p) = 10 $

Since $ n \cdot p = 15 $, we substitute $ 15 $ for $ n \cdot p $ in the variance formula:

$ 15 \cdot (1 - p) = 10 $

Solving for $ p $:

$ 1 - p = \frac{10}{15} = \frac{2}{3} $

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

Now, substituting $ p $ back into the mean formula:

$ n \cdot \left( \frac{1}{3} \right) = 15 $

$ n = 15 \times 3 = 45 $

Therefore, the value of $ n $ is 45.

Total number of balls $=100$

odd numbered balls $\rightarrow 50$

Even numbered balls $\rightarrow 50$

Sum of three number will be odd when

(i) two numbers are even and one odd.

(ii) all three numbers are odd.

$\begin{aligned} & \text { Required probability }=\frac{{ }^{50} C_2 \cdot{ }^{50} C_1+{ }^{50} C_3}{{ }^{100} C_3} \\ &=\frac{\frac{50 \cdot 49}{2 \cdot 1} \cdot 50+\frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1}}{\frac{100 \cdot 99 \cdot 98}{3 \cdot 2 \cdot 1}} \\ &=\frac{50 \cdot 49 \cdot 50 \cdot 3+50 \cdot 49 \cdot 48}{100 \cdot 99 \cdot 98} \\ &=\frac{50 \cdot 49[50 \times 3+48]}{100 \cdot 99 \cdot 98} \\ &=\frac{6[25+8]}{2 \cdot 99 \cdot 2}=\frac{6 \cdot 33}{2 \cdot 99 \cdot 2}=\frac{1}{2} \end{aligned}$

Total number of tickets $=35$

Number of tickets bearing a prize $=10$

$\begin{gathered} P \text { (not getting a prize })=1-P \text { (getting a prize) } \\ =1-\frac{10}{35}=\frac{25}{35}=\frac{5}{7} \end{gathered}$

Number of green balls $=7$

Number of black balls $=5$

Total number of balls $=7+5=12$

While drawing, ball is not replaced.

Required probability

$\begin{aligned} & =\frac{{ }^7 C_1 \cdot{ }^6 C_1 \cdot{ }^5 C_1}{{ }^{12} C_1{ }^{11} C_1 \cdot{ }^{10} C_1} \\ & =\frac{7 \cdot 6 \cdot 5}{12 \cdot 11 \cdot 10}=\frac{7}{44} \end{aligned}$

There are 12 possible values of $x y$.

$x \in\{1,2,3,4\}, y \in\{5,6,7\}$

Values of $x y$ are :

$\begin{aligned} & 1 \times 5=5 \\ & 1 \times 6=6 \end{aligned}$

$\begin{aligned} & 1 \times 7=7 \\ & 2 \times 5=10 \\ & 2 \times 6=12 \\ & 2 \times 7=14 \\ & 3 \times 5=15 \\ & 3 \times 6=18 \\ & 3 \times 7=21 \\ & 4 \times 5=20 \\ & 4 \times 6=24 \\ & 4 \times 7=28 \end{aligned}$

$P(x y \text { is even })=\frac{8}{12}=\frac{2}{3}$

Comparing with the standard binomial $(n, p)$

distribution, variance $=n p(1-p)$

$\begin{aligned} & =16 \times 0.25(1-0.29) \\ & =4 \times 0.75=3 \end{aligned}$

Comparing with the standard poisson ($\lambda$) distribution, variance $=\lambda=2$

$\therefore$ Sum of the variance of $X$ and $Y$

$=\operatorname{Var}(X)+\operatorname{Var}(Y)=3+2=5$

Probability distribution: $P(X=x)=\frac{e^{-\lambda} x^x}{x!}$

$\lambda=$ mean = 6

$\begin{aligned} & p(X \geq 3) =1-[P(X=0)+p(X=1)+p(X=2)] \\ & =1-\left[\frac{e^{-6} \cdot 6^0}{0!}+\frac{e^{-6} \cdot 6^1}{1!}+\frac{e^{-6} \cdot 6^2}{2!}\right] \\ & =1-\frac{1}{e^6}[1+6+, 18] \\ & =1-\frac{25}{e^6} \end{aligned}$

Let P be the probability of getting an even number. Then by hypothesis the probability of getting an odd number is 3P. Since the events of getting an even number and an odd number are mutually exclusive and exhaustive.

$\therefore$ P + 3P = 1 $\Rightarrow$ P = 1/4

Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$. If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd.

$\therefore$ Required probability = $\frac{3}{4}$ $\times$ $\frac{3}{4}$ + $\frac{1}{4}$ $\times$ $\frac{1}{4}$ = $\frac{10}{16}$ = $\frac{5}{8}$

Let x be the number of heads obtained by A, and y be the number of heads obtained by B.

Note that x and y are binomial variable with parameters n = 3 and p = ${1 \over 2}$

$\therefore$ Probability that both A and B obtained the same number of heads is

$ = P(x = 0)\,.\,P(y = 0) + P(x = 1)\,.\,P(y = 1) + P(x = 2)\,.\,P(y = 2) + P(x = 3)\,.\,P(y = 3)$

$ = {\left[ {{3_{{C_0}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_1}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_2}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_3}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2}$

$ = {\left( {{1 \over 2}} \right)^6}\left[ {1 + 9 + 9 + 1} \right]$

$ = {{20} \over {64}}$

$ = {5 \over {16}}$