Probability

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 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}}$