Probability Practice Questions | Binomial Poisson Normal

Module IV: Statistical Techniques – II
Practice Questions with Solutions

Question 1 – Addition & Multiplication Laws

In a class of 50 students, 30 passed in Mathematics, 25 passed in Physics and 15 passed in both. A student is selected at random. Find the probability that the student
(a) passed in Mathematics or Physics
(b) passed only in Mathematics
(c) passed in neither subject.

Solution

Let $M$ = Mathematics, $P$ = Physics.

P(M)=\frac{30}{50}=\frac{3}{5},\quad P(P)=\frac{25}{50}=\frac{1}{2},\quad P(M\cap P)=\frac{15}{50}=\frac{3}{10}

(a)

P(M\cup P)=P(M)+P(P)-P(M\cap P)=\frac{3}{5}+\frac{1}{2}-\frac{3}{10}=\frac{6+5-3}{10}=\frac{8}{10}=0.8

(b) Only Mathematics:

P(M-P)=P(M)-P(M\cap P)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10}=0.3

P(M^c\cap P^c)=1-P(M\cup P)=1-0.8=0.2

Question 2 – Conditional Probability & Bayes’ Theorem

A factory has two machines A and B. Machine A produces 60% of the items and Machine B produces 40%. The defective rates are 2% for A and 5% for B. An item is drawn at random and found to be defective. Find the probability that it was produced by Machine B.

Solution

Let $A$ = produced by A, $B$ = produced by B, $D$ = defective.

P(A)=0.6,\quad P(B)=0.4

P(D|A)=0.02,\quad P(D|B)=0.05

By Bayes’ theorem:

P(B|D)=\frac{P(D|B)P(B)}{P(D|A)P(A)+P(D|B)P(B)} =\frac{0.05\times0.4}{0.02\times0.6+0.05\times0.4} =\frac{0.02}{0.012+0.02}=\frac{0.02}{0.032}=0.625

Question 3 – Random Variable (Discrete)

A discrete random variable $X$ has the following probability distribution:

$x$	0	1	2	3	4
$P(X=x)$	0.1	0.2	0.3	0.25	0.15

Find $E(X)$ , $E(X^2)$ and $\operatorname{Var}(X)$ .

Solution

E(X)=\sum x\,P(X=x)=0(0.1)+1(0.2)+2(0.3)+3(0.25)+4(0.15)=2.15

E(X^2)=0+1(0.2)+4(0.3)+9(0.25)+16(0.15)=0.2+1.2+2.25+2.4=6.05

\operatorname{Var}(X)=E(X^2)-[E(X)]^2=6.05-(2.15)^2=6.05-4.6225=1.4275

Question 4 – Binomial Distribution

The probability that a student passes an examination is $\frac{2}{3}$ . Five students appear. Find the probability that
(a) exactly 3 pass
(b) at least 3 pass
(c) at most 2 pass.

Solution

$X\sim B(n=5,p=\frac{2}{3})$ ,

P(X=x)=\binom{5}{x}\left(\frac{2}{3}\right)^x\left(\frac{1}{3}\right)^{5-x}

(a)

P(X=3)=\binom{5}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=10\cdot\frac{8}{27}\cdot\frac{1}{9}=\frac{80}{243}

(b)

P(X\ge3)=P(3)+P(4)+P(5)=\frac{80}{243}+\binom{5}{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)+\binom{5}{5}\left(\frac{2}{3}\right)^5 =\frac{80}{243}+\frac{80}{243}+\frac{32}{243}=\frac{192}{243}=\frac{64}{81}

(c)

P(X\le2)=1-P(X\ge3)=1-\frac{64}{81}=\frac{17}{81}

Question 5 – Poisson Distribution

If the mean of a Poisson distribution is 2, find
(a) $P(X=0)$
(b) $P(X\ge1)$
(c) $P(1\le X\le3)$

Solution

$X\sim P(\lambda=2)$

P(X=x)=e^{-2}\frac{2^x}{x!}

(a)

P(X=0)=e^{-2}\approx0.1353

(b)

P(X\ge1)=1-P(X=0)=1-e^{-2}\approx0.8647

(c)

\begin{align*} P(X=1)&=e^{-2}\cdot2\approx0.2707\\ P(X=2)&=e^{-2}\cdot\frac{4}{2}=2e^{-2}\approx0.2707\\ P(X=3)&=e^{-2}\cdot\frac{8}{6}=\frac{4}{3}e^{-2}\approx0.1804 \end{align*}

P(1\le X\le3)\approx0.2707+0.2707+0.1804=0.7218

Question 6 – Normal Distribution

The heights of students are normally distributed with mean 160 cm and standard deviation 5 cm. Find the probability that a randomly selected student has height
(a) less than 155 cm
(b) between 155 cm and 165 cm
(c) more than 170 cm.

Solution

$X\sim N(\mu=160,\sigma=5)$

Standardize: $Z=\dfrac{X-160}{5}$

(a)

P(X<155)=P\left(Z<\frac{155-160}{5}\right)=P(Z<-1)=\Phi(-1)=1-\Phi(1)=1-0.8413=0.1587

(b)

P(155<X<165)=P(-1<Z<1)=2\Phi(1)-1=2(0.8413)-1=0.6826

(c)

P(X>170)=P\left(Z>\frac{170-160}{5}\right)=P(Z>2)=1-\Phi(2)=1-0.9772=0.0228

Question 7 – Expectation & Variance (Continuous)

A continuous random variable $X$ has pdf

f(x)=\begin{cases} kx(1-x), & 0\le x\le1\\ 0, & \text{otherwise} \end{cases}

Find $k$ , $E(X)$ and $\operatorname{Var}(X)$ .

Solution

\int_0^1 kx(1-x)\,dx=1\implies k\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=1\implies k\left(\frac12-\frac13\right)=1\implies\frac{k}{6}=1\implies k=6

E(X)=\int_0^1 x\cdot6x(1-x)\,dx=6\int_0^1(x^2-x^3)\,dx=6\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1=6\left(\frac13-\frac14\right)=\frac12

E(X^2)=6\int_0^1 x^2(x-x^2)\,dx=6\int_0^1(x^3-x^4)\,dx=6\left[\frac{x^4}{4}-\frac{x^5}{5}\right]_0^1=6\left(\frac14-\frac15\right)=\frac{3}{10}

\operatorname{Var}(X)=\frac{3}{10}-\left(\frac12\right)^2=\frac{3}{10}-\frac{1}{4}=\frac{1}{20}=0.05

Question 8 – Mixed (Binomial to Poisson)

The probability of a defective item produced by a machine is 0.01. Find the probability that in a sample of 100 items there are exactly 2 defectives, using
(a) Binomial distribution
(b) Poisson approximation.

Solution

(a) Binomial: $n=100$ , $p=0.01$

P(X=2)=\binom{100}{2}(0.01)^2(0.99)^{98}\approx0.1849

(b) Poisson approximation: $\lambda=np=1$

P(X=2)=e^{-1}\frac{1^2}{2!}=\frac{1}{2e}\approx0.1839

Practice Tips

Always verify that probabilities sum to 1 for a discrete distribution.
For Bayes’ theorem, clearly identify the events and write the formula first.
In Normal distribution problems, draw a rough sketch and standardize carefully.
Use Poisson approximation when $n$ is large and $p$ is small ( $np$ moderate).