By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is 0.99 . The probability of an healthy person diagnosed to have TB is 0.001 . In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
$$\begin{aligned} & \text { Let } E_1=\text { Event that person has TB } \\ & E_2=\text { Event that person does not have TB } \\ & E=\text { Event that the person is diagnosed to have TB } \\ & \therefore \quad P\left(E_1\right)=\frac{1}{1000}=0.001, P\left(E_2\right)=\frac{999}{1000}=0.999 \\ & \text { and } \quad P\left(E / E_1\right)=0.99 \text { and } P\left(E / E_2\right)=0.001 \\ & \therefore \quad P\left(E_1 / E\right)=\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)} \\ & =\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.001} \\ & =\frac{0.000990}{0.000990+0.000999} \\ & =\frac{990}{1989}=\frac{110}{221} \end{aligned}$$
An item is manufactured by three machines $A, B$ and $C$. Out of the total number of items manufactured during a specified period, $50 \%$ are manufactured on $A, 30 \%$ on $B$ and $20 \%$ on $C$. $2 \%$ of the items produced on $A$ and $2 \%$ of items produced on $B$ are defective and $3 \%$ of these produced on $C$ are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine $A$ ?
Let $E_1=$ Event that item is manufactured on $A$, $E_2=$ Event that an item is manufactured on $B$,
$E_3=$ Event that an item is manufactured on $C$,
Let $E$ be the event that an item is defective.
$$\therefore \quad P\left(E_1\right)=\frac{50}{100}=\frac{1}{2}, P\left(E_2\right)=\frac{30}{100}=\frac{3}{10} \text { and } P\left(E_3\right)=\frac{20}{100}=\frac{1}{5}$$
$P\left(\frac{E}{E_1}\right)=\frac{2}{100}=\frac{1}{50}, P\left(\frac{E}{E_2}\right)=\frac{2}{100}=\frac{1}{50}$ and $P\left(\frac{E}{E_3}\right)=\frac{3}{100}$
$$\begin{aligned} \therefore \quad P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{50}}{\frac{1}{2} \cdot \frac{1}{50}+\frac{3}{10} \cdot \frac{1}{50}+\frac{1}{5} \cdot \frac{3}{100}} \\ & =\frac{\frac{1}{100}}{\frac{1}{100}+\frac{3}{500}+\frac{3}{500}}=\frac{\frac{1}{100}}{\frac{5+3+3}{500}}=\frac{5}{11} \end{aligned}$$
Let $X$ be a discrete random variable whose probability distribution is defined as follows.
$$P(X=x)= \begin{cases}k(x+1), & \text { for } x=1,2,3,4 \\ 2 k x, & \text { for } x=5,6,7 \\ 0, & \text { otherwise }\end{cases}$$
where, $k$ is a constant. Calculate
(i) the value of $k$.
(ii) $E(X)$.
(iii) standard deviation of $X$.
$$\begin{aligned} &P(X=x)= \begin{cases}k(x+1), & \text { for } x=1,2,3,4 \\ 2 k x, & \text { for } x=5,6,7 \\ 0, & \text { otherwise }\end{cases}\\ &\text { Thus, we have following table } \end{aligned}$$
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Otherwise |
|---|---|---|---|---|---|---|---|---|
| $P(X)$ | $2k$ | $3k$ | $4k$ | $5k$ | $10k$ | $12k$ | $14k$ | 0 |
| $XP(X)$ | $2 k$ | $6 k$ | $12 k$ | $20 k$ | $50 k$ | $72 k$ | $98k$ | 0 |
| $X^2P(X)$ | $2 k$ | $12 k$ | $36 k$ | $80k$ | $250 k$ | $432 k$ | $686k$ | 0 |
(i) Since, $\Sigma P_j=1$
$$\Rightarrow \quad k(2+3+4+5+10+12+14)=1 \Rightarrow k=\frac{1}{50}$$
$$\begin{aligned} \text{(ii)}\quad \because \quad E(X) & =\Sigma X P(X) \\ \therefore \quad E(X) & =2 k+6 k+12 k+20 k+50 k+72 k+98 k+0=260 k \\ & =260 \times \frac{1}{50}=\frac{26}{5}=5.2\quad \left[\because k=\frac{1}{50}\right] \quad\text{.... (i)} \end{aligned}$$
(iii) We know that,
$$\begin{array}{rlr} \operatorname{Var}(X) & =\left[E\left(X^2\right)\right]-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma\{X P(X)\}]^2 & \\ & =[2 k+12 k+36 k+80 k+250 k+432 k+686 k+0]-[5.2]^2 & \text { [using Eq. (i) }] \\ & =[1498 k]-27.04=\left[1498 \times \frac{1}{50}\right]-27.04 & {\left[\because k=\frac{1}{50}\right]} \\ & =29.96-27.04=2.92 & \end{array}$$
We know that, standard deviation of $X=\sqrt{\operatorname{Var}(X)}=\sqrt{2.92}=1.7088=1.7$ (approx)
The probability distribution of a discrete random variable $X$ is given as under
| $X$ | 1 | 2 | 4 | 2A | 3A | 5A |
|---|---|---|---|---|---|---|
| $P(X)$ | $\frac{1}{2}$ | $\frac{1}{5}$ | $\frac{3}{25}$ | $\frac{1}{10}$ | $\frac{1}{25}$ | $\frac{1}{25}$ |
Calculate
(i) the value of $A$, if $E(X)=2.94$.
(ii) variance of $X$.
(i) We have, $\begin{aligned} \Sigma X P(X) & =\frac{1}{2}+\frac{2}{5}+\frac{12}{25}+\frac{2 A}{10}+\frac{3 A}{25}+\frac{5 A}{25} \\ & =\frac{25+20+24+10 A+6 A+10 A}{50}=\frac{69+26 A}{50}\end{aligned}$
$$\begin{array}{ll} \text { Since, } & E(X)=\sum X P(X) \\ \Rightarrow & 2.94=\frac{69+26 A}{50} \end{array}$$
$$\begin{array}{ll} \Rightarrow & 26 A=50 \times 2.94-69 \\ \Rightarrow & A=\frac{147-69}{26}=\frac{78}{26}=3 \end{array}$$
$$\begin{aligned} &\text { (ii) We know that, }\\ &\begin{aligned} \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2 \\ & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{4 A^2}{10}+\frac{9 A^2}{25}+\frac{25 A^2}{25}-[E(X)]^2 \\ & =\frac{25+40+96+20 A^2+18 A^2+50 A^2}{50}-[E(X)]^2 \\ & =\frac{161+88 A^2}{50}-[E(X)]^2=\frac{161+88 \times(3)^2}{50}-[E(X)]^2 \quad[\because A=3] \\ & =\frac{953}{50}-[2.94]^2 \quad[\because E(X)=2.94] \\ & =19.0600-8.6436=10.4164 \end{aligned} \end{aligned}$$
The probability distribution of a random variable $x$ is given as under
$$P(X=x)=\left\{\begin{array}{l} k x^2, x=1,2,3 \\ 2 k x, x=4,5,6 \\ 0, \text { otherwise } \end{array}\right.$$
where, $k$ is a constant. Calculate
(i) $E(X)$
(ii) $E\left(3 X^2\right)$
(iii) $P(X \geq 4)$
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 | Otherwise |
|---|---|---|---|---|---|---|---|
| $P(X)$ | k | 4k | 9k | 8k | 10k | 12k | 0 |
$$\begin{aligned} &\text { We know that, } \Sigma P_i=1\\ &\Rightarrow \quad 44 k=1 \Rightarrow k=\frac{1}{44} \end{aligned}$$
$$\begin{aligned} \therefore \quad \Sigma X P(X) & =k+8 k+27 k+32 k+50 k+72 k+0 \\ & =190 k=190 \times \frac{1}{44}=\frac{95}{22} \end{aligned}$$
(i) So, $E(X)=\Sigma X P(X)=\frac{95}{22}=4.32$
(ii) Also, $\quad E\left(X^2\right)=\Sigma X^2 P(X)=k+16 k+81 k+128 k+250 k+432 k$
$$\begin{array}{ll} =908 k=908 \times \frac{1}{44} & {\left[\because k=\frac{1}{44}\right]} \\ =20.636=20.64 \text { (approx) } \end{array}$$
$$ \begin{aligned} &\therefore \quad E\left(3 X^2\right)=3 E\left(X^2\right)=3 \times 20.64=61.92=61.9\\ &\begin{aligned} \text { (iii) }\quad P(X \geq 4) & =P(X=4)+P(X=5)+P(X=6) \\ & =8 k+10 k+12 k=30 k=30 \cdot \frac{1}{44}=\frac{15}{22} \end{aligned} \end{aligned}$$