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, $\quad E\left(X^2\right)=\Sigma X^2 P(X)=k+16 k+81 k+128 k+250 k+432 k$
$$\begin{array}{ll} =908 k=908 \times \frac{1}{44} & {\left[\because k=\frac{1}{44}\right]} \\ =20.636=20.64 \text { (approx) } \end{array}$$
$$ \begin{aligned} &\therefore \quad E\left(3 X^2\right)=3 E\left(X^2\right)=3 \times 20.64=61.92=61.9\\ &\begin{aligned} \text { (iii) }\quad P(X \geq 4) & =P(X=4)+P(X=5)+P(X=6) \\ & =8 k+10 k+12 k=30 k=30 \cdot \frac{1}{44}=\frac{15}{22} \end{aligned} \end{aligned}$$
A bag contains $(2 n+1)$ coins. It is known that $n$ of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\frac{31}{42}$, then determine the value of $n$.
Given, $n$ coins have head on both sides and $(n+1)$ coins are fair coins.
Let $E_1=$ Event that an unfair coin is selected
$E_2=$ Event that a fair coin is selected
$E=$ Event that the toss results in a head
$$\begin{array}{ll} \therefore & P\left(E_1\right)=\frac{n}{2 n+1} \text { and } P\left(E_2\right)=\frac{n+1}{2 n+1} \\ \text { Also, } & P\left(\frac{E}{E_1}\right)=1 \text { and } P\left(\frac{E}{E_2}\right)=\frac{1}{2} \end{array}$$
$$\begin{array}{ll} \therefore & P(E)=P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)=\frac{n}{2 n+1} \cdot 1+\frac{n+1}{2 n+1} \cdot \frac{1}{2} \\ \Rightarrow & \frac{31}{42}=\frac{2 n+n+1}{2(2 n+1)} \Rightarrow \frac{31}{42}=\frac{3 n+1}{4 n+2} \\ \Rightarrow & 124 n+62=126 n+42 \\ \Rightarrow & 2 n=20 \Rightarrow n=10 \end{array}$$