Suppose that $6 \%$ of the people with blood group 0 are left handed and $10 \%$ of those with other blood groups are left handed, $30 \%$ of the people have blood group 0 . If a left handed person is selected at random, what is the probability that he/she will have blood group 0 ?
| Blood group 'O' | Other than blood group 'O' | |
|---|---|---|
| I. Number of people | 30% | 70% |
| II. Percentage of left handed people | 6% | 10% |
$$\begin{aligned} & E_1=\text { Event that the person selected is of blood group O } \\ & E_2=\text { Event that the person selected is of other than blood group O } \\ & \left(E_3\right)=\text { Event that selected person is left handed } \\ & \therefore \quad \begin{aligned} P\left(E_1\right)=0.30, P\left(E_2\right)=0.70 \\ P\left(E_3 / E_1\right)=0.06 \text { and } P\left(E_3 / E_2\right)=0.10 \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { By using Baye's theorem, }\\ &\begin{aligned} P\left(E_1 / E_3\right) & =\frac{P\left(E_1\right) \cdot P\left(E_3 / E_1\right)}{P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right)} \\ & =\frac{0.30 \times 0.06}{0.30 \cdot 0.06+0.70 \cdot 0.10} \\ & =\frac{0.0180}{0.0180+0.0700} \\ & =\frac{0.0180}{0.0880}=\frac{180}{880}=\frac{9}{44} \end{aligned} \end{aligned}$$
If two natural numbers $r$ and $s$ are drawn one at a time, without replacement from the set $S=\{1,2,3, \ldots n\}$, then find $P(r \leq p / s \leq p)$, where $p \in S$.
$$\begin{aligned} & \because \text { Set } S=\{1,2,3, \ldots, n\} \\ & \therefore \quad P(r \leq p / S \leq p)=\frac{P(p \cap S)}{P(S)} \\ & \\ \end{aligned}$$
$=\frac{p-1}{n} \times \frac{n}{n-1}=\frac{p-1}{n-1}$
Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.
Let $X$ is the random variable score obtained when a die is thrown twice.
$$\begin{array}{lrl} \therefore & X & =1,2,3,4,5,6 \\ \text { Here, } & S & =\{(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3), \ldots,(6,6)\} \\ \therefore & P(X & =1)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36} \\ & & P(X=2)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{3}{36} \\ & P(X & =3)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{5}{36} \end{array}$$
$$\begin{aligned} \text{Similarly,}\quad & P(X=4)=\frac{7}{36} \\ & P(X=5)=\frac{9}{36} \\ & P(X=6)=\frac{11}{36} \end{aligned}$$
So, the required distribution is,
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $P(X)$ | 1/36 | 3/36 | 5/36 | 7/36 | 9/36 | 11/36 |
$$\begin{aligned} &\text { Also, we know that, Mean }\{E(X)\}=\Sigma X P(X)\\ &=\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36}=\frac{161}{36} \end{aligned}$$
The random variable $X$ can take only the values $0,1,2$. If
$$P(X=0)=P(X=1)=p \text { and } E\left(X^2\right)=E[X]$$
then find the value of $p$.
Since, $X=0,1,2$ and $P(X)$ at $X=0$ and 1 is $p$, let at $X=2, P(X)$ is $x$.
$$\begin{aligned} \Rightarrow & \quad p+p+x =1 \\ \Rightarrow & \quad x =1-2 p \end{aligned}$$
We get, the following distribution.
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $p$ | $p$ | $1-2p$ |
$$\begin{aligned} \therefore\quad E[X] & =\Sigma X P(X) \\ & =0 \cdot p+1 \cdot p+2(1-2 p) \\ & =p+2-4 p=2-3 p \end{aligned}$$
$$\begin{aligned} &\text { and }\\ &\begin{aligned} & E\left(X^2\right)=\Sigma X^2 P(X) \\ & =0 \cdot p+1 \cdot p+4 \cdot(1-2 p) \\ & =p+4-8 p=4-7 p \end{aligned} \end{aligned}$$
$$\begin{array}{lrl} \text { Also, } & \text { given that } E\left(X^2\right) & =E[X] \\ \Rightarrow & 4-7 p & =2-3 p \\ \Rightarrow & 4 p & =2 \Rightarrow p=\frac{1}{2} \end{array}$$
Find the variance of the following distribution.
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $P(X)$ | $\frac{1}{6}$ | $\frac{5}{18}$ | $\frac{2}{9}$ | $\frac{1}{6}$ | $\frac{1}{9}$ | $\frac{1}{18}$ |
We have,
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $P(X)$ | $\frac{1}{6}$ | $\frac{5}{18}$ | $\frac{2}{9}$ | $\frac{1}{6}$ | $\frac{1}{9}$ | $\frac{1}{18}$ |
| $XP(X)$ | 0 | $\frac{5}{18}$ | $\frac{4}{9}$ | $\frac{1}{2}$ | $\frac{4}{9}$ | $\frac{5}{18}$ |
| $X^2P(X)$ | 0 | $\frac{5}{18}$ | $\frac{8}{9}$ | $\frac{3}{2}$ | $\frac{16}{9}$ | $\frac{25}{18}$ |
$$\begin{aligned} \therefore \quad \text { Variance } & =E\left(X^2\right)-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\left[0+\frac{5}{18}+\frac{8}{9}+\frac{3}{2}+\frac{16}{9}+\frac{25}{18}\right]-\left[0+\frac{5}{18}+\frac{4}{9}+\frac{1}{2}+\frac{4}{9}+\frac{5}{18}\right]^2 \\ & =\left[\frac{5+16+27+32+25}{18}\right]-\left[\frac{5+8+9+8+5}{18}\right]^2 \\ & =\frac{105}{18}-\frac{35 \cdot 35}{18 \cdot 18}=\frac{18 \cdot 105-35 \cdot 35}{18 \cdot 18} \\ & =\frac{35}{18 \cdot 18}[54-35]=\frac{19 \cdot 35}{324}=\frac{665}{324} \end{aligned}$$