In a dice game, a player pays a stake of ₹ 1 for each throw of a die. She receives ₹ 5 , if the die shows a 3 , ₹ 2 , if the die shows a 1 or 6 and nothing otherwise, then what is the player's expected profit per throw over a long series of throws?
Let X is the random variable of profit per throw.
| X | $-1$ | 1 | 4 |
|---|---|---|---|
| P(X) | $\frac{1}{2}$ |
$\frac{1}{3}$ | $\frac{1}{6}$ |
Since, she loss ₹ 1 on getting any of 2,4 or 5.
So, at $X=-1\quad,$ $$ P(X)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$$
Similarly, at $X=1\quad$, $P(X)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\quad$ [if die shows of either 1 or 6 ]
and at $X=4, \quad P(X)=\frac{1}{6}$ [if die shows a 3]
$\therefore$ Player's expected profit $=E(X)=\Sigma X P(X)$
$=-1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}$
$=\frac{-3+2+4}{6}=\frac{3}{6}=\frac{1}{2}=$ ₹ 0.50
Three dice are thrown at the same time. Find the probability of getting three two's, if it is known that the sum of the numbers on the dice was six.
On a throw of three dice, we have sample space $[n(S)]=6^3=216$
Let $E_1$ is the event when the sum of numbers on the dice was six and $E_2$ is the event when three two's occurs.
$$\begin{array}{r} \Rightarrow \quad E_1=\{(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2),(2,3,1),(3,1,2), \\ (3,2,1),(4,1,1)\} \end{array}$$
$$\begin{array}{l} \Rightarrow & n\left(E_1\right) =10 \text { and } E_2=\{2,2,2\} \\ \Rightarrow & n\left(E_2\right)=1 \\ \text { Also, } & \left(E_1 \cap E_2\right)=1 \\ \therefore & P\left(E_2 / E_1\right)=\frac{P \cdot\left(E_1 \cap E_2\right)}{P\left(E_1\right)}=\frac{1 / 216}{10 / 216}=\frac{1}{10} \end{array}$$
Suppose 10000 tickets are sold in a lottery each for ₹ 1 . First prize is of ₹ 3000 and the second prize is of ₹ 2000 . There are three third prizes of ₹ 500 each. If you buy one ticket, then what is your expectation?
Let X is the random variable for the prize.
| X | 0 | 500 | 2000 | 3000 |
|---|---|---|---|---|
| P(X) | $\frac{9995}{10000}$ |
$\frac{3}{10000}$ | $\frac{1}{10000}$ | $\frac{1}{10000}$ |
$$\begin{aligned} \text { Since, } \quad E(X) & =\Sigma X P(X) \\ \therefore \quad E(X) & =0 \times \frac{9995}{10000}+\frac{1500}{10000}+\frac{2000}{10000}+\frac{3000}{10000} \\ & =\frac{1500+2000+3000}{10000} \end{aligned}$$
$=\frac{6500}{10000}=\frac{13}{20}=$ ₹ 0.65
A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.
Here, $W_1=\{4$ white balls $\}$ and $B_1=\{5$ black balls $\}$
and $W_2=\{9$ white balls $\}$ and $B_2=\{7$ black balls $\}$
Let $E_1$ is the event that ball transferred from the first bag is white and $E_2$ is the event that the ball transferred from the first bag is black.
Also, $E$ is the event that the ball drawn from the second bag is white.
$$\begin{array}{ll} \therefore & P\left(E / E_1\right)=\frac{10}{17}, P\left(E / E_2\right)=\frac{9}{17} \\ \text { and } & P\left(E_1\right)=\frac{4}{9} \text { and } P\left(E_2\right)=\frac{5}{9} \\ \therefore & P(E)=P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) \end{array}$$
$$\begin{aligned} & =\frac{4}{9} \cdot \frac{10}{17}+\frac{5}{9} \cdot \frac{9}{17} \\ & =\frac{40+45}{153}=\frac{85}{153}=\frac{5}{9} \end{aligned}$$
Bag I contains 3 black and 2 white balls, bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
$$\begin{aligned} &\text { Bag I }=\{3 B, 2 W\}, \text { Bag } I I=\{2 B, 4 W\}\\ &\text { Let } \quad E_1=\text { Event that bag } I \text { is selected }\\ &E_2=\text { Event that bag II is selected }\\ &\text { and }\\ &E=\text { Event that a black ball is selected } \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad P\left(E_1\right)=1 / 2, P\left(E_2\right) & =\frac{1}{2}, P\left(E / E_1\right)=\frac{3}{5}, P\left(E / E_2\right)=\frac{2}{6}=\frac{1}{3} \\ \therefore \quad P(E) & =P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) \\ & =\frac{1}{2} \cdot \frac{3}{5}+\frac{1}{2} \cdot \frac{2}{6}=\frac{3}{10}+\frac{2}{12} \\ & =\frac{18+10}{60}=\frac{28}{60}=\frac{7}{15} \end{aligned}$$