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.
$$\begin{aligned} &\text { Bag I }=\{3 B, 2 W\}, \text { Bag } I I=\{2 B, 4 W\}\\ &\text { Let } \quad E_1=\text { Event that bag } I \text { is selected }\\ &E_2=\text { Event that bag II is selected }\\ &\text { and }\\ &E=\text { Event that a black ball is selected } \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad P\left(E_1\right)=1 / 2, P\left(E_2\right) & =\frac{1}{2}, P\left(E / E_1\right)=\frac{3}{5}, P\left(E / E_2\right)=\frac{2}{6}=\frac{1}{3} \\ \therefore \quad P(E) & =P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) \\ & =\frac{1}{2} \cdot \frac{3}{5}+\frac{1}{2} \cdot \frac{2}{6}=\frac{3}{10}+\frac{2}{12} \\ & =\frac{18+10}{60}=\frac{28}{60}=\frac{7}{15} \end{aligned}$$
A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then, another ball is drawn at random. What is the probability of second ball being blue?
$A$ box $=\{5$ blue, 4 red $\}$
Let $E_1$ is the event that first ball drawn is blue, $E_2$ is the event that first ball drawn is red and $E$ is the event that second ball drawn is blue.
$$\begin{aligned} \therefore \quad P(E) & =P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right) \\ & =\frac{5}{9} \cdot \frac{4}{8}+\frac{4}{9} \cdot \frac{5}{8}=\frac{20}{72}+\frac{20}{72}=\frac{40}{72}=\frac{5}{9} \end{aligned}$$
Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are king?
Let $E_1, E_2, E_3$ and $E_4$ are the events that the first, second, third and fourth card is king, respectively.
$$\begin{aligned} \therefore \quad P\left(E_1 \cap E_2 \cap E_3 \cap E_4\right) & =P\left(E_1\right) \cdot P\left(E_2 / E_1\right) \cdot P\left(E_3 / E_1 \cap E_2\right) \cdot P\left[E_4 /\left(E_1 \cap E_2 \cap E_3 \cap E_4\right)\right] \\ & =\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49}=\frac{24}{52 \cdot 51 \cdot 50 \cdot 49} \\ & =\frac{1}{13 \cdot 17 \cdot 25 \cdot 49}=\frac{1}{270725} \end{aligned}$$