**Example1:** A fair coin and a fair die are thrown. Find the probabilities of

**1)** Head on the coin and the number 6 on the die.

**2)** Head on the coin and even number on the die.

**Solution :**

Let $A$ : The coin shows head.

$B$ : The die shows the number 6.

$C $: The die shows even number.

$P(A)$ = $\frac{1}{2}$, $P(B)$ = $\frac{1}{6}$, $P(C)$ = $\frac{3}{6}$ = $\frac{1}{2}$

**1)** Events A and B are independent.

Therefore, $P$ [Head on the coin and 6 on the die] = $P(A \cap B)$

= $P(A)$ . $P(B)$

= $\frac{1}{2}$ . $\frac{1}{6}$

= $\frac{1}{12}$

= 0.0833

**2)** Events $A$ and $C$ are independent.

Therefore, $P$ [Head on the coin and even number on the die] = $P(A \cap C)$

= $P(A)$ . $P(C)$

= $\frac{1}{2}$ . $\frac{1}{2}$

= $\frac{1}{4}$

= 0.25

**Example 2:** A bag contains 25 chocolates, among them 17 are Dairymilk silk chocolates and 8 of the chocolates are Bournville.

**1)** What is the probability of getting a Bournville?

**2)** What is the probability of getting a Dairymilk silk?

**3)** What is the probability of getting Bournville and then Dairymilk silk?

**Solution : **

__Given :__

A bag contains 25 chocolates, 17 are Dairymilk silk chocolates and 8 of them are Bournville.

**1)** Probability of getting a Bournville

As there are 8 Bournville chocolates among 25,

Probability of getting a bournville will be $\frac{8}{25}$

**2)** Probability of getting a Dairymilk silk

As there are 17 Dairymilk silk chocolates among 25,

Probability of getting a Dairymilk silk will be $\frac{17}{25}$.

**3)** Probability of getting Bournville and then Dairymilk silk

= $P(B)$ $\times$ $P(D)$

= $\frac{8}{25}$ $\times$ $\frac{17}{25}$

= $\frac{136}{625}$

= 0.2176

**Example 3:** A box has 5 white and 2 red balls. Another box has 3 white and 4 red balls. One ball is randomly selected from the first box and it is transferred to the second box. After that, one ball is randomly drawn from the second box. Find the probability that it is red.

**Solution:**

Let $A$ : The transferred ball is white.

$B$ : The transferred ball is red.

$C$ : The ball drawn from the second box is red.

$P(A)$ = $\frac{5}{7}$ and $P(B)$ = $\frac{2}{7}$

$P$ $(\frac{C}{A})$ = $\frac{4}{8}$ and $P$ $(\frac{C}{B})$ = $\frac{5}{8}$

Event $C $can occur along with $A$ or along with $B$. Therefore,

$P$ [Ball from second box is red] = $P (A \cap C) \cup P (B \cap C)$

= $P(A \cap C)$ + $P(B \cap C)$

= $P(A)$ . $P$ $(\frac{C}{A})$ + $P(B)$ . $P$ $(\frac{C}{B})$

= $\frac{5}{7}$ $\times$ $\frac{4}{8}$ $+$ $\frac{2}{7}$ $\times$ $\frac{5}{8}$

= $\frac{15}{28}$

= 0.5357