**Letâ€™s take a look at a couple of problems based on approaches:**

**Example 1:**

Three unbiased coins are tossed, what is the probability of obtaining:

**1)** All heads

**2)** Two heads

**3)** Atleast one head

**4)** Atleast two heads

**5)** All tails

**Solution:**

Random Experiment: Three unbiased coins are tossed.

Sample space $S$ = {HHH, HHT, HTT, TTT, TTH, THH, THT, HTH} So $n$ = 8.

Probability = $\frac{Number\ of\ ways\ a\ certain\ outcome\ can\ occur}{Total\ Possible\ outcomes}$ = $\frac{m}{n}$

**1)** $A$ : Getting all heads

$A$ = {HHH}, i.e. $m$ = 1

$P(A)$ = $\frac{1}{8}$

**2)** $B$: Getting two heads, i.e. $m$ = 3

$B$ = {(HHT), (HTH), (THH)}

$P(B)$ = $\frac{3}{8}$

**3)** $C$: Atleast one head

$C$ = {(HHH), (HHT), (HTT),(TTH), (THH), (THT), (HTH)}, $m$ = 7

$P(C)$ = $\frac{7}{8}$

**4)** $D$: Getting all tails

$D$ = { TTT }, i.e. $m$ = 1

$P(D)$ = $\frac{1}{8}$

**5)** $E$: Getting atleast two heads

$E$ = {(HHH), (HHT), (THH), (HTH)}, i.e. $m$ = 4

$P(E)$ = $\frac{4}{8}$

**Example 2:** Two fair dice are rolled. Find the probability that

**1)** Both the dice show the number 6.

**2)**The sum of the numbers obtained is 10.

**3)** The sum is divisible by 3.

**Solution:** Random Experiment: Two fair dice are rolled.

$S$ = {(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3,1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4,3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6,1), (5, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

So, $n$ = 36

Probability = $\frac{Number\ of\ ways\ a\ certain\ outcome\ can\ occur}{Total\ Possible\ outcomes}$ = $\frac{m}{n}$**1)** $A$: Both the dice show the number 6.

$P(A)$ = $\frac{1}{36}$

**2)** $B$: The sum of the numbers obtained is 10.

$B$ = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9,1)}

$P(B)$ = $\frac{9}{36}$

**3)** $C$: The sum is divisible by 3.

$C$ = {(1, 5), (1, 2), (2, 1), (2, 4), (3, 3), (6, 3), (4, 5), (5, 4), (6, 3), (6, 6), (5, 1), (4, 2)}, $m$ = 12

$P(C)$ = $\frac{12}{36}$

**Example
3:** There are 17 balls, numbered from 1 to 17 in a bag. If a person
selects one at random, what is the probability that the number printed
on the ball will be an even number greater than 9?

**Solution:**

Random Experiment: Selecting a ball from a bag containing 17 balls at random.

One ball can be selected from 17 balls in $17C_{1}$ ways. $n$ = 17

$A$: Getting an even number greater than 9.

Selecting one of the following balls which are {10, 12, 14, 16}

$m$ = 4$C_{1}$

Therefore, $P(A)$ = $\frac{4}{17}$ = 0.23