Two coins are tossed once. Find the probability of

(i) Getting 2 heads

(ii) Getting at least 1 head

(iii) Getting no head

(iv) Getting 1 head and 1 tail

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Solution

When two coins are tossed once, the sample space is given by
$S = {H H, H T, T H, T T}$ and, therefore, $n (S) = 4$

(i) Let $E_{1} =$ event of getting 2 heads. Then,

$E_{1} = {H H}$ and, therefore, $n (E_{1}) = 1$

$∴$ P(getting 2 heads) $= P (E_{1}) = \frac{n (E_{1})}{n (S)} = \frac{1}{4}$

(ii) Let $E_{2} =$ event of getting at least 1 head. Then,

$E_{2} = {H T, T H, H H}$ and, therefore, $n (E_{2}) = 3$

$∴$ P(getting at least 1 head) $= P (E_{2}) = \frac{n (E_{2})}{n (S)} = \frac{3}{4}$

(iii) Let $E_{3} =$ event of getting no head. Then,

$E_{3} = {T T}$ and, therefore, $n (E_{3}) = 1$

$∴$ P(getting no head) $= P (E_{3}) = \frac{n (E_{3})}{n (S)} = \frac{1}{4}$

(iv) Let $E_{4} =$ event of getting 1 head and 1 tail. Then,

$E_{4} = {H T, T H}$ and, therefore, $n (E_{4}) = 2$

$∴$ P(getting 1 head and 1 tail) $= P (E_{4}) = \frac{n (E_{4})}{n (S)} = \frac{2}{4} = \frac{1}{2}$

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