Question

Two coins are tossed simultaneously. Find the probability of getting
(i) exactly 1 head
(ii) at most 1 head
(iii) at least 1 head

Answer 1

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH and TT.
Total number of possible outcomes = 4

(i) Let E be the event of getting exactly one head.

Then, the favourable outcomes are HT and TH.
​Number of favourable outcomes = 2

∴ P(getting exactly 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{4}=\frac{1}{2}$

(ii) Let E be the event of getting at most one head.

Then, the favourable outcomes are HT, TH and TT.
​Number of favourable outcomes = 3

∴ P (getting at most 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$

(iii) Let E be the event of getting at least one head.

Then, the favourable outcomes are HT, TH and HH
​Number of favourable outcomes = 3

∴ P (getting at least 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$