Question

Two coins were tossed simultaneously 80 times and the outcomes were recorded as under:

Outcome	Two heads	One head	No head
Frequency	24	36	20

If two coins are tossed at random, find the probability of getting
(i) two heads
(ii) one head
(iii) no head

Answer 1

Total number of trials = 80
Number of times 2 heads appear = 24
Number of times 1 head appears = 36
Number of times 0 head appears = 20

In a random toss of two coins, let E₁, E₂, E₃ be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

(i) P(getting 2 heads) = P(E₁) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}2\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{24}{80}=\frac{3}{10}=0.3$

(ii) P(getting 1 head) = P(E₂) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{36}{80}=\frac{9}{20}=0.45$

(iii) P(getting 0 head) = P(E₃) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}0\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{20}{80}=\frac{1}{4}=0.25$

Clearly, when two coins are tossed, the only possible outcomes are E₁, E₂ and E₃ and P(E₁) + P(E₂) + P(E₃) = (0.3 + 0.45 + 0.25) = 1