Question

Two coins are tossed 400 times and we get
two heads: 112 times; one head: 160 times; 0 head: 128 times.
When two coins are tossed at random, what is the probability of getting
(i) 2 heads?
(ii) 1 head?
(iii) 0 head?

Answer 1

_Total number of tosses = 400
Number of times 2 heads appear = 112
Number of times 1 head appears = 160
Number of times 0 head appears = 128

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) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}2\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{112}{400}=0.28$

(ii) P( getting 1 head) = P(E₂) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{160}{400}=0.4$

(iii) P( getting 0 head) = P(E₃) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}0\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{128}{400}=0.32$

Remark: Clearly, when two coins are tossed, the only possible outcomes are E₁, E₂ and E₃ and P(E₁) + P(E₂) + P(E₃) = (0.28 + 0.4 + 0.32) = 1