Question

Three coins are tossed 200 times and we get
three heads: 39 times; two heads: 58 times;
one head: 67 times; 0 head: 36 times.
When three coins are tossed at random, what is the probability of getting
(i) 3 heads?
(ii) 1 head?
(iii) 0 head?
(iv) 2 heads?

Answer 1

Total number of tosses = 200
Number of times 3 heads appear = 39
Number of times 2 heads appear = 58
Number of times 1 head appears = 67
Number of times 0 head appears = 36

In a random toss of three coins, let E₁, E₂, E₃ and E₄ be the events of getting 3 heads, 2 heads, 1 head and 0 head, respectively. Then;

(i) P(getting 3 heads) = P(E₁) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}3\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{39}{200}=0.195$


(ii) P(getting 1 head) = P(E₂) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{67}{200}=0.335$

(iii) P(getting 0 head) = P(E₃) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}0\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{36}{200}=0.18$

(iv) P(getting 2 heads) = P(E₄) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}2\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{58}{200}=0.29$

Remark: Clearly, when three coins are tossed, the only possible outcomes are E₁, E₂, E_{3 and} E₄ and P(E₁) + P(E₂) + P(E₃) + P(E₄) = (0.195 + 0.335 + 0.18 + 0.29) = 1