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

ANSWER:
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 E 1 , E 2 , E 3 and E 4 be the events of getting 3 heads, 2 heads, 1 head and 0
head, respectively. Then;
(i) P(getting 3 heads) = P(E 1 ) = Number of times 3 heads appear/Total number of trials = $\frac{39}{200}$ = 0.195
(ii) P(getting 1 head) = P(E 2 ) = Number of times 1 head appears/Total number of trials = $\frac{67}{200}$ = 0.335
(iii) P(getting 0 head) = P(E 3 ) = Number of times 0 head appears/Total number of trials = $\frac{36}{200}$ = 0.18
(iv) P(getting 2 heads) = P(E 4 ) = Number of times 2 heads appear/Total number of trials = $\frac{58}{200}$ = 0.29
Remark: Clearly, when three coins are tossed, the only possible outcomes are E 1 , E 2 , E 3 and E 4 and P(E 1 )
+ P(E 2 ) + P(E 3 ) + P(E 4 ) = (0.195 + 0.335 + 0.18 + 0.29) = 1