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?
Solution:
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 $)=P\left(E_{1}\right)=\frac{\text { Number of times } 3 \text { heads appear }}{\text { Total number of trials }}=\frac{39}{200}=0.195$
(ii) $P($ getting 1 head $)=P\left(E_{2}\right)=\frac{\text { Number of times } 1 \text { head appears }}{\text { Total number of trials }}=\frac{67}{200}=0.335$
(iii) $P$ (getting 0 head) $=P\left(E_{3}\right)=\frac{\text { Number of times } 0 \text { head appears }}{\text { Total number of trials }}=\frac{36}{200}=0.18$
(iv) $P$ (getting 2 heads) $=P\left(E_{4}\right)=\frac{\text { Number of times } 2 \text { heads appear }}{\text { Total number of trials }}=\frac{58}{200}=0.29$
Remark: Clearly, when three coins are tossed, the only possible outcomes are E1, E2, E3 and E4 and P(E1) + P(E2) + P(E3) + P(E4) = (0.195 + 0.335 + 0.18 + 0.29) = 1
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