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?
Solution:
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 $)=P\left(E_{1}\right)=\frac{\text { Number of times } 2 \text { heads appear }}{\text { Total number of trials }}=\frac{112}{400}=0.28$
(ii) $P($ getting 1 head $)=P\left(E_{2}\right)=\frac{\text { Number of times } 1 \text { head appears }}{\text { Total number of trials }}=\frac{160}{400}=0.4$
(iii) $P($ getting 0 head $)=P\left(E_{3}\right)=\frac{\text { Number of times } 0 \text { head appears }}{\text { Total number of trials }}=\frac{128}{400}=0.32$
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
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