Question:
Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(i) $\frac{1}{2}$
(ii) $\frac{1}{4}$
(iii) $\frac{3}{8}$
(iv) $\frac{3}{4}$
Solution:
(C) $\frac{3}{8}$
Explanation:
When 3 coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, THT, HTT, TTH and TTT.
Total number of possible outcomes = 8
Let E be the event of getting exactly two heads.
Then, the favourable outcomes are HHT, THH, and HTH.
Number of favourable outcomes = 3
$\therefore$ Probability of getting exactly 2 heads $=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{3}{8}$