Question

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes-no head $-14$, one head$-38$, two heads$-36$, three heads$-12$.
If the three coins are simultaneously tossed again, compute he probability of (i) getting more heads than tails,
(ii) getting more tails than heads,

Answer 1

Since 3 coins are tossed simultaneously 100 times, then Total number of possible outcomes = 100
(i).
Number of times 2 Heads means 2 Heads and 1 Tail appears = 36
Number of times 3 Heads means 3 Heads and 0 Tail appears = 12
Number of favourable outcomes = 36 + 12 = 48
P(getting more Heads than Tails) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$
$=\frac{48}{100}=\frac{12}{25}$
b. Number of times 2 Tails means 2 Tails and 1 Head appears = 38
Number of times 3 Tails means 3 Tails and 0 Head appears = 14
Number of favourable outcomes = 38 + 14 = 52
P(getting more Tails than Heads)=$\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$
=$\frac{52}{100}=\frac{13}{25}$