Three fair coins are tossed. If both heads and tails appears, then probability that exactly one head

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14.Probability

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Three fair coins are tossed. If both heads and tails appears, then the probability that exactly one head appears, is

A

$\frac{3}{8}$

B

$\frac{1}{6}$

C

$\frac{1}{2}$

D

$\frac{1}{3}$

Solution

(c) Since both heads and tails appears, so

$n(S) = \{ HHT,\,HTH,\,THH,\,HTT,\,THT,\,TTH\} $

$n(E) = \{ HTT,\,THT,\,TTH\} $

Hence required probability $ = \frac{3}{6} = \frac{1}{2}.$

Standard 11

Mathematics

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