An unbiased coin is tossed to get 2 points for turning up a head and one point for the tail. If three unbiased coins are tossed simultaneously, then the probability of getting a total of odd number of points is

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$\frac{1}{8}$

$\frac{3}{8}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

3 Coins are tossed

$S = \{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}$

no of points $= {6, 5, 4, 3}$

$n (S) = 2^{3} = 8 E =$ getting odd no. of points

$n (E) = 4 \Rightarrow P (E) = \frac{4}{8} = \frac{1}{2}$

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