A coin is tossed three tiimes. Find the probability of getting exactly two heads.

\frac{5}{8}

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\frac{2}{8}

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\frac{1}{2}

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\frac{3}{8}

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Solution

The correct option is D $\frac{3}{8}$
Let B be the event of getting $2$ head.

$B = [H H T, H T H, T H H]$

$n (B) = 3$
Probability of getting two heads $= P (B)$

$= \frac{n (B)}{n (S)}$

$= \frac{3}{2^{3}}$

$= \frac{3}{8}$

Hence, the correct option is D. $3 / 8$ .

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A coin is tossed three tiimes. Find the probability of getting exactly two heads.

The correct option is D 38Let B be the event of getting 2 head.B=[HHT,HTH,THH]n(B)=3Probability of getting two heads =P(B)=n(B)n(S)=323=38Hence, the correct option is D. 3/8.

The correct option is D $\frac{3}{8}$
Let B be the event of getting $2$ head.

$B = [H H T, H T H, T H H]$

$n (B) = 3$
Probability of getting two heads $= P (B)$

$= \frac{n (B)}{n (S)}$

$= \frac{3}{2^{3}}$

$= \frac{3}{8}$

Hence, the correct option is D. $3 / 8$ .