If a fair coin is tossed 4 times, what is the probability that two heads and two tails will result?

\frac{3}{8}

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\frac{1}{2}

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\frac{5}{8}

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\frac{3}{4}

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Solution

The correct option is A $\frac{3}{8}$
Using Binomial Distribution

Let $p = P (H) = \frac{1}{2},$

$q = P (T) = \frac{1}{2}, n = 4,$
x:Number of Heads

Probability that exactly two heads is
$P (x = 2) =^{4} C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{2} = \frac{3}{8}$

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If a fair coin is tossed 4 times, what is the probability that two heads and two tails will result?

The correct option is A 38Using Binomial Distribution Let p=P(H)=12, q=P(T)=12, n=4, x:Number of Heads Probability that exactly two heads is P(x=2)= 4C2(12)2(12)2=38

The correct option is A $\frac{3}{8}$
Using Binomial Distribution

Let $p = P (H) = \frac{1}{2},$

$q = P (T) = \frac{1}{2}, n = 4,$
x:Number of Heads

Probability that exactly two heads is
$P (x = 2) =^{4} C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{2} = \frac{3}{8}$