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

My calculation is. no. of ways of getting exactly 2 head and 2 tails .will be $6$ out of $8$. Eg $$HHTT,THHT,TTHH,HTTH,HTHT,THTH,HHHT,TTTH$$

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    $\begingroup$ You mean, 6 out of 16, right? $\endgroup$
    – Did
    Jul 3, 2016 at 14:53
  • $\begingroup$ There are $16$ equally probable possible throws. Of these $\binom 42=6$ will have exactly two $H's$. $\endgroup$
    – lulu
    Jul 3, 2016 at 14:54
  • $\begingroup$ ... don't forget $HHTH, HTHH, THHH, TTHT, THTT, HTTT, HHHH, TTTT$ $\endgroup$
    – Joffan
    Jul 3, 2016 at 15:01
  • $\begingroup$ $$HHHH,HHHT,HHTH,HTHH,THHH , TTTT , TTTH , TTHT ,THTT,HTTT,\color{red}{HHTT,HTHT,HTTH,TTHH,THTH,THHT}$$ $\endgroup$ Jul 3, 2016 at 15:02

2 Answers 2

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Let the random variable $X$ be the number of heads that come up when a fair coin is tossed $4$ times. Then $X \sim B\left(4, \frac{1}{2}\right)$.

We want there to be exactly two heads (forcing the other two tosses to be tails), so $$\mathbb{P}(X = 2) = \binom{4}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^2 = \frac{3}{8}.$$

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There are $2^4=16$ possible outcomes:

$HHHH$ $\ \ $ $HHHT$ $\ \ $ $HHTH$ $\ \ $ $HTHH$ $\ \ $ $THHH$ $\ \ $ $\color{red}{HHTT}$ $\ \ $ $\color{red}{HTHT}$ $\ \ $ $\color{red}{THHT}$ $\ \ $ $\color{red}{HTTH}$ $\ \ $ $\color{red}{THHT}$ $\ \ $ $\color{red}{TTHH}$ $\ \ $ $HTTT$ $\ \ $ $THTT$ $\ \ $ $TTHT$ $\ \ $ $TTTH$ $\ \ $ $TTTT$ $\ \ $

6 of them have 2 tails and 2 heads. Thus the probability of getting 2 heads and 2 tails is

$\frac {\text{No. of favorable outcomes}}{\text{No. of possible outcomes}}=\frac{6}{16}=\frac38$

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