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+7 votes

If a fair coin is tossed four times. What is the probability that two heads and two tails will result?

  1. $\frac{3}{8}$

  2. $\frac{1}{2}$

  3. $\frac{5}{8}$

  4. $\frac{3}{4}$

asked in Probability by Veteran (59.5k points) | 1.1k views

Binomial distribution:-
P(H) = 1/2
P(T) = 1/2
4C2(1/2)2(1/2)2 = 6/16 = 3/8.

for better explanation

given that a coin is tossed 4 times and asking the probability of getting 2heads and 2tails?

if we apply binomial distribution, then as given in the above explanation ^

n=4  in one toss p(head)=1/2=p(tail)

now if we want just 2 heads out of 4 tosses then it means that if we just need 2 heads only then definitely other 2 will be tails ,,right .  thus fixing  2 heads implicitly means other 2 are tails which is what asked i.e 2 heads and 2 tails.

thus n=4, p(h)=1/2=p(t) rest is all above.

3 Answers

+9 votes
Best answer
Answer - $A$

Out of $4$ times $2$ times head should be present

No. of ways of selecting these $2$ places $= ^{4}C_{2}$

probability of getting $2$ heads and $2$ tails $= \left(\dfrac{1}{2^2}\right).\left(\dfrac{1}{2^2}\right)$

probability $=\dfrac{^{4}C_{2}}{2^{4}} =\dfrac{3}{8}.$
answered by Loyal (9k points)
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+10 votes
As fair coin is tossed four times then sample space will contain total 2*2*2*2=16 elements

Number of permutation for HHTT


So,probability = 6/16

answered by (375 points)
+4 votes
Another approach is tat in d sample space of Heads and Tails there are 10 placements of Head and Tail which don't contain Two Heads and Two Tails.They are given as follows:

(T,T,T,T),(H,H,H,H),(T,T,T,H),(T,H,T,T),(T,T,H,T),(H,T,T,T),(H,H,H,T),(H,T,H,H),(H,H,T,H),(T,H,H,H).After having removed these options 6 will be left.Therefore,probability is given by 6/16 which evaluates to 3/8.:)
answered by Boss (13.3k points)

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