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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. $\frac{3}{4}$

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 ^

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.

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 $=\dfrac{^{4}C_{2}}{2^{4}} =\dfrac{3}{8}.$
edited
1/22 ??? who edited this ??
Thanks for reporting.
As fair coin is tossed four times then sample space will contain total 2*2*2*2=16 elements

Now,
Number of permutation for HHTT
=4!/2!*2!

=6

So,probability = 6/16

=3/8
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.:)