GATE CSE 2004 | Question: 25

Question

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

• A. $\frac{3}{8}$

• B. $\frac{1}{2}$

• C. $\frac{5}{8}$

• D. $\frac{3}{4}$

Comments

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

---

total outcomes with 4 tosses = 2^4 =16

number of ways in which we can get 2H or 2T  = out of 4 outcomes, any 2 are same =  4c2 = 6

Pr(we get 2H or 2T on tossing 4 times) = 6 / 16 = 3/8.

Answer 1

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 $=\dfrac{^{4}C_{2}}{2^{4}} =\dfrac{3}{8}.$

Comments

1/22 ??? who edited this ??

---

Thanks for reporting.

Answer 2

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

Comments

perfect

Answer 3

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.:)

Answer 4

Outcome of one coin doesn't  affect other outcomes, thus it is an Independent Event 

For 2 heads and 2 tails, sequences are - {HHTT,HTTH,TTHH,THHT,THTH,HTHT}

As it is and Independent event, 

Similary, for other sequences also it is  

Therefore, P(Two heads and Two Tails) =

Just one another approach :)