5,620 views
16 votes
16 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. $\frac{3}{4}$

3 Answers

Best answer
21 votes
21 votes
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 $=\dfrac{^{4}C_{2}}{2^{4}} =\dfrac{3}{8}.$
edited by
47 votes
47 votes
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
7 votes
7 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.:)
Answer:

Related questions

38 votes
38 votes
4 answers
4
Kathleen asked Sep 18, 2014
8,310 views
The following finite state machine accepts all those binary strings in which the number of $1$’s and $0$’s are respectively: divisible by $3$ and $2$odd and evene...