A coin is tossed 4 times. What is the probability that the

Solution:
We need the probability of: HHTT.

\(P(HHTT)=\frac{4!}{2!*2!}*(\frac{1}{2})^2*(\frac{1}{2})^2=\frac{3}{8}\).

Explanation:
\(\frac{4!}{2!*2!}=6\) is number of all possible cases for our wining scenarios HHTT to appear (which is number of permutation of 4 letters HHTT where 2 H's and 2T's are identical: HHTT, HTHT, HTTH, TTHH, THTH, THHT), \((\frac{1}{2})^2\) is the probability of HH and \((\frac{1}{2})^2\) is the probability of TT.

Theory behind the concept above:
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

General theory on this topic:
Probability: math-probability-87244.html
Combinatorics: math-combinatorics-87345.html

Hope it helps.