Problem Solving (PS) | Re: An unfair coin lands on heads with a probability of 1/4. When tossed

Probability of getting exactly 2 Heads on n tosses =
1/4 * 1/4 *3/4 * 3/4 * .... * Number of arrangements of 2 Heads and rest tails

Probability of getting exactly 3 Heads on n tosses =
1/4 * 1/4 *1/4 * 3/4 * .... * Number of arrangements of 3 Heads and rest tails

In the case of 2 Heads, there is an extra 3 in the numerator. If both these probabilities are to be the same, the number of arrangements in case of 3 heads should have an extra 3 in the numerator.

Number of arrangements of 2 Heads and rest tails\( =\frac{ n!}{2!*(n -2)!}\)
Number of arrangements of 3 Heads and rest tails\( =\frac{ n!}{3!*(n -3)!}\)

\( =\frac{ n!}{3!*(n -3)!} = 3 *\frac{ n!}{2!*(n -2)!}\)

\( n =11\)

Answer(D)