An unfair coin lands on heads with a probability of 1/4. When tossed

\(P(2 heads) = C(2, n) * (\frac{1}{4})^2 * (\frac{3}{4})^{(n-2)}\)
\(P(3 heads) = C(3, n) * (\frac{1}{4})^3 * (\frac{3}{4})^{(n-3)}\)

Let the 2 equal and we get:
\(C(2, n) * (\frac{1}{4})^2 * (\frac{3}{4})^{(n-2)}=C(3, n) * (\frac{1}{4})^3 * (\frac{3}{4})^{(n-3)}\)
That is:
\(C(2, n) * 3^{(n-2)} = C(3,n) * 3^{(n-3)}\)
That is:
\(3 * C(2,n) = C(3, n)\)
That is:
\(3 * \frac{n*(n-1)}{2} = \frac{n*(n-1)*(n-2)}{(3 * 2 * 1)}\)
That is:
\(n-2=9\)
\(n=11\)
Thus D.

Probability of getting 2 Heads= nC2*(1/4)^2 *(3/4)^n-2
={n(n-1)/2} * 3^(n-2) * {1/(4^n) }
Probability of getting 2 Heads= nC3 * (1/4)^3 * (3/4)^n-3
={n(n-1)(n-2)/6} * 3^(n-3) * {1/(4^n) }
Now as the above 2 probabilities are equal
{n(n-1)/2} * 3^(n-2) * {1/(4^n) } = {n(n-1)(n-2)/6} * 3^(n-3) * {1/(4^n) }
3= (n-2)/3
n-2=9
n=11

IMO D

we have prob of head = 1/4
So probab of tail = 3/4

take value of n from options and starting solving

n= 5

we have prob of exactly 2 head = 5C2*(1/4)^2*(3/4)^3
and probab of exaclty 3 heads = 5C3*(1/4)^3*(3/4)^2

we see only n = 11 will satisfy the given condition

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)

­Thank you for the explanation. 

Ma'am what is the cue to know that the following question will require a combination formula? I have seen that the previous answer is multiplying nC2 and nC3 to the equation. Can not we work only with the probabilty?

The key is that with multiple tosses more than one pattern could satisfy the condition, for example, 5 total tosses and requiring 2 heads can be accomplished:

HHTTT, HTHTT, TTTHHH, etcetera

Each of these has the same probability so identifying the probability of all of these means counting the number of different patterns.

That is where the combination approach for counting is required, which is really just the number of permutations of in the example 5 things:

5!

But because this treats HH as two results when it's actually one pattern, for example, the result above needs to be divided by the permutations of identical things. So then:

5!/2!3! = 10

Would be multiplied against the base probability initially calculated.

This is what is meant in her response about the number of different arrangements.

Posted from my mobile device

­Both approaches are exactly the same. Since we view Heads and Tails results as HTHTTT etc., we talk in terms of arranging some Heads and some Tails. 
If we have n tosses and exactly 2 heads, we can select 2 positions for the 2 Heads in \(nC2 = \frac{n!}{2!*(n-2)!}\) ways

If we have n tosses and exactly 2 heads, we can arrange them as \(\frac{n!}{2!*(n-2)! }\) because the 2 Heads are identical and the (n-2) tails are identical. 

These are just two different perspectives on the same thing. The combinations formula is used for "selection".

Check more on it here:
https://youtu.be/tUPJhcUxllQ
 ­

­why we are multiplying an extra 3 in 2 Heads case, wouldn't that change the probability for the case­

­Look at the highlighted part above. We have an additional 3 in the first case because we have only 2 Heads so there is an additional Tails. So it is accounted for in the calculation and only then the probabilities are equal. 

\(\frac{1}{4} * \frac{1}{4 }* \frac{3}{4} * \frac{3}{4} * .... *\frac{n!}{2!*(n - 2)!} = \frac{1}{4} * \frac{1}{4} * \frac{1}{4} * \frac{3}{4} * .... *\frac{n!}{3!*(n - 3)!}\)­