devashish2407 wrote:
KarishmaB wrote:
Bunuel wrote:
An unfair coin lands on heads with a probability of 1/4. When tossed n > 1 times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of n?
(A) 5
(B) 8
(C) 10
(D) 11
(E) 13
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.
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