GATE 2018 IN Q.5: solution and answer

GATE 2018 · INQ.5 · 1 mark · MCQ

Consider a sequence of tossing of a fair coin where the outcomes of tosses are independent. The probability of getting the head for the third time in the fifth toss is

(A)
$\frac{5}{16}$
(B)
$\frac{3}{16}$
(C)
$\frac{3}{5}$
(D)
$\frac{9}{16}$

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Answer and solution

ANSWER

The probability that the third head appears on the fifth toss is $\frac{3}{16}$ , option (B).

Official answer: BSource: Official answer key, GATE 2018 IN (IIT Guwahati)

"Head for the third time on the fifth toss" means two things happen together: the fifth toss is a head, and the first four tosses contain exactly 2 heads.
The tosses are independent, so the probability that both happen is the product of their probabilities.
Binomial probability for the first four tosses: $P (exactly 2 heads in 4) = (2 4) (\frac{1}{2})^{2} (\frac{1}{2})^{2} = \frac{6}{16} = \frac{3}{8}$ .
The fifth toss is a head with probability $\frac{1}{2}$ .
Multiply. This is the negative binomial probability of the $r$ -th success on the $n$ -th trial, $(r - 1 n - 1) p^{r} (1 - p)^{n - r}$ , with $r = 3$ , $n = 5$ , $p = \frac{1}{2}$ : $P = \frac{3}{8} \times \frac{1}{2} = (2 4) (\frac{1}{2})^{5} = \frac{6}{32} = \frac{3}{16}$
Answer: option (B), $\frac{3}{16}$ .

KEY IDEA

For the $r$ -th success to land exactly on trial $n$ , trial $n$ must be a success and the first $n - 1$ trials must hold exactly $r - 1$ successes. That gives the negative binomial probability $(r - 1 n - 1) p^{r} (1 - p)^{n - r}$ . Here it is $(2 4) (\frac{1}{2})^{5} = \frac{3}{16}$ .

WATCH OUT FOR

Using $(3 5) (\frac{1}{2})^{5} = \frac{10}{32} = \frac{5}{16}$ (option A). That is the chance of exactly 3 heads somewhere in 5 tosses. It includes sequences such as HHHTT, where the third head comes before the fifth toss. The fifth toss has to be fixed as the third head, so only the first four tosses are free to arrange, giving $(2 4)$ .

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GATE 2018 IN Q.5: solution and answer

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