JEE Advanced 2023 — Paper 2 — MATH-I solution

JEE Advanced2023Paper 2MATH-I

Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is $\frac{1}{3}$ , then the probability that the experiment stops with head is.

A
$\frac{1}{3}$
B
$\frac{5}{21}$
C
$\frac{4}{21}$
D
$\frac{2}{7}$

Solution

Let $p = P (H) = \frac{1}{3}$ and $q = P (T) = \frac{2}{3}$ . The experiment stops with head when the sequence ends in HH and no two consecutive identical outcomes appear before the final pair.

Such sequences either begin with H (HH, HTHH, HTHTHH, ...) or with T (THH, THTHH, THTHTHH, ...).

Group 1 (begin with H, end with HH): $p^{2} + p^{2} (q p) + p^{2} (q p)^{2} + \dots = \frac{p ^{2}}{1 - q p} = \frac{1/9}{1 - 2/9} = \frac{1}{7} .$

Group 2 (begin with T, end with HH): $q p \cdot p + q p \cdot p (q p) + \dots = \frac{q p ^{2}}{1 - q p} = \frac{2/27}{7/9} = \frac{2}{21} .$

Adding: $\frac{1}{7} + \frac{2}{21} = \frac{3}{21} + \frac{2}{21} = \frac{5}{21} .$

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