JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Probability XII

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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) $\frac{5}{21}$ ✓ Correct
(C) $\frac{4}{21}$
(D) $\frac{2}{7}$

Answer: (B)

Solution

Let

P (H) = \frac{1}{3}

and

P (T) = \frac{2}{3}

.

The experiment stops with head when we get HH. This can happen as:

HH (directly)
HTHH (after one HT cycle)
HTHTHH (after two HT cycles), etc.

Or starting with T:

THH
THTHH, etc.

Probability =

P (H H or H T H H or \dots) + P (T H H or T H T H H or \dots)

= \frac{\frac{1}{3} \cdot \frac{1}{3}}{1 - \frac{2}{3} \cdot \frac{1}{3}} + \frac{\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}}{1 - \frac{2}{3} \cdot \frac{1}{3}}

= \frac{\frac{1}{9}}{\frac{7}{9}} + \frac{\frac{2}{27}}{\frac{7}{9}} = \frac{1}{7} + \frac{2}{21} = \frac{3}{21} + \frac{2}{21} = \frac{5}{21}

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Single correct option

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Probability XII

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