JEE 2024 — Mathematics PYQ

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Question

JEE 2024 — Mathematics PYQ

JEE | Mathematics | 2024

A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a = P (X = 3)$ , $b = P (X \geq 3)$ and $C = P(X > 6 \mid X > 3)$ . Then $\frac{b + c}{a}$ is equal to

Choose the correct answer:

Explanation

$P (A) = P (getting a six) = \frac{1}{6}$

$P (X = 3) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216} = a$

$P (X \geq 3) = (\frac{5}{6})^{2} (\frac{1}{6}) + (\frac{5}{6})^{3} (\frac{1}{6}) + \dots$

$= \frac{25}{216} \div (1 - \frac{5}{6}) = \frac{25}{36}$

$First term = \frac{25}{216}, r = \frac{5}{6}$

$\frac{P ( X \geq 6 )}{P ( X \geq 3 )} = \frac{P ( X \geq 6 )}{P ( X \geq 3 )}$

$= \frac{( \frac{5}{6} ) ^{5} ( \frac{1}{6} ) + ( \frac{5}{6} ) ^{6} ( \frac{1}{6} ) + \dots}{( \frac{5}{6} ) ^{2} ( \frac{1}{6} ) + ( \frac{5}{6} ) ^{3} ( \frac{1}{6} ) + \dots}$

$= \frac{( \frac{5}{6} ) ^{3} ( \frac{1}{6} ) + ( \frac{5}{6} ) ^{4} ( \frac{1}{6} ) + \dots}{( \frac{5}{6} ) ^{2} ( \frac{1}{6} ) + ( \frac{5}{6} ) ^{3} ( \frac{1}{6} ) + \dots}$

$= \frac{25}{36} = C$

$So, \frac{b + c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}} = \frac{50}{36} \times \frac{216}{25} = 12$