A fair die is tossed repeatedly until a six is obtained. Let denote the number of tosses required and let and Then is equal to

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Solution

Q.
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 $c = P (X \geq 6 ∣ X > 3) .$ Then $\frac{b + c}{a}$ is equal to _______ . [2024]

Ans.
(12)

Probability of getting six $= \frac{1}{6}$

Probability of not getting a six $= \frac{5}{6}$

Let $X$ denote the number of tosses required. Then,

$a = P (X = 3) = \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{5^{2}}{6^{3}} = \frac{25}{216}$

$b = P (X \geq 3) = 1 - [P (X = 1) + P (X = 2)]$ $= 1 - [\frac{1}{6} + \frac{5}{6} \times \frac{1}{6}] = \frac{25}{36}$

$c = P (X \geq 6 ∣ X > 3) = \frac{P [(X \geq 6) \cap (X > 3)]}{P (X > 3)} = \frac{P (X \geq 6)}{P (X > 3)}$

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

$= \frac{{(\frac{5}{6})}^{5} \times \frac{1}{6} [1 + \frac{5}{6} + {(\frac{5}{6})}^{2} + \dots]}{\frac{25}{36} - \frac{25}{216}}$ $= \frac{25}{36}$

$∴$ $\frac{b + c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}} = 12$

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