Probability Question 3
Question 3 - 2024 (27 Jan Shift 1)
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 $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to
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Answer (12)
Solution
$a=P(X=3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{25}{216}$
$b=P(X \geq 3)=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{3} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\ldots \ldots$
$=\frac{\frac{25}{216}}{1-\frac{5}{6}}=\frac{25}{216} \times \frac{6}{1}=\frac{25}{36}$
$P(X \geq 6)=\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{6} \cdot \frac{1}{6}+\ldots \ldots$.
$=\frac{\left(\frac{5}{6}\right)^{5} \cdot \frac{1}{6}}{1-\frac{5}{6}}=\left(\frac{5}{6}\right)^{5}$
$c=\frac{\left(\frac{5}{6}\right)^{5}}{\left(\frac{5}{6}\right)^{3}}=\frac{25}{36}$
$\frac{b+c}{a}=\frac{\left(\frac{5}{6}\right)^{2}+\left(\frac{5}{6}\right)^{2}}{\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}}=12$
