Probability
Geometric Distribution — Conditional Probability
nta_pyq_2024_jan
Grade 12
Question:
A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$, $b=P(X\geq3)$ and $c=P(X\geq6\mid X>3)$. Then $\dfrac{b+c}{a}$ is equal to
Step-by-Step Solution
Key Concept: $P(X=k)=(5/6)^{k-1}(1/6)$. $a=P(X=3)=(5/6)^2(1/6)=25/216$. $b=P(X\geq3)=\sum_{k=3}^\infty(5/6)^{k-1}(1/6)=(5/6)^2/(1-5/6)=25/36$. $c=P(X\geq6|X>3)$: by memoryless property of geometric distribution, $c=P(X\geq3)=25/36$.
$a=25/216$, $b=c=25/36$. $\frac{b+c}{a}=\frac{50/36}{25/216}=12$.
Correct Answer: 12