CBSE Class 12 Mathematics sample paper 2024-25: question 31(b)

Hints

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Step-by-step solution

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1. $P(X=r)=k\cdot \frac{1}{5^r}$, $r=0,1,2,\dots$, where $k$ is a non-zero constant.

   ½ mark

2. So $P(X=0)=k$, $P(X=1)=\frac{k}{5}$, $P(X=2)=\frac{k}{5^2}$, $P(X=3)=\frac{k}{5^3},\dots$

   ½ mark

3. Sum of probabilities is 1: $k\left(1+\frac{1}{5}+\frac{1}{5^2}+\dots \right)=1$.

   ½ mark

4. Infinite G.P.: $k\cdot \frac{1}{1-\frac{1}{5}}=1\Rightarrow k=\frac{4}{5}$.

   ½ mark

5. $P(X<3)=P(0)+P(1)+P(2)=\frac{4}{5}\left(1+\frac{1}{5}+\frac{1}{25}\right)=\frac{4}{5}\cdot \frac{31}{25}=\frac{124}{125}$.

   1 mark

Answer: $P(X<3)=\frac{124}{125}$

Revise the chapter

• Chapter: Probability (notes, formulas, where marks are lost and every past question)