Probability
Probability
nta_pyq_2025_jan
Grade 12
Question:
Bag $1$ contains $4$ white balls and $5$ black balls, and Bag $2$ contains $n$ white balls and $3$ black balls. One ball is drawn randomly from Bag $1$ and transferred to Bag $2$. A ball is then drawn randomly from Bag $2$. If the probability that the ball drawn is white is $\dfrac{29}{45}$, then $n$ is equal to:
Step-by-Step Solution
Key Concept: Condition on the colour transferred. $P(\text{final white})=P(\text{W transferred})\cdot\dfrac{n+1}{n+4}+P(\text{B transferred})\cdot\dfrac{n}{n+4}.$
$P(\text{transfer W})=\dfrac{4}{9},\ P(\text{transfer B})=\dfrac{5}{9}.$
$P(\text{final W})=\dfrac{4}{9}\cdot\dfrac{n+1}{n+4}+\dfrac{5}{9}\cdot\dfrac{n}{n+4}=\dfrac{9n+4}{9(n+4)}=\dfrac{29}{45}.$
$45(9n+4)=29\cdot 9(n+4)\Rightarrow 405n+180=261n+1044\Rightarrow 144n=864\Rightarrow n=6.$
Correct Answer: 1