Probability
Probability
Allen Star Batch
Grade 12
Question:
A machine containing $n$ different balls, when switched on, can throw up any number of balls one by one. The probability of throwing $r$ balls is directly proportional to $r$. Given that a particular ball is the first ball to pop up, the probability that machine has thrown up all the balls is:
$\frac{2}{n+1}$
$\frac{1}{n+1}$
$\frac{2}{n}$
None of these
Step-by-Step Solution
Key Concept: The probability P(r) = kr is proportional to r balls thrown. Using normalization ∑P(r) = 1 gives k = 2/[n(n+1)]. Then apply conditional probability: P(all n balls | first ball thrown) = P(all n balls AND first ball thrown)/P(first ball thrown) using Bayes' theorem.
Using the law of total probability, $P(E) = \sum_{r=1}^{n} P(E_r)P(E/E_r)$. Since $\sum_{r=1}^{n} P(E_r) = 1$, we get $k = \frac{2}{n(n+1)}$. The conditional probability $P(E/E_r) = \frac{n-1}{^nC_r \cdot r!} \cdot \frac{1}{^nC_r \cdot r!} = \frac{1}{n}$. Therefore, $P(E) = \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{2r}{n(n+1)} = \frac{1}{n}$, and $P(E_n/E) = \frac{2}{n+1}$.
Correct Answer: 1