Probability
Probability
Allen Star Batch
Grade 12
Question:
A number is chosen at random from the numbers 10 to 99. By seeing the number a man will laugh if product of the digits is 12. If he choose three numbers with replacement then the probability that he will laugh at least once is:
$1 - \left(\frac{3}{5}\right)^3$
$\left(\frac{43}{45}\right)^3$
$1 - \left(\frac{4}{25}\right)^3$
$1 - \left(\frac{43}{45}\right)^3$
Step-by-Step Solution
Key Concept: The law of total probability combined with Bayes' theorem requires computing the weighted average of conditional probabilities across all partitioning events.
Three events partition the sample space: $E_1$ (A draws white), $E_2$ (A draws black), $E_3$ (A draws green) with probabilities $P(E_1) = \frac{2}{12}$, $P(E_2) = \frac{4}{12}$, $P(E_3) = \frac{6}{12}$ respectively. Event $E$ is that B draws two green balls. We compute $P(E|E_i)$ for each case, then use the law of total probability: $P(E_2|E) = \frac{P(E_2)P(E|E_2)}{\sum_{i=1}^{3} P(E_i)P(E|E_i)} = \frac{\frac{4}{12} \cdot \frac{^6C_2}{^{11}C_2}}{\frac{2}{12} \cdot \frac{^5C_2}{^{11}C_2} + \frac{4}{12} \cdot \frac{^6C_2}{^{11}C_2} + \frac{6}{12} \cdot \frac{^7C_2}{^{11}C_2}} = \frac{2}{5}$.
Correct Answer: 4