The probabilities that John and James pass an examination are 3 4 and 3 5 respectively. Find the probability…

Analysis

To solve this problem, we need to understand the concept of independent events in probability. Since John passing or failing does not affect James passing or failing, their probabilities are independent.

Concept

1. Complementary Events: The probability of an event happening ($P(A)$) and the probability of it not happening ($P(A')$) must sum to 1. • $P(\text{Fail}) = 1 - P(\text{Pass})$
2. Independent Events: For two independent events $A$ and $B$, the probability of both occurring is $P(A \cap B) = P(A) \times P(B)$.

Step-by-Step Solution

Step 1: Find the probability of each boy failing.

• Probability that John passes, $P(J_p) = \frac{3}{4}$
• Probability that John fails, $P(J_f) = 1 - \frac{3}{4} = \frac{1}{4}$

• Probability that James passes, $P(M_p) = \frac{3}{5}$
• Probability that James fails, $P(M_f) = 1 - \frac{3}{5} = \frac{2}{5}$

Step 2: Find the probability of both failing. Since these are independent events, we multiply the probability of John failing by the probability of James failing:

• $P(\text{Both fail}) = P(J_f) \times P(M_f)$
• $P(\text{Both fail}) = \frac{1}{4} \times \frac{2}{5}$

Step 3: Simplify the fraction.

• $P(\text{Both fail}) = \frac{1 \times 2}{4 \times 5} = \frac{2}{20}$
• Reducing the fraction: $\frac{2}{20} = \frac{1}{10}$

Conclusion

The probability that both boys fail the examination is $\frac{1}{10}$.

Exam Tip: Always remember that "failing" is the complement of "passing." If you are ever asked for the probability of "at least one passing," it is often easier to calculate $1 - P(\text{both failing})$ rather than adding up the individual probabilities of different success scenarios!