A company has two plants \( \mathrm{P_1} \) and \( \mathrm{P_2} \) which produces 60% and 40% of pens respectively. At plant \( \mathrm{P_1} \) 80% of pens are rated of grade A and at plant \( \mathrm{P_2} \) 90% of pens are rated of grade A. A pen is picked up at random and is found to be of grade A. The probability that it has manufactured from plant \( \mathrm{P_1} \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{4}{7} \)
Let \( E_1 \), \( E_2 \) be the events that the pen is from \( P_1 \), \( P_2 \), and A the event that it is of grade A.
\( P(E_1)=0.6,\ P(E_2)=0.4 \)
\( P(A|E_1)=0.8,\ P(A|E_2)=0.9 \)
By Bayes’ theorem,
\( \displaystyle P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)} \)
\( \displaystyle =\frac{0.6\times 0.8}{0.6\times 0.8+0.4\times 0.9} \)
\( \displaystyle =\frac{0.48}{0.48+0.36} \)
\( \displaystyle =\frac{0.48}{0.84} \)
\( \displaystyle =\frac{4}{7} \)
Shortcut: Compare the two “grade A” shares directly: 60 × 80 = 4800 and 40 × 90 = 3600, in the ratio 4 : 3. So \( P(E_1|A)=\frac{4}{4+3}=\frac{4}{7} \).
Why the other options are incorrect
Option (2) \( \displaystyle\frac{3}{7} \) — Incorrect. \( \frac{3}{7} \) is \( P(E_2|A) \), the probability that the pen came from plant \( P_2 \), not \( P_1 \).
Option (3) \( \displaystyle\frac{56}{83} \) — Incorrect. The value is \( \frac{4}{7} \), not \( \frac{56}{83} \).
Option (4) \( \displaystyle\frac{17}{83} \) — Incorrect. The value is \( \frac{4}{7} \), not \( \frac{17}{83} \).