GATE 2020 IN Q.32: solution and answer

GATE 2020 · INQ.32 · 2 marks · NAT

Consider two identical bags B1 and B2 each containing 10 balls of identical shapes and sizes. Bag B1 contains 7 Red and 3 Green balls, while bag B2 contains 3 Red and 7 Green balls. A bag is picked at random and a ball is drawn from it, which was found to be Red. The probability that the Red ball came from bag B1 (rounded off to one decimal place) is ____

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Answer and solution

ANSWER

The probability that the Red ball came from bag B1 is 0.7.

Official answer: 0.7Source: Official answer key, GATE 2020 IN (IIT Delhi)

Priors: the bag is picked at random, so $P (B_{1}) = P (B_{2}) = \frac{1}{2}$ .
Likelihoods: each bag holds 10 balls, so $P (R ∣ B_{1}) = \frac{7}{10} = 0.7$ and $P (R ∣ B_{2}) = \frac{3}{10} = 0.3$ .
Total probability theorem: $P (R) = P (B_{1}) P (R ∣ B_{1}) + P (B_{2}) P (R ∣ B_{2}) = \frac{1}{2} (0.7) + \frac{1}{2} (0.3) = 0.5$ .
Bayes' theorem: $P (B_{1} ∣ R) = \frac{P ( B _{1} ) P ( R ∣ B _{1} )}{P ( R )} = \frac{0.5 \times 0.7}{0.5} = 0.7$
Check: $P (B_{2} ∣ R) = \frac{0.5 \times 0.3}{0.5} = 0.3$ , and $0.7 + 0.3 = 1$ . Rounded to one decimal place, the answer is $0.7$ .

KEY IDEA

This is Bayes' theorem. The two bags are equally likely to be picked, so the priors cancel. The posterior $P (B_{1} ∣ R)$ is then bag B1's share of all the red balls: $\frac{7}{7 + 3} = 0.7$ .

WATCH OUT FOR

The common slip is to answer $P (R) = 0.5$ , the overall chance of drawing a red ball, or to multiply out only the numerator, $P (B_{1}) P (R ∣ B_{1}) = 0.35$ , and forget to divide by $P (R)$ . Both confuse the joint or total probability with the conditional probability $P (B_{1} ∣ R)$ .

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GATE 2020 IN Q.32: solution and answer

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