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Question

For any two events A and B, if P(A|B) = 0.6 and P(B|A) = 0.4, then what is the value of P(A)/P(B) ?

This question was previously asked in
SSC CGL 2023 (Tier-II) Paper 1 Previous Year Paper (26-Oct-2023) (Shift-1)
The correct answer is

1.5

We are given \(P(A\mid B) = 0.6\) and \(P(B\mid A) = 0.4\), and we need to find \(\dfrac{P(A)}{P(B)}\).

By the definition of conditional probability:

\(P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}\) and \(P(B\mid A) = \dfrac{P(A\cap B)}{P(A)}\).

Dividing the first by the second, the common factor \(P(A\cap B)\) cancels:

\(\dfrac{P(A\mid B)}{P(B\mid A)} = \dfrac{P(A\cap B)/P(B)}{P(A\cap B)/P(A)} = \dfrac{P(A)}{P(B)}\).

Therefore \(\dfrac{P(A)}{P(B)} = \dfrac{0.6}{0.4} = 1.5\).

Hence, the correct answer is 1.5.

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Important Questions from Conditional Probability

  1. A and B are two events such that A̅ and B̅ are mutually exclusive. If P(A) = 0.5 and P(B) = 0.6, then what is the value of P(A|B)?

  2. For two dependent events A and B, it is given that P(A) = 0.2 and P(B) = 0.5. If A ⊆ B, then the values of conditional probabilities P(A|B) and P(B|A) are respectively

  3. In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?

  4. Let X1 and X2 be independent random variables each having geometric distribution qk p ; k = 0, 1, 2, …. Then the conditional distribution of X1 given X1 + X2 is

  5. In a batch Of 100 resistors Of 1kiloohm resistance, 80 numbers are within the required tolerance values and II numbers are below the required tolerance values, the remaining are above the required tolerance values. If two resistors are drawn one after the other without replacement, the probability of the first one drawn is below and the second one drawn is above the required tolerance value is

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