A box contains 10 cards: 4 blue cards numbered 1, 2, 3, 5 and 6 red cards numbered 1, 2, 3… — GRE General Test

A box contains $10$ cards: $4$ blue cards numbered $1, 2, 3, 5$ and $6$ red cards numbered $1, 2, 3, 4, 6, 8$ . Two cards are drawn sequentially at random without replacement from the box. Let $A$ be the event that the first card drawn is blue, and let $B$ be the event that the sum of the numbers on the two drawn cards is an even number. What is the value of the conditional probability $P(A \mid B)$ ?

Answer: 0.4

Answer

0.4 (or 2/5)

The conditional probability

P(A \mid B)

represents the likelihood that the first card drawn was blue given that the sum of the two drawn cards is even. There are 40 total outcome pairs resulting in an even sum (20 where both are odd and 20 where both are even). Among these 40 outcomes, exactly 16 start with a blue card (12 starting with a blue odd card and 4 starting with a blue even card). Therefore,

P(A \mid B) = \frac{16}{40} = 0.4

Step-by-Step Solution

Classify the sample space of cards by color and number parity.

Blue cards consist of 3 odds (1, 3, 5) and 1 even (2). Red cards consist of 2 odds (1, 3) and 4 evens (2, 4, 6, 8). Across all 10 cards, there are 5 odd cards and 5 even cards.

Categorizing by parity is essential because the sum of two integers is even if and only if both numbers share the same parity (both odd or both even).

Calculate the total number of sequential draw outcomes belonging to event

B

(sum is even).

Number of (Odd, Odd) outcomes =

5 \times 4 = 20

. Number of (Even, Even) outcomes =

5 \times 4 = 20

. Total outcomes for event

B

N(B) = 20 + 20 = 40

Since draws are without replacement, drawing a card reduces the available count of that parity by 1 for the second draw.

Calculate the number of outcomes belonging to the joint event

A \cap B

(first card is blue AND sum is even).

Subcase 1 (Blue Odd 1st, Odd 2nd):

3 \times 4 = 12

outcomes. Subcase 2 (Blue Even 1st, Even 2nd):

1 \times 4 = 4

outcomes. Total outcomes for

A \cap B

N(A \cap B) = 12 + 4 = 16

To satisfy both event

A

(first card blue) and event

B

(even sum), the second card must match the parity of the selected blue card.

Compute the conditional probability

P(A \mid B)

P(A \mid B) = \frac{N(A \cap B)}{N(B)} = \frac{16}{40} = \frac{2}{5} = 0.4

By the definition of conditional probability,

P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{N(A \cap B)}{N(B)}

when all outcomes in the reduced sample space are equally likely.

Key Concept

Conditional Probability and Sequential Dependent Sampling

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