A container holds 5 red marbles, 4 blue marbles, and 3 green marbles. Two marbles are draw… — GRE General Test

A container holds 5 red marbles, 4 blue marbles, and 3 green marbles. Two marbles are drawn sequentially at random without replacement. Let event $A$ be the event that at least one of the drawn marbles is red, and let event $B$ be the event that the second marble drawn is green. What is the conditional probability $P(B \mid A)$ ?

$\frac{1}{6}$ Answer
B
$\frac{1}{4}$
C
$\frac{3}{22}$
D
$\frac{3}{19}$
E
$\frac{5}{22}$

Answer

The conditional probability

P(B \mid A)

\frac{1}{6}

To find

P(B \mid A)

, we evaluate

\frac{P(A \cap B)}{P(A)}

. The probability of at least one red marble

P(A)

1 - P(\text{no red}) = 1 - \frac{7 \times 6}{12 \times 11} = \frac{90}{132}

. The intersection event

A \cap B

requires the second marble to be green and at least one marble to be red, which means the first marble must be red and the second green. The probability of this is

\frac{5 \times 3}{12 \times 11} = \frac{15}{132}

. Taking the ratio

\frac{15/132}{90/132}

yields

\frac{15}{90} = \frac{1}{6}

Step-by-Step Solution

Calculate the total number of ordered outcomes and the probability of event A using the complementary event.

Total outcomes drawing 2 marbles from 12 without replacement is

12 \times 11 = 132

. The complement

A^c

(no red marbles selected from the 7 non-red marbles) has

7 \times 6 = 42

outcomes. Thus,

P(A^c) = \frac{42}{132} = \frac{7}{22}

, which means

P(A) = 1 - \frac{7}{22} = \frac{15}{22} = \frac{90}{132}

Using the complement rule is the most efficient way to compute 'at least one' probabilities.

Determine the intersection event

A \cap B

and calculate its probability.

Event

B

specifies that the second marble is green. For event

A

(at least one marble is red) to also occur, the first marble must be red. Thus,

A \cap B

is equivalent to 'the first marble is red AND the second marble is green'. The number of favorable outcomes is

5 \times 3 = 15

. So

P(A \cap B) = \frac{15}{132}

Mutual exclusivity between red and green on the second draw simplifies the intersection logic.

Apply the conditional probability formula

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

P(B \mid A) = \frac{\frac{15}{132}}{\frac{90}{132}} = \frac{15}{90} = \frac{1}{6}

Evaluating the ratio yields the exact conditional probability requested.

Key Concept

Conditional Probability of Dependent Events

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