A glass bowl contains 6 red candies and N green candies, where N is a positive integer. If… — GMAT

A glass bowl contains 6 red candies and $N$ green candies, where $N$ is a positive integer. If two candies are selected at random one after another without replacement, the probability that both candies selected are red is $\frac{1}{3}$ . What is the value of $N$ ?

Answer: 4

Answer

The correct value of

N

is 4. Because the selection is performed without replacement, the probability of drawing the first red candy is

\frac{6}{N+6}

and the second red candy is

\frac{5}{N+5}

. Setting their product equal to

\frac{1}{3}

yields

\frac{30}{(N+6)(N+5)} = \frac{1}{3}

, which expands to

N^2 + 11N - 60 = 0

. Factoring gives

(N-4)(N+15) = 0

, yielding

N = 4

as the only positive solution.

Step-by-Step Solution

Define total outcomes and event probabilities for dependent selection

Initial total candies =

N + 6

. Probability of first red =

\frac{6}{N+6}

. Probability of second red given first is red =

\frac{5}{N+5}

Since selection is without replacement, the total count and red count both decrease by 1 after the first draw, making the events dependent.

Set up the probability multiplication equation

P(\text{both red}) = \frac{6}{N+6} \times \frac{5}{N+5} = \frac{30}{(N+6)(N+5)} = \frac{1}{3}

The probability of joint dependent events

A

and

B

occurring is

P(A \cap B) = P(A) \times P(B|A)

Solve the equation for

N

(N+6)(N+5) = 90 \Rightarrow N^2 + 11N - 60 = 0 \Rightarrow (N-4)(N+15) = 0

Cross-multiplying gives a quadratic equation in terms of

N

Select the valid positive integer root

N = 4

The count of candies must be a positive number, so

N = -15

is discarded.

Key Concept

Probability of Dependent Events (Sampling without Replacement)

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