A container holds 6 red tokens and n blue tokens, where n is a positive integer. Two token… — GMAT

A container holds $6$ red tokens and $n$ blue tokens, where $n$ is a positive integer. Two tokens are drawn at random from the container one after another without replacement. If the probability that the first token drawn is red and the second token drawn is blue is equal to $\frac{9}{35}$ , what is the value of $n$ ?

Answer: 9

Answer

The value of

n

9

By applying the multiplication rule for dependent events without replacement, the probability is

\frac{6}{6+n} \times \frac{n}{5+n} = \frac{9}{35}

. Simplifying and solving the resulting quadratic equation

3n^2 - 37n + 90 = 0

yields

n = 9

as the only integer solution.

Step-by-Step Solution

Formulate the joint probability for dependent events.

The probability of drawing a red token first then a blue token second is

\frac{6}{6+n} \times \frac{n}{5+n}

Because sampling is done without replacement, the total number of remaining tokens decreases from

6+n

5+n

for the second draw.

Set up the algebraic equation with the given probability.

\frac{6n}{(6+n)(5+n)} = \frac{9}{35}

Equate the theoretical probability expression to the given value

\frac{9}{35}

Simplify and expand the equation.

Divide both sides of the numerator by

3

: \frac{2n}{n^2 + 11n + 30} = \frac{3}{35}. Cross-multiplying yields

70n = 3n^2 + 33n + 90

Simplifying ratios makes polynomial expansion easier.

Solve the quadratic equation for

n

Rearrange to

3n^2 - 37n + 90 = 0

, which factors into

(3n - 10)(n - 9) = 0

Find the roots of the quadratic equation.

Select the valid integer solution.

Since

n

must be an integer count of tokens,

n = 9

(rejecting

n = \frac{10}{3}

The number of tokens

n

must be a whole number.

Key Concept

Probability of Dependent Events (Sampling Without Replacement)

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