Question

A bag contains 6 red, 5 blue and 4 yellow marbles. Two marbles are drawn, but the first marble drawn is not replaced. Find P(red, then blue).

• A. $\frac{6}{15}$
• B. $\frac{5}{14}$
• C. $\frac{1}{7}$

Answer 1

The correct option is C $\frac{1}{7}$

There are 6 red marbles.

There are a total of 15 marbles.

P(red) $=\frac{6}{15}$

The result of the first draw affected the probability of the second draw.

There are 5 blue marbles.

There are a total of 14 marbles left.

P(blue after red) $=\frac{5}{14}$

P(red, then blue) = P(red) $\times$ P(blue after red) $=\frac{6}{15}\times \frac{5}{14}=\frac{1}{7}$

The probability of drawing a red marble and then a blue marble is $\frac{1}{7}$.