A box contains 6 red balls and 4 blue balls. A ball is drawn at random, its color is recor… — JAMB UTME

A box contains $6$ red balls and $4$ blue balls. A ball is drawn at random, its color is recorded, and it is then returned to the box. A second ball is subsequently drawn at random. What is the probability of selecting a red ball on the first draw and a blue ball on the second draw?

$\frac{6}{25}$ Answer
B
$\frac{4}{15}$
C
$\frac{1}{10}$
D
$1$

Answer

\frac{6}{25}

Because the first ball is replaced after being recorded, the two draws are independent compound events. The probability of drawing a red ball first is

\frac{6}{10} = \frac{3}{5}

, and the probability of drawing a blue ball second is

\frac{4}{10} = \frac{2}{5}

. Multiplying these probabilities yields

P(\text{Red and Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

Step-by-Step Solution

Calculate the probability of drawing a red ball on the first draw

P(\text{Red}) = \frac{6}{6 + 4} = \frac{6}{10} = \frac{3}{5}

There are

6

red balls out of a total of

10

balls.

Determine independence and calculate the probability of drawing a blue ball on the second draw

P(\text{Blue}) = \frac{4}{10} = \frac{2}{5}

Because the first ball is replaced before the second draw, the total number of balls and their composition remain unchanged.

Apply the multiplication law for independent compound events

P(\text{Red and then Blue}) = P(\text{Red}) \times P(\text{Blue}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}

For independent events

A

and

B

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

Key Concept

Multiplication Law of Probability for Independent Events

Estimated Time:45s

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