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Question: A box contains 6 red balls, 4 white balls and 5 blue balls. Three balls are drawn successively from ...

A box contains 6 red balls, 4 white balls and 5 blue balls. Three balls are drawn successively from the box. Find the probability that they are drawn in the order red, white and blue if each ball is not replaced.

Answer

\frac{4}{91}

Explanation

Solution

Total number of balls in the box = 6 (red) + 4 (white) + 5 (blue) = 15 balls.

We want to find the probability of drawing the balls in the order red, white, and blue without replacement.

Probability of drawing a red ball first: P(Red first)=Number of red ballsTotal number of balls=615P(\text{Red first}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{6}{15}

After drawing one red ball, there are 14 balls remaining in the box (5 red, 4 white, 5 blue).

Probability of drawing a white ball second, given the first was red and not replaced: P(White second | Red first)=Number of white ballsTotal number of remaining balls=414P(\text{White second | Red first}) = \frac{\text{Number of white balls}}{\text{Total number of remaining balls}} = \frac{4}{14}

After drawing one red and one white ball, there are 13 balls remaining in the box (5 red, 3 white, 5 blue).

Probability of drawing a blue ball third, given the first was red and the second was white and neither was replaced: P(Blue third | Red first and White second)=Number of blue ballsTotal number of remaining balls=513P(\text{Blue third | Red first and White second}) = \frac{\text{Number of blue balls}}{\text{Total number of remaining balls}} = \frac{5}{13}

The probability of drawing the balls in the order red, white, and blue is the product of these probabilities: P(Red then White then Blue)=P(Red first)×P(White second | Red first)×P(Blue third | Red first and White second)P(\text{Red then White then Blue}) = P(\text{Red first}) \times P(\text{White second | Red first}) \times P(\text{Blue third | Red first and White second}) P(Red then White then Blue)=615×414×513P(\text{Red then White then Blue}) = \frac{6}{15} \times \frac{4}{14} \times \frac{5}{13}

Simplify the fractions: 615=2×35×3=25\frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5} 414=2×27×2=27\frac{4}{14} = \frac{2 \times 2}{7 \times 2} = \frac{2}{7}

Substitute the simplified fractions back into the probability calculation: P=25×27×513P = \frac{2}{5} \times \frac{2}{7} \times \frac{5}{13}

Multiply the numerators and the denominators: P=2×2×55×7×13P = \frac{2 \times 2 \times 5}{5 \times 7 \times 13}

Cancel out the common factor of 5 in the numerator and denominator: P=2×27×13P = \frac{2 \times 2}{7 \times 13} P=491P = \frac{4}{91}

The probability of drawing the balls in the order red, white, and blue is 491\frac{4}{91}.