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A box contains 5 red, 4 white and 6 blue balls. If three balls are drawn at random without replacement, determine the probability that none of the balls selected is red.

Answer :

Answer:

The probability that none of the balls selected is red ²⁴/₉₁

Step-by-step explanation:

Given;

number of red balls, R = 5

number of white balls, W = 4

number of blue balls, B = 6

Total number of balls, = 5 + 4 + 6 = 15

Probability of selecting 3 none red balls = P(WWW) or  P(BBB) or P(WWB) or  P(WBW) or P(BWW) or P(BBW) or P(BWB) or P(WBB)

[tex]P = (\frac{4}{15} \times \frac{3}{14} \times \frac{2}{13} ) + (\frac{6}{15} \times \frac{5}{14} \times \frac{4}{13} ) + (\frac{4}{15} \times \frac{3}{14} \times \frac{6}{13} ) + (\frac{4}{15} \times \frac{6}{14} \times \frac{3}{13} ) + \\\\(\frac{6}{15} \times \frac{4}{14} \times \frac{3}{13} ) + (\frac{6}{15} \times \frac{5}{14} \times \frac{4}{13} ) + (\frac{6}{15} \times \frac{4}{14} \times \frac{5}{13} ) + (\frac{4}{15} \times \frac{6}{14} \times \frac{5}{13} )[/tex]

[tex]P = \frac{24}{2730} + \frac{120}{2730} + \frac{72}{2730} +\frac{72}{2730} + \frac{72}{2730} + \frac{120}{2730} + \frac{120}{2730} + \frac{120}{2730} \\\\P = \frac{720}{2730} \\\\P = \frac{24}{91} \\\\[/tex]

Therefore, the probability that none of the balls selected is red ²⁴/₉₁

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