Question #234797

A box contains 24 light bulbs, of which four are defective. If a person selects four bulbs from the box at
random, without replacement, what is the probability that all four bulbs will be defective?

Expert's answer

Number of ways to select r items from n, Crn=n!(r!×(nr)!)C^n_r = \frac{n!}{(r! \times (n-r)!)}

Number of bulbs = 24

Number of defective bulbs = 4

Number of working bulbs = 24 - 4 = 20

P(all four bulbs will be defective) = Number of ways to select 4 defective bulbs / Number of ways to select any 4 bulbs

=C44C424=110626= \frac{C^4_4 }{ C^{24}_4} \\ = \frac{1}{10626}


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