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A box contains 24 light bulbs, of which two are defective. If a person selects 10 bulbs at random, without replacement, what is the probability that both defective bulbs will be selected?

Short Answer

Expert verified

The probability that both defective bulbs will be selected is 0.16

Step by step solution

01

Given information

Total light bulbs:24

No. of total defective bulbs: 2

No. of non-defective bulbs:22

02

To compute the probability

For selecting r items from n items, we have the formula\(^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)

The probability of selecting 10 bulbs at random with two defective bulbs

\(\begin{aligned}{l} &= \frac{{^2{C_2}{ \times ^{22}}{C_8}}}{{^{24}{C_{10}}}}\\ &= \frac{{10 \times 9}}{{24 \times 23}}\end{aligned}\)

\(\begin{aligned}{l} = \frac{{90}}{{552}}\\ = 0.16\end{aligned}\)

Thus, the required probability is 0.16.

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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?

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