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Question
From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.
Solution
Let X denote the number of defective bulbs in a sample of 2 bulbs drawn from a lot of 10 bulbs containing 3 defectives and 7 non-defectives.Then X can take values 0, 1, 2.
Now,
\[P\left( X = 0 \right) = P\left( \text{ no defective bulb }\right) = \frac{{}^7 C_2}{{}^{10} C_2} = \frac{7}{15}\]
\[P\left( X = 1 \right) = P\left( 1 \text{ defective bulb } \right) = \frac{{}^3 C_1 \times^7 C_1}{{}^{10} C_2} = \frac{7}{15}\]
\[P\left( X = 2 \right) = P\left( 2 \text{ defective bulbs } \right) = \frac{{}^3 C_2}{{}^{10} C_2} = \frac{1}{15}\]
Thus, the probability distribution of X is given below,
| X | P(X) |
| 0 |
\[\frac{7}{15}\]
|
| 1 |
\[\frac{7}{15}\]
|
| 2 |
\[\frac{1}{15}\]
|
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