A container has 50 electronic components, of which 10 are defective.

If 8 components are drawn at random from the container, what is the probability that exactly 3 of them are defective?

0.1468 = 0.147 ;

Answered by eurydike | 2024-06-12

(10/50) * (3/8) = .075 .075 * 100 = 7.5% chance of this happening

Answered by Coding8with8Valor | 2024-06-24

The probability of drawing exactly 3 defective out of 8 components from a total of 50 with 10 defective ones is calculated using the hypergeometric distribution formula. After calculating the necessary combinations, the final probability is approximately 0.147. This indicates there is about a 14.7% chance of this occurring.
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Answered by eurydike | 2024-11-04