3 bulbs are drawn at random from a lot of 50 | vedclass

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(D) Total number of bulbs = $50$. Number of defective bulbs = $5$. Number of non-defective bulbs = $50 - 5 = 45$. We need to draw $3$ bulbs without re

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$\frac{1419}{1960}$

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(D) Total number of bulbs = $50$. Number of defective bulbs = $5$. Number of non-defective bulbs = $50 - 5 = 45$. We need to draw $3$ bulbs without replacement. The probability that the first bulb is non-defective is $P(E_1) = \frac{45}{50}$. After drawing one non-defective bulb,there are $44$ non-defective bulbs left out of $49$ total bulbs. The probability that the second bulb is non-defective is $P(E_2|E_1) = \frac{44}{49}$. After drawing two non-defective bulbs,there are $43$ non-defective bulbs left out of $48$ total bulbs. The probability that the third bulb is non-defective is $P(E_3|E_1 \cap E_2) = \frac{43}{48}$. The probability that all $3$ bulbs are non-defective is $P = \frac{45}{50} 	imes \frac{44}{49} 	imes \frac{43}{48}$. $P = \frac{9}{10} 	imes \frac{44}{49} 	imes \frac{43}{48} = \frac{9 	imes 11 	imes 43}{10 	imes 49 	imes 12} = \frac{3 	imes 11 	imes 43}{10 	imes 49 	imes 4} = \frac{1419}{1960}$.

$3$ bulbs are drawn at random from a lot of $50$ bulbs,$5$ of which are defective. Find the probability of the event that all $3$ bulbs are non-defective (drawn without replacement).

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