In a box of 10 electric bulbs,two are defecti | vedclass

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(B) Total number of bulbs = $10$. Number of defective bulbs = $2$. Number of non-defective bulbs = $10 - 2 = 8$. Since the first bulb is replaced befo

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$\frac{16}{25}$

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(B) Total number of bulbs = $10$. Number of defective bulbs = $2$. Number of non-defective bulbs = $10 - 2 = 8$. Since the first bulb is replaced before the second selection,the events are independent. The probability of selecting a non-defective bulb in one draw is $P = \frac{8}{10} = \frac{4}{5}$. Since the selection is done with replacement,the probability of selecting a non-defective bulb in the second draw remains $P = \frac{4}{5}$. The probability that both bulbs are without defect is $P(	ext{both non-defective}) = P(	ext{first is non-defective}) 	imes P(	ext{second is non-defective}) = \frac{4}{5} 	imes \frac{4}{5} = \frac{16}{25}$.

In a box of $10$ electric bulbs,two are defective. Two bulbs are selected at random one after the other from the box. The first bulb after selection is put back in the box before making the second selection. The probability that both the bulbs are without defect is

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