In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is(A) 10-1 (B) `(1/2)^5` (C) `(9/(10))^5` (D) `9/(10)`

In a box containing 100 bulbs, 10 are defective. The probability that

1.	In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is(A) 10-1 (B) `(1/2)^5` (C) `(9/(10))^5` (D) `9/(10)`
Answer» Probablity of getting first non-defective bulb `= 90/100` Probablity of getting second non-defective bulb `= 89/99` Probablity of getting third non-defective bulb `= 88/98` Probablity of getting fourth non-defective bulb `= 87/97` Probablity of getting fifth non-defective bulb `= 86/96` So, probablity of getting all non-defective bulbs in a sample of 5 bulbs = `90/10089/9988/9887/9786/96` Closest option to this is option `C`.

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In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is(A) 10-1 (B) `(1/2)^5` (C) `(9/(10))^5` (D) `9/(10)`

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