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Of the 100 bulbs in a box, 5 are defective. If two bulbs are selected
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14 Feb 2022, 10:10
14 Feb 2022, 10:10
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Of the 100 bulbs in a box, 5 are defective. If two bulbs are selected at random, what is the probability that both of the bulbs are fine?
(A) \(\frac{94+95}{100*99}\)
(B) \(\frac{94*95}{100*99}\)
(C) \(1-\frac{4*5}{100*99}\)
(D) \(\frac{95}{100}*\frac{94}{100}\)
(E) \(\frac{95}{100}+\frac{94}{100}\)
(A) \(\frac{94+95}{100*99}\)
(B) \(\frac{94*95}{100*99}\)
(C) \(1-\frac{4*5}{100*99}\)
(D) \(\frac{95}{100}*\frac{94}{100}\)
(E) \(\frac{95}{100}+\frac{94}{100}\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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Re: Of the 100 bulbs in a box, 5 are defective. If two bulbs are selected
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14 Feb 2022, 10:30
14 Feb 2022, 10:30
Carcass wrote:
Of the 100 bulbs in a box, 5 are defective. If two bulbs are selected at random, what is the probability that both of the bulbs are fine?
(A) \(\frac{94+95}{100*99}\)
(B) \(\frac{94*95}{100*99}\)
(C) \(1-\frac{4*5}{100*99}\)
(D) \(\frac{95}{100}*\frac{94}{100}\)
(E) \(\frac{95}{100}+\frac{94}{100}\)
(A) \(\frac{94+95}{100*99}\)
(B) \(\frac{94*95}{100*99}\)
(C) \(1-\frac{4*5}{100*99}\)
(D) \(\frac{95}{100}*\frac{94}{100}\)
(E) \(\frac{95}{100}+\frac{94}{100}\)
P(both bulbs work) = P(1st bulb works AND 2nd bulb works)
= P(1st bulb works) x P(2nd bulb works)
= 95/100 x 94/99
= (95)(94)/(100)(99)
Answer: B
gmatclubot
Re: Of the 100 bulbs in a box, 5 are defective. If two bulbs are selected [#permalink]
14 Feb 2022, 10:30
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