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Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, nu

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Joined: 02 Sep 2009
Posts: 53069
Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, nu  [#permalink]

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28 Jan 2018, 22:35
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Difficulty:

25% (medium)

Question Stats:

76% (01:34) correct 24% (01:41) wrong based on 83 sessions

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Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, numbered 5 - 9. What is the probability that the sum of a card pulled randomly from Bowl X and a card pulled randomly from Bowl Y will equal 8?

A. 1/20
B. 1/10
C. 3/20
D. 2/16
E. 3/16

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Joined: 20 Apr 2017
Posts: 9
Re: Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, nu  [#permalink]

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28 Jan 2018, 22:51

Total number of possibilities is 20 out of which 3 will yield sum of 8

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Joined: 11 Sep 2015
Posts: 3446
Re: Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, nu  [#permalink]

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13 Feb 2018, 09:55
Top Contributor
Bunuel wrote:
Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, numbered 5 - 9. What is the probability that the sum of a card pulled randomly from Bowl X and a card pulled randomly from Bowl Y will equal 8?

A. 1/20
B. 1/10
C. 3/20
D. 2/16
E. 3/16

NOTE: In order to get a sum of 8, the card selected from Bowl X must be 1, 2 or 3.
Here's why: Cards 1, 2 and 3 all have corresponding cards in Bowl Y that add to get 8 (i.e., 1+7 = 8, 2+6 = 8 and 3+5 = 8)
However, card 4 (from Bowl X) does NOT have a corresponding card in Bowl Y that add to get 8.

So P(sum is 8) = P(card selected from Bowl X is 1, 2 or 3 AND card selected from Bowl Y is the one card that works with the first card to get a sum of 8)
= P(card selected from Bowl X is 1, 2 or 3) x P(card selected from Bowl Y is the one card that works with the first card to get a sum of 8)
= 3/4 x 1/5
= 3/20

Cheers,
Brent
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Re: Bowl X has 4 cards in it, numbered 1 - 4. Bowl Y has 5 cards in it, nu   [#permalink] 13 Feb 2018, 09:55
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