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A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduati
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18 Oct 2021, 05:30
18 Oct 2021, 05:30
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90% (01:24) correct
9% (02:13) wrong based on 11 sessions
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A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduation cards. If three cards are purchased at random from the shop, what is the probability that the three cards will be of the same type?
A. 6/91
B. 5/93
C. 4/95
D. 3/97
E. 2/99
A. 6/91
B. 5/93
C. 4/95
D. 3/97
E. 2/99
Re: A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduati
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18 Oct 2021, 06:01
18 Oct 2021, 06:01
The first card can be any card, so the probability to draw the first one is 1.
The second card must be the same type of the first one, the number of the same typed-card is 4 and total number of cards is 14, so the probability is 4/14.
The third card must be the same type of the first and second one, the number of the same typed-card is 3 and total number of cards is 13, so the probability is 3/13.
=> The probability to draw all of them in the same type is 1 x 4/14 x 3/13 = 6/91
The second card must be the same type of the first one, the number of the same typed-card is 4 and total number of cards is 14, so the probability is 4/14.
The third card must be the same type of the first and second one, the number of the same typed-card is 3 and total number of cards is 13, so the probability is 3/13.
=> The probability to draw all of them in the same type is 1 x 4/14 x 3/13 = 6/91
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Re: A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduati
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18 Oct 2021, 06:48
18 Oct 2021, 06:48
1
Carcass wrote:
A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduation cards. If three cards are purchased at random from the shop, what is the probability that the three cards will be of the same type?
A. 6/91
B. 5/93
C. 4/95
D. 3/97
E. 2/99
A. 6/91
B. 5/93
C. 4/95
D. 3/97
E. 2/99
Let's first rewrite our probability and then apply probability rules.
P(All 3 cards the same type) = P(1st card is ANY type AND 2nd card matches type of 1st card AND 3rd card matches type of 1st card)
= P(1st card is ANY type) x P(2nd card matches type of 1st card) x P(3rd card matches type of 1st card)
= 1 x 4/14 x 3/13
= 6/91
Answer: A
ASIDE
P(1st card is ANY type) =1 because the first selection can be any type
P(2nd card matches type of 1st card) = 4/14, because once the 1st card is selected, there are 14 cards remaining, and there are 4 cards left that are the same type as the first card
P(3rd card matches type of 1st card) = 3/13, because once cards 1 and 2 have been selected, there are 13 cards remaining, and only 3 of them are the same type as the first card
Cheers,
Brent
