A card shop contains 5 birthday cards, 5 holiday cards...

[BTGmoderatorLU]

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

The OA is A.

I'm confused by this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

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Hi LUANDATO,

We're told that a card shop contains 5 birthday cards, 5 holiday cards, and 5 graduation cards and that three cards are purchased at random from the shop. We're asked for the probability that the three cards will be of the SAME type.

To start, the first card can be ANY of the 15 cards - it's the second and third cards that will have to match the first.

Probability that the second card matches the first = (4/14)
Probability that the third card matches the first = (3/13)

(4/14)(3/13) = 12/182 = 6/91

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[Jeff@TargetTestPrep]

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

The number of ways to select 3 birthday cards is 5C3 = (5 x 4 x 3)/(3 x 2) = 10, which is also the number of ways to select 3 holiday cards and the number of ways to select 3 graduation cards.

The total number ways to select 3 cards from 15 is:

15C3 = (15 x 14 x 13)/(3 x 2) = 5 x 7 x 13 = 455

So the probability of selecting 3 birthday cards or 3 holiday cards or 3 graduation cards is:

10/455 + 10/455 + 10/455 = 30/455 = 6/91

Answer: A

[BTGmoderatorRO]

There are 5 Birthday cards (B), 5 Holiday cards (H) and 5 graduation cards (G).
Total number of cards=15
prob. of purchasing three cards of the same type can be pr(BBB), pr(HHH) or pr(GGG).
Therefore,
$$pr\left(BBB\right)+pr\left(HHH\right)+pr\left(GGG\right)$$
$$pr\left(\frac{5}{15}\cdot\frac{4}{14}\cdot\frac{3}{13}\right)+pr\left(\frac{5}{15}\cdot\frac{4}{14}\cdot\frac{3}{13}\right)+pr\left(\frac{5}{15}\cdot\frac{4}{14}\cdot\frac{3}{13}\right)$$
$$pr\left(\frac{1}{3}\cdot\frac{2}{7}\cdot\frac{3}{13}\right)+pr\left(\frac{1}{3}\cdot\frac{2}{7}\cdot\frac{3}{13}\right)+pr\left(\frac{1}{3}\cdot\frac{2}{7}\cdot\frac{3}{13}\right)$$
$$\frac{6}{273}+\frac{6}{273}+\frac{6}{273}=\frac{18}{273}=\frac{6}{91}$$
Hence, option A is the correct answer

[Brent@GMATPrepNow]

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

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