A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduati

Question

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

BrentGMATPrepNow · Accepted Answer

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

chetan2u · Answer

Hi,
 Bunuel , It might require to be written as "5 different birthday cards, 5 different holiday cards, and 5 different graduation cards" rather than "5 birthday cards, 5 holiday cards, and 5 graduation cards"..
 If the cards are same there will be ONLY 1 way to pick from each type and thus overall 3..
so numerator has to be 3, or 1 if 3 cancels out..

If I take all as different cards--
 ways to pick 3 cards of same type = 5C3 * 3= 30
ways to pick 3 cards out of 15 = 15C3 = 455
 
Prob = 30/455 = 6/91 
A

aekukula · Answer

Hi There,

if there are 5 Birthday Cards (B), 5 Holiday Cards (H) and 5 Graduation cards (G),

the probablity that all the purchased 3 cards is as follows :

It's all B : 5 x 4 x 3 = 60
or 
It's all H : 5 x 4 x 3 = 60
or 
It's all G : 5 x 4 x 3 = 60

all possible outcomes is = 15 x 14 x 13

so the probability is :

(60+60+60) / (15 x 14 x 13) = 6/91

TeamGMATIFY · Answer

This can be broken into following steps:

1. Select a category of card = 3C1 = 3
2. Select 3 cards from the selected category = 5C3 = 10

Total ways of selecting 3 cards of same category = 3*10 = 30
Total ways f selecting 3 cards from the whole lot = 15C3 = 455

P(selecting 3 cards of same category from the whole lot) = 30/455 = 6/91
Option A

lucalimoncelli · Answer

It doesn't matter which type of card you choose, so:

Probability of choosing 1st card: 1
Probability of choosing 2nd card of the same type: 4/14
Probability of choosing 3rd card of the same type: 3/13

Multiply and you get 6/91

Answer A.

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A card shop contains 5 birthday cards, 5 holiday cards, and 5 graduati

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