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# Simultaneous Probability

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Intern
Joined: 06 Apr 2012
Posts: 34

Kudos [?]: 32 [0], given: 48

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14 Apr 2012, 09:31
Hi - was wondering if anyone could provide guidance to answering the questions below:

1) what is the probability of picking two cards off the top of a deck of cards and getting a pair?

2) what is the probability of getting at least one pair if four fair six sided dice are rolled simultaneously?

Thanks,

Kudos [?]: 32 [0], given: 48

Intern
Joined: 06 Apr 2012
Posts: 34

Kudos [?]: 32 [0], given: 48

Re: Simultaneous Probability [#permalink]

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14 Apr 2012, 11:30
randomguy2012 wrote:
1. Define terms.

2. 1/36

1/6*1/6

52 cards with 13 sets of 4 matching cards. If possible please guide me through the solutions (don't just provide an answer), because I am not clear on the thinking behind the answers here, and perhaps a better way to understand how to solve this type of problems.

Question 2 answer is off... Again whether right or wrong is not as important as a step-by-step vision.

Thanks much,

Kudos [?]: 32 [0], given: 48

Intern
Joined: 27 Oct 2013
Posts: 39

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Schools: Erasmus '17
Re: Simultaneous Probability [#permalink]

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04 Jul 2016, 05:47
Ans 2: Total possible combination 6^4

combination w/o any pair: 6*5*4*3.

probability of no pair. = 6*5*4*3/6^4

probality of atleast 1 pair: 1- 6*5*4*3/6^4 = 26/36 = 13/18

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Re: Simultaneous Probability [#permalink]

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04 Jul 2016, 11:58
Hi All,

While this is an old set of posts, the questions themselves are not too difficult to deal with (although you will need to know how probability 'math' works).

1) With a standard deck of cards, there are 4 suits with 13 cards in each suit. The probability of selecting a pair of cards (with the same face value) off the top of a deck of cards can be calculated like this....

The first card could be ANY of the 52 cards, so it doesn't really matter what it is; it's the second card that ultimately matters (it will either match the first card or it won't). Once the first card is flipped over, there are only 3 remaining cards in the deck that match the first card (and there are 51 cards remaining).

(1)(3/51) = 3/51 = 1/17 = the probability that the first two cards are a matching pair.

2) In probability, there are only two possible outcomes - what you "want" and what you "don't want"; the sum of those two outcomes always totals 1. When a probability question uses the phrase "at least", it's often easiest to calculate the probability of NOT getting what you "want" and subtracting that fraction from the number 1. In that way, you'll have the probability of what you DO want.

Here, the wording of the question is a bit vague, but I'm going to interpret it to mean "... the probability of getting at least one duplicate value among the 4 dice rolls..." With this question, I'm going to calculate the probability of not having any duplicates...

The first roll can be any of the 6 possible numbers.

First die: 1
Second die (can't match the first die): 5/6
Third die (can't match either of the first two dice): 4/6
Fourth die (can't match any of the first three dice): 3/6

(1)(5/6)(4/6)(3/6) =
(1)(5/6)(2/3)(1/2) =
10/36 =
5/18 = the probability of NO matching values

1 - 5/18 = 13/18 = the probability of at least one duplicate value

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Re: Simultaneous Probability   [#permalink] 04 Jul 2016, 11:58
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# Simultaneous Probability

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