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A deck of cards contains 2 blue cards and 5 red cards. A

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A deck of cards contains 2 blue cards and 5 red cards. A [#permalink]

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New post 19 Sep 2005, 13:10
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A deck of cards contains 2 blue cards and 5 red cards. A card is randomly selected from the deck. Whatever color card is selected, a card of the other color is inserted into the deck. For instance, a blue card is drawn and a red card is reinserted in the deck. If another card is now selected, what is the probability that the card is red?

A. 6/7
B. 11/15
C. 32/49
D. 4/7
E. 12/49
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New post 19 Sep 2005, 13:15
The wording of that question is very confusing. Where is it from?

I see two ways of interpreting it:

1) A blue card is pulled - what's the probability that the next card selected is red?

The answer to this is simply 6/7, since after pulling a blue card and adding a red card, the deck would contain 1 blue card and 6 red cards. So the answer is 6/7 or A.

2) A random card is selected first - what's the probability of the second card selected being red?

This is a bit more complex as we have two cases to consider: first card blue, and first card red.

If the first card pulled is blue, there's a 6/7 probability that the second card pulled will be red.

If the first card pulled is red, there's a 4/7 probability that the second card pulled will be red.

There's a 2/7 probability of the first card pulled being blue, and a 5/7 probability of the first card pulled being red.

So the probability should be (2/7)(6/7) + (5/7)(4/7) = 32/49 or C.
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Re: PS: Probability [#permalink]

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New post 19 Sep 2005, 13:20
popew626 wrote:
A deck of cards contains 2 blue cards and 5 red cards. A card is randomly selected from the deck. Whatever color card is selected, a card of the other color is inserted into the deck. For instance, a blue card is drawn and a red card is reinserted in the deck. If another card is now selected, what is the probability that the card is red?

A. 6/7
B. 11/15
C. 32/49
D. 4/7
E. 12/49


Probabiliy of selecting Red = 5/7 and Blue = 2/7

5/7 probability that the next set will be 3 Blue and 4 Red (in which case probabilityof red is 4/7

2/7 probability taht the next set will be 1 Blue and 6 Red (in which case the probability of red is 6/7

so the overall probability of red is (5/7*4/7)+(2/7*6/7) = 32/49
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New post 19 Sep 2005, 13:22
I get C) 32/49
Required Probabilty = (Probability First Card Red)(Probability Second Card Red) + (Probability First Card Blue)(Probability Second Card Red)

Probability First Card Red = 5/7
Probability Second Card Red = 4/7

Probability First Card Blue = 2/7
Probability Second Card Red = 6/7

So P = (5/7)(4/7) + (2/7)(6/7) = (20+12)/49 = 32/49
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New post 19 Sep 2005, 15:02
I interpreted the question as saying what is the probability of getting B,R or R,R. So,

P(B,R or R,R)= 2/7*6/7 + 5/7*4/7 = 32/49 or C
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  [#permalink] 19 Sep 2005, 15:02
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