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Probability question

This topic has 9 expert replies and 4 member replies

Probability question

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Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33

the MGMAT explanation doesn't make sense to me:
"Third card: Since there have been no pairs so far, there are two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. There are 10 cards left in the deck.

Probability of turning over a third card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 8/10."

At this point, if 2 of the cards in the deck contain the same values as the two cards already dealt, then why does it say that the prob of a third card cannot form a pair?

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mikepamlyla wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33

We can solve this question using probability rules.

First, recognize that P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33
= C

Cheers,
Brent

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Hi mikepamlyla,

When it comes to probability questions, sometimes the easiest way to calculate the probability of what you "want" is to calculate the probability of what you "DON'T want" and subject that fraction from 1. (since, by definition, the probability of what you "want" + the probability of what you "don't want" always add up to 1).

Here, the question asks for "at least" one matching pair of cards when you draw 4 cards from this deck. It's far easier to calculate the probability of NO matching pairs than to calculate the probability of at least one matching pair.

1st card: Can be ANY of the 12 cards (and it honestly doesn't matter which one).

2nd card: Right now, there is only 1 card left that matches the 1st card and 10 that don't match. The probability of NOT matching the first card is 10/11.

3rd card: Now there are 2 cards left that match either of the first two cards you pulled and 8 that don't match. The probability of NOT matching either of the first two cards is 8/10.

4th card: Now there are 3 cards left that match any of the first three cards you pulled and 6 that don't match. The probability of NOT matching either of the first two cards is 6/9.

We have 1(10/11)(8/10)(6/9) = 480/990 = 48/99 = 16/33 This is the probability of getting NO matching pairs.

1 - 16/33 = 17/33 This is the probability of getting at least one matching pair.

Final Answer: C

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Quote:
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33
We can also solve the question using counting techniques.
First recognize that P(at least one pair) = 1-P(no pairs)

We'll find P(no pairs)

First, the number of possible outcomes.
We have 12 cards and we select 4.
This is accomplished in 12C4 ways = 495

Aside: If anyone is interested, we have a free video on calculating combinations (like 12C4) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

Now we count the number of ways to select 4 different cards with no pairs. In other words, we want 4 different card values.
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways.
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2=16 possibilities.

So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

So, the probability that there are no pairs = 240/495 = 16/33

So, P(at least one pair) = 1- 16/33 = 17/33 = C

Cheers,
Brent

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experts plz tell , whats wrong in it

12/12 * 1/11 * 10/10 * 8/10(one pair) + 12/12 * 1/11 * 10/10 * 1/9(two pair) = 9/11

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The 4th fraction should be not 8/10 but 8/9, as shown below:

vipulgoyal wrote:
experts plz tell , whats wrong in it

12/12 * 1/11 * 10/10 * 8/9(one pair) + 12/12 * 1/11 * 10/10 * 1/9(two pair) = 9/11
Case 1: exactly one matching pair
The expression in red accurately represents the probability that only the FIRST TWO CARDS MATCH.
Now we must account for ALL OF THE WAYS to get exactly one matching pair.
From the 4 cards, any COMBINATION OF 2 could form a matching pair.
The number of combinations of 2 that can be formed from 4 cards = 4C2 = (4*3)/(2*1) = 6.
Since there are 6 combinations of 2 that could form the matching pair, the expression in red must be multiplied by 6:
12/12 * 1/11 * 10/10 * 8/9 * 6 = 16/33.

Case 2: exactly 2 matching pairs
The expression in blue accurately represents the probability that the FIRST TWO CARDS MATCH and the LAST TWO CARDS MATCH.
Now we must account for ALL OF THE WAYS to get exactly 2 matching pairs:
1st and 2nd match, 3rd and 4th match
1st and 3rd match, 2nd and 4th match
1st and 4th match, 2nd and 3rd match.
Since there are 3 ways to get exactly 2 matching pairs, the expression in blue must be multiplied by 3:
12/12 * 1/11 * 10/10 * 1/9 * 3 = 1/33.

Since a favorable outcome is yielded by Case 1 OR Case 2, we ADD the probabilities::
16/33 + 1/33 = 17/33.

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Brent,

For this question, why is the probability of selecting the first card 12/12 instead of 1/12?

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clsrocks wrote:
Brent,

For this question, why is the probability of selecting the first card 12/12 instead of 1/12?
We want to find P(select any 1st card)
Notice that we aren't trying to select a certain card; we are selecting ANY card.
There are 12 cards to choose from and ALL 12 of them are acceptable selections.
So, P(select any 1st card) = 12/12 = 1

Cheers,
Brent

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clsrocks wrote:
Brent,

For this question, why is the probability of selecting the first card 12/12 instead of 1/12?
Another way to look at it:
We have a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card.

So, P(select the black card with a 2 on it) = 1/12 since there is black card with a 2 on it among the 12 cards. In other words, among the 12 cards, only 1 card meets the required outcome.

P(select ANY black card) = 6/12 since there are 6 black cards among the 12 cards. In other words, among the 12 cards, there are 6 cards that meet the required outcome.

P(select ANY card) = 12/12 because, among the 12 cards, ALL of them meet the required outcome.

Does that help?

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Yes that helps. Thank you so much!

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For another probability question with similar logic, see here: http://www.beatthegmat.com/probability-if-4-fair-dice-are-thrown-simultaneously-t108134.html

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Good morning everyone,

I'd like to ask about my solution. It's surely wrong but I'd like to know what is wrong with it.

I tend to calculate the P of having one pair and 2 pairs, which is equal to P=(6C1x10C2+6C2)/12C4.

The result is 19/33...

I am very confused! Please help me resolve this!

Thank you in advance!!!!

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Hi yenngoc2409,

I'm going to give you a hint so that you can adjust your calculation. Your result is too high because you have not accounted for the 'duplicate' entries. For example, if the 4 cards were 2233, then you have an option that has two pair... but you also have two of the options that have one pair... (11xy and 22xy)... it just so happens that the two additional cards also form a pair.

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mikepamlyla wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

A. 8/33
B. 62/165
C. 17/33
D. 103/165
E. 25/33
Since there 12 cards, the number of ways to choose 4 cards is 12C4 = 12!/(4! x 8!) = (12 x 11 x 10 x 9)/(4 x 3 x 2) = 11 x 5 x 9 = 495. Now we need to determine, out of these 495 sets of 4 cards, how many have at least one pair of cards that have the same value. We can divide this into 2 cases:

Case 1: Two pairs of the same value

For example, we can have {1, 2, 1, 2}. Let the bold 1 and 2 denote the black cards and the italic 1 and 2 the red cards. To determine the number of such sets, let’s consider the numbers from the red cards are dependent on the black cards (since they have to match the ones on the black cards), then the number of such sets becomes the number of ways of choosing 2 black cards from 6 black cards, which is 6C2 = (6 x 5)/2 = 15.

Case 2: Exactly one pair of the same value

For example, we can have {1, 1, 2, 3}. Again, let’s assume red cards are dependent on black cards. So we have 6C1 x 1 x 5C2 = 6 x 10 = 60 ways of choosing 3 black cards and 1 red card with the red card matching one of black cards. However, the set {1, 1, 2, 3} can also be {1, 1, 2, 3} or {1, 1, 2, 3}.

For sets such as {1, 1, 2, 3}, we have 6C1 x 1 x 5C1 x 4C1 = 6 x 5 x 4 = 120 ways of choosing 2 black cards and 2 red cards with one of the two red cards matching one of the two black cards.

For sets such {1, 1, 2, 3}, we have 6C1 x 1 x 5C2 = 60 ways of choosing 1 black cards and 3 red cards with one of the two red cards matching the black card.

Therefore, there are a total of 60 + 120 + 60 = 240 sets in this case.

Therefore, the probability that Bill finds at least one pair of cards that have the same value is (15 + 240)/495 = 255/495 = 17/33.

Answer: C
.

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