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probabaility......where am I wrong?

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probabaility......where am I wrong?

by cubicle_bound_misfit » Fri Apr 17, 2009 4:12 am
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?

8/33

62/165

17/33

103/165

25/33


My approach was (1 - no 4 card selected are pair) for that I used event of 2 cards selected from one deck and event of 2 cards selected from another deck which are not pair two previously selected cards/total event space
--------->
6c2 * 4C2/12C4.............(A)

and answer = 1- (A)

What's wrong with my approach?
Please help
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by AlaaQasem » Fri Apr 17, 2009 5:32 am
Is the answer C, i.e. 17/33?

Here is how I approached it.

Since the question is asking for the probability of at least one pair of cards we know that we want to find the probability of finding no matching cards and subtract it from 1.

To find the probability of no matching cards:
The probablity of selecting card 1 = 1 (any card can be withdrawn)
The probability of selecting card 2 which does not matche card 1 = 10/11 (That's because from the remaining 11 cards one will surely match card 1)
The probability of selecting card 3 which does not match cards 1 or 2 = 8/10 (That's because from the remaining 10 cards there are two cards, one will surely match card 1 and the other will surely match card 2)
The probability of selecting card 4 which does not match cards 1, 2 or 3 = 6/9 (That's because from the remaining 9 cards there are three cards, one will surely match card 1, the other will surely match card 2 and the last will surely match card 3)

Now multiply all these probabilities since this is an AND event:
1*(10/11)*(8/10)*(6/9) = 16/33

As I mentioned above to find the probability of selecting at least one matching pair you will need to subtract the result above from 1:
1-16/33 = 17/33
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by AlaaQasem » Fri Apr 17, 2009 6:11 am
Back to your approach. You were in the right path but you didn't consider all the possilities of how the four cards can be selected. You calculated the probability of Bill selecting 2 cards from the first deck and then different ones from the second deck. However the question states that Bill shuffled both decks of cards so the 4 cards he selects can come from one of the following possibilities:
1. All four cards from deck 1
2. Three cards from deck 1 and 1 from deck 2
3. Two cards from deck 1 and 2 from deck 2
4. One card from deck 1 and 3 from deck 2
5. All four cards from deck 2

Now let's calculate the combinations of cards in these possibilities taking into consideration that we don't want to have any matching pairs in these combination.

1. All four cards from deck 1
6C4 = 15
No two cards will be matching since they all come from the same deck.

2. Three cards from deck 1 and 1 from deck 2
6C3*3C1 = 20*3 = 60
Because we selected 3 cards from deck 1 we have to be careful to exclude the three matching cards in deck 2 when we calculate the combinations for selecting the last card from deck 2.

3. Two cards from deck 1 and 2 from deck 2
6C2*4C2 = 15*6 = 90
Because we selected 2 cards from deck 1 we have to be careful to exclude the two matching cards in deck 2 when we calculate the combinations for selecting the other two cards from deck 2.

4. One card from deck 1 and 3 from deck 2
6C1*5C3 = 6*10 = 60
Because we selected 1 cards from deck 1 we have to be careful to exclude the matching card in deck 2 when we calculate the combinations for selecting the other three cards from deck 2.

5. All four cards from deck 2
6C4 = 15
No two cards will be matching since they all come from the same deck.

To calculate the probability of not selecting any matching pairs, you will need to add all the calculations above and divide them over all the possibilities as follows:

(6C4 + (6C3*3C1) + (6C2*4C2) + (6C1*5C3) + 6C4) / 12C4

= 15+60+90+60+15/495 = 240/495 = 16/33 -----> (A)

Finally to find the probability of selecting at least a pair of matching cards you need to subtract the results of A from 1.

1-16/33 = 17/33
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by cubicle_bound_misfit » Fri Apr 17, 2009 7:00 am
AlaaQasem

Thanks a bunch. I am acquiring cubicle bound dumbness as it shows from my missing all the other possibilities
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by schumi_gmat » Fri Apr 17, 2009 5:30 pm
This is how I solved.

no of ways 4 cards are drawn from 12 cards is 12x11x10x9

Assume that no cards drawn is a pair

first card drawn is 12 ways
second card drawn in 10 ways............because the pair card of suit that is already select cannot be consider.

For eg. u select 5 from suit 1 then u can select 5 from other suit. so u have only 10 cards left to choose from for the 2nd card.

3rcard is selected in 8 ways

4th in 6

total no of ways no card is drawn = 12x10x8x6

prob = 16/33

prop of atleast 1 = 1 - 16/33 = 17/33
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by sanju09 » Sat Apr 18, 2009 2:26 am
AlaaQasem!! Excellent!!!
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by ketkoag » Mon Apr 20, 2009 10:39 am
nice explanation!!
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