I think I solved it after thinking about it for awhile:

You can either get a pair with the 2nd, 3 or 4th card.

P(2) = 1/1 * 1/11
P(3) = 1/1 * 10/11 * 2/10
P(4) = 10/11 * 8/10 * 3/9


P(2) + P(3) + P(4) = 17/33


Let me know if there is a quicker strategy.

canuckclint wrote:I think I solved it after thinking about it for awhile:

You can either get a pair with the 2nd, 3 or 4th card.

P(2) = 1/1 * 1/11
P(3) = 1/1 * 10/11 * 2/10
P(4) = 10/11 * 8/10 * 3/9


P(2) + P(3) + P(4) = 17/33


Let me know if there is a quicker strategy.

the Quickest way would be

Atleast 1 = 1- None

1- p(0)
1- 10/11*8/10*6/9
= 17/33

hope that helps...

sudhir3127 wrote:

canuckclint wrote:I think I solved it after thinking about it for awhile:

You can either get a pair with the 2nd, 3 or 4th card.

P(2) = 1/1 * 1/11
P(3) = 1/1 * 10/11 * 2/10
P(4) = 10/11 * 8/10 * 3/9


P(2) + P(3) + P(4) = 17/33


Let me know if there is a quicker strategy.

the Quickest way would be

Atleast 1 = 1- None

1- p(0)
1- 10/11*8/10*6/9
= 17/33

hope that helps...

Beautiful minds....RESPECT! :!:

This question confused me because I found 17/33 as AT LEAST one PAIR probability , then I was trying to find out the probability of having TWO PAIRS and add them up...

canuckclint wrote:I think I solved it after thinking about it for awhile:

You can either get a pair with the 2nd, 3 or 4th card.

P(2) = 1/1 * 1/11
P(3) = 1/1 * 10/11 * 2/10
P(4) = 10/11 * 8/10 * 3/9


P(2) + P(3) + P(4) = 17/33


Let me know if there is a quicker strategy.

Great thinking canuckclint!

SO what would you say about having the second pair ?

logitech wrote:

canuckclint wrote:I think I solved it after thinking about it for awhile:

You can either get a pair with the 2nd, 3 or 4th card.

P(2) = 1/1 * 1/11
P(3) = 1/1 * 10/11 * 2/10
P(4) = 10/11 * 8/10 * 3/9


P(2) + P(3) + P(4) = 17/33


Let me know if there is a quicker strategy.

Great thinking canuckclint!

SO what would you say about having the second pair ?

P(2) is simply the probabilty of not getting the 1st card as a pair 1/1
AND 2nd one as a pair. Since there is AND, you multiply.

P(3) = not getting 1st as pair, not getting 2nd as pair and getting 3rd as pair with 1st or 2nd. For the second card there are 11 cards left and you have to NOT get the same as 1st card: 10/11. Finally 3rd card, you
have 2/10 chances of getting a pair with the 1st or 2nd card.

Does that makes sense?


You get the idea.

canuckclint wrote:

logitech wrote:

canuckclint wrote:I think I solved it after thinking about it for awhile:

You can either get a pair with the 2nd, 3 or 4th card.

P(2) = 1/1 * 1/11
P(3) = 1/1 * 10/11 * 2/10
P(4) = 10/11 * 8/10 * 3/9


P(2) + P(3) + P(4) = 17/33


Let me know if there is a quicker strategy.

Great thinking canuckclint!

SO what would you say about having the second pair ?

P(2) is simply the probabilty of not getting the 1st card as a pair 1/1
AND 2nd one as a pair. Since there is AND, you multiply.

P(3) = not getting 1st as pair, not getting 2nd as pair and getting 3rd as pair with 1st or 2nd. For the second card there are 11 cards left and you have to NOT get the same as 1st card: 10/11. Finally 3rd card, you
have 2/10 chances of getting a pair with the 1st or 2nd card.

Does that makes sense?


You get the idea.

No I am clear on that. It sure does make sense too.

But the question is asking: " What is the chance that Bill finds at least one pair of cards that have the same value? "

AT LEAST ONE PAIR means: We have to find P(one pair) + P(two pairs)

Or in other words for example:

You will either have

1 1 3 4 - ONE PAIR

OR

3 4 3 4 - TWO PAIR

Your solution is for having one pair ? Am I making any sense?

Logitech you are so right! Good stuff.
Here's the OA.



The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1.

First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt).

Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. We have 11 cards left in the deck.

Probability of NO pairs so far = 10/11.

Third card: Since we have gotten no pairs so far, we have 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. We have 10 cards left in the deck.

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

Cumulative probability of avoiding a pair BOTH on the second card AND on the third card = product of the two probabilities above = (10/11) (8/10) = 8/11.

Fourth card: Now we have three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. We have 9 cards left in the deck.

Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that we have NO pairs so far = 6/9.

Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.

Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 - 16/33 = 17/33.

The correct answer is C.

It was just lucky for me to get the answer using my flawed logic!
I'll take it!