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
Ans . C.
Please help me solving this..
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Rahul@gurome
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Solution:
From 12 playing cards, any 4 cards can be selected in 12C4 or 45*11 = 495 ways.
The chance that Bill finds at least one pair of cards that have the same value is = 1 - (chance that Bill finds no pair of cards that have the same value).
Let us calculate the chance that Bill finds no pair of cards that have the same value.
There can be 5 cases.
Note that when we are selecting from second pack, we cannot select from among those cards that have the same value as that selected from first pack of cards. This is to avoid having pairs.
Case (1) All 4 cards selected are from first pack of cards. No. of ways is 6C4 = 15.
Case (2) 3 selected are from first pack and 1 from second pack. No. of ways is 6C3*3C1 = 60.
Case (3) 2 are selected from first pack and 2 from second pack. No. of ways is 6C2*4C2 = 90.
Case (4) 1 is selected from first pack and 3 from second pack. No. of ways is 6C1*5C3 = 60.
Case (5) All 4 cards slected are from second pack. No of ways is 6C4 = 15.
So total number of ways is 15+60+90+60+15 = 240.
So the chance that Bill finds no pair of cards that have the same value is 240/495 = 16/33.
Or the chance that Bill finds at least one pair of cards that have the same value is 1- (16/33) = 17/33.
The correct answer is (C) .
From 12 playing cards, any 4 cards can be selected in 12C4 or 45*11 = 495 ways.
The chance that Bill finds at least one pair of cards that have the same value is = 1 - (chance that Bill finds no pair of cards that have the same value).
Let us calculate the chance that Bill finds no pair of cards that have the same value.
There can be 5 cases.
Note that when we are selecting from second pack, we cannot select from among those cards that have the same value as that selected from first pack of cards. This is to avoid having pairs.
Case (1) All 4 cards selected are from first pack of cards. No. of ways is 6C4 = 15.
Case (2) 3 selected are from first pack and 1 from second pack. No. of ways is 6C3*3C1 = 60.
Case (3) 2 are selected from first pack and 2 from second pack. No. of ways is 6C2*4C2 = 90.
Case (4) 1 is selected from first pack and 3 from second pack. No. of ways is 6C1*5C3 = 60.
Case (5) All 4 cards slected are from second pack. No of ways is 6C4 = 15.
So total number of ways is 15+60+90+60+15 = 240.
So the chance that Bill finds no pair of cards that have the same value is 240/495 = 16/33.
Or the chance that Bill finds at least one pair of cards that have the same value is 1- (16/33) = 17/33.
The correct answer is (C) .
Rahul Lakhani
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P(at least 1 pair of the same value) = 1 - P(all different values)
Probability that each pick is a different value:
1st pick = 12/12 (because any of the 12 cards can be chosen)
2nd pick = 10/11 (because of the 11 cards remaining, we can't use the mate of the 1st pick, leaving us 11-1=10 good choices)
3rd pick = 8/10 (because of the 10 cards remaining, we can't use the mates of the first 2 picks, leaving us 10-2=8 good choices)
4th pick = 6/9 (because of the 9 cards remaining, we can't use the mates of the first 3 picks, leaving us 9-3=6 good choices)
Since we want all of these events to happen together, we multiply the fractions:
P(all different values) = 12/12 * 10/11 * 8/10 * 6/9 = 16/33
So P(at least 1 pair of the same value) = 1 - 16/33 = 17/33
The correct answer is C.
Probability that each pick is a different value:
1st pick = 12/12 (because any of the 12 cards can be chosen)
2nd pick = 10/11 (because of the 11 cards remaining, we can't use the mate of the 1st pick, leaving us 11-1=10 good choices)
3rd pick = 8/10 (because of the 10 cards remaining, we can't use the mates of the first 2 picks, leaving us 10-2=8 good choices)
4th pick = 6/9 (because of the 9 cards remaining, we can't use the mates of the first 3 picks, leaving us 9-3=6 good choices)
Since we want all of these events to happen together, we multiply the fractions:
P(all different values) = 12/12 * 10/11 * 8/10 * 6/9 = 16/33
So P(at least 1 pair of the same value) = 1 - 16/33 = 17/33
The correct answer is C.
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Hi Rahul,Rahul@gurome wrote:Solution:
From 12 playing cards, any 4 cards can be selected in 12C4 or 45*11 = 495 ways.
The chance that Bill finds at least one pair of cards that have the same value is = 1 - (chance that Bill finds no pair of cards that have the same value).
Let us calculate the chance that Bill finds no pair of cards that have the same value.
There can be 5 cases.
Note that when we are selecting from second pack, we cannot select from among those cards that have the same value as that selected from first pack of cards. This is to avoid having pairs.
Case (1) All 4 cards selected are from first pack of cards. No. of ways is 6C4 = 15.
Case (2) 3 selected are from first pack and 1 from second pack. No. of ways is 6C3*3C1 = 60.
Case (3) 2 are selected from first pack and 2 from second pack. No. of ways is 6C2*4C2 = 90.
Case (4) 1 is selected from first pack and 3 from second pack. No. of ways is 6C1*5C3 = 60.
Case (5) All 4 cards slected are from second pack. No of ways is 6C4 = 15.
So total number of ways is 15+60+90+60+15 = 240.
So the chance that Bill finds no pair of cards that have the same value is 240/495 = 16/33.
Or the chance that Bill finds at least one pair of cards that have the s ame value is 1- (16/33) = 17/33.
The correct answer is (C) .
Can u provide clarification on this.
For having pairs::
3 selected are from first pack and 1 from second pack. No. of ways is 6C3*3C1
2 are selected from first pack and 2 from second pack. No. of ways is 6C2*2C1(For having a pair)*5C1 (Any other card from pack 2)+6C2*2C2(Case for two pairs)
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(3 selected are from first pack and 1 from second pack. No. of ways is 6C3*3C1)
Explanation of above statement:
We are calculating the chance that Bill finds no pair of cards that have the same value.
When we select three from first pack, there will be corresponding three cards of the same value in the second pack. Since there cannot be any pairs, when we carry out selection process from second pack, we cannot select from among those 3 cards. So from second pack selection is out of (6-3) or 3 cards.
Or no. of selections such that 3 selected are from first pack and 1 from second pack is 6C3*3C1.
I think if you understand this, you will understand other cases as well.
Hope it is clearer.
Explanation of above statement:
We are calculating the chance that Bill finds no pair of cards that have the same value.
When we select three from first pack, there will be corresponding three cards of the same value in the second pack. Since there cannot be any pairs, when we carry out selection process from second pack, we cannot select from among those 3 cards. So from second pack selection is out of (6-3) or 3 cards.
Or no. of selections such that 3 selected are from first pack and 1 from second pack is 6C3*3C1.
I think if you understand this, you will understand other cases as well.
Hope it is clearer.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
