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# On a game show, there are five boxes, each of which contains a prize.

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Math Expert
Joined: 02 Sep 2009
Posts: 46305
On a game show, there are five boxes, each of which contains a prize. [#permalink]

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23 Nov 2016, 06:29
2
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Difficulty:

75% (hard)

Question Stats:

55% (01:48) correct 45% (02:09) wrong based on 92 sessions

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On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. 2/5
B. 11/21
C. 13/21
D. 2/3
E. 16/21

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Math Expert
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Posts: 5952
On a game show, there are five boxes, each of which contains a prize. [#permalink]

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23 Nov 2016, 19:31
1
1
Bunuel wrote:
On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. 2/5
B. 11/21
C. 13/21
D. 2/3
E. 16/21

Hi
You can do this in two ways..

FIRST by finding prob of opening only onebox and add to it prob of opening two boxes..

But the SECOND way is better and less error prone..
Since we are looking for ATLEAST one, find prob when none open and subtract from 1..
So prob that none opens..
1) first key can be any so prob =10/10=1
2) second key can be any except the pair of first key so any 8 out of remaining 9, prob = 8/9
3) third can be any except the pair of first two, so we can choose 6 out of 8..prob =6/8
4) fourth similarly can be any except pair of first three, so we can choose 4 out of 6..prob=4/6..

Prob none opens =$$\frac{10}{10}*\frac{8}{9}*\frac{6}{8}*\frac{4}{7}=\frac{8}{21}$$..
Ans =1-8/21=13/21
C
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On a game show, there are five boxes, each of which contains a prize. [#permalink]

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24 Nov 2016, 01:01
chetan2u wrote:
Bunuel wrote:
On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. 2/5
B. 11/21
C. 13/21
D. 2/3
E. 16/21

Hi
You can do this in two ways..

FIRST by finding prob of opening only onebox and add to it prob of opening two boxes..

But the SECOND way is better and less error prone..
Since we are looking for ATLEAST one, find prob when none open and subtract from 1..
So prob that none opens..
1) first key can be any so prob =10/10=1
2) second key can be any except the pair of first key so any 8 out of remaining 9, prob = 8/9
3) third can be any except the pair of first two, so we can choose 6 out of 8..prob =6/8
4) fourth similarly can be any except pair of first three, so we can choose 4 out of 6..prob=4/6..

Prob none opens =$$\frac{10}{10}*\frac{8}{9}*\frac{6}{8}*\frac{4}{7}=\frac{8}{21}$$..
Ans =1-8/21=13/21
C

Hello Chetan,
Could you please tell me where im going wrong?
to get at least 1 prize you have to select 2 keys right out of 10
1-(non event) can be done when all the 4 keys don't match or 1 key matches and 3 doesn't because there are 2 locks to open to claim a prize
you can select all keys wrong in 8C4 ways = 14*5= 70
total number of key selection is 10C4= 30*7= 210
could you help me proceed further? Because i think i'm messing the next part
number of ways to select 1 key right and 3 keys wrong = 1*8C3*3! = 56 * 6= 336
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Re: On a game show, there are five boxes, each of which contains a prize. [#permalink]

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24 Nov 2016, 06:57
Bunuel wrote:
On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. 2/5
B. 11/21
C. 13/21
D. 2/3
E. 16/21

Alternatively, we can also divide them into pairs

5c1*4c2*2c1*2c1/10c4 + 5c2/10c4

130/210 = 12/21
Ans C
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Joined: 02 Aug 2009
Posts: 5952
Re: On a game show, there are five boxes, each of which contains a prize. [#permalink]

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24 Nov 2016, 19:49
Vinayak Shenoy wrote:
chetan2u wrote:
Bunuel wrote:
On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. 2/5
B. 11/21
C. 13/21
D. 2/3
E. 16/21

Hi
You can do this in two ways..

FIRST by finding prob of opening only onebox and add to it prob of opening two boxes..

But the SECOND way is better and less error prone..
Since we are looking for ATLEAST one, find prob when none open and subtract from 1..
So prob that none opens..
1) first key can be any so prob =10/10=1
2) second key can be any except the pair of first key so any 8 out of remaining 9, prob = 8/9
3) third can be any except the pair of first two, so we can choose 6 out of 8..prob =6/8
4) fourth similarly can be any except pair of first three, so we can choose 4 out of 6..prob=4/6..

Prob none opens =$$\frac{10}{10}*\frac{8}{9}*\frac{6}{8}*\frac{4}{7}=\frac{8}{21}$$..
Ans =1-8/21=13/21
C

Hello Chetan,
Could you please tell me where im going wrong?
to get at least 1 prize you have to select 2 keys right out of 10
1-(non event) can be done when all the 4 keys don't match or 1 key matches and 3 doesn't because there are 2 locks to open to claim a prize
you can select all keys wrong in 8C4 ways = 14*5= 70
total number of key selection is 10C4= 30*7= 210
could you help me proceed further? Because i think i'm messing the next part
number of ways to select 1 key right and 3 keys wrong = 1*8C3*3! = 56 * 6= 336

Hi Vinayak,
The approach you have taken would be right if we were talking of just one box..

Here it will never happen that all keys are wrong, because same set of 4 keys will be used to open the 5 boxes.
So each key will fit in for some box....

I hope you are getting the reasoning
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On a game show, there are five boxes, each of which contains a prize. [#permalink]

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23 Feb 2017, 00:28
1
Bunuel wrote:
On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. $$\frac{2}{5}$$

B. $$\frac{11}{21}$$

C. $$\frac{13}{21}$$

D. $$\frac{2}{3}$$

E. $$\frac{16}{21}$$

Official solution from Veritas Prep.

For this probability problem, it is first important to interpret the situation. To win a prize, the player needs to find two keys that "match," fitting the same box. So the question is actually asking "what is the probability of getting at least one pair when drawing from 5 pairs of keys?"

To calculate that, you'll want to use good "at least" probability strategy, in which you first determine the probability of not getting any pairs and then subtracting that from 100%. In that case:

On the first key there is a 100% chance that a player won't get a pair - with only one key, it's of course impossible to have two matching keys.

On the second draw, there are 9 remaining keys, 8 of which will not match the first one drawn. So the probability of not a match is $$\frac{8}{9}$$.

On the third draw, there are 8 remaining keys, but now two keys to "avoid" matching. So the probability of not matching is $$\frac{6}{8}$$.

And then on the fourth draw, there are 7 remaining keys but three already-chosen keys to avoid, so only 4 keys help avoid a match. Therefore the probability is $$\frac{4}{7}$$.

Therefore, the probability of NOT getting a pair is $$\frac{8}{9}∗\frac{6}{8}∗\frac{4}{7}$$ which reduces to $$\frac{8}{21}$$. But remember - you need to subtract that from 1, making the correct answer $$\frac{13}{21}$$.
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Re: On a game show, there are five boxes, each of which contains a prize. [#permalink]

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09 May 2018, 03:26
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Re: On a game show, there are five boxes, each of which contains a prize.   [#permalink] 09 May 2018, 03:26
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