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Math Expert V
Joined: 02 Sep 2009
Posts: 59725
On a game show, there are five boxes, each of which contains a prize.  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 53% (02:52) correct 47% (02:54) wrong based on 100 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 V
Joined: 02 Aug 2009
Posts: 8320
On a game show, there are five boxes, each of which contains a prize.  [#permalink]

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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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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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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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Math Expert V
Joined: 02 Aug 2009
Posts: 8320
Re: On a game show, there are five boxes, each of which contains a prize.  [#permalink]

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

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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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

Beautiful problem.

Let´s do by complementarity, i.e., we start by finding the probability of choosing 4 keys in such a way that no locks will open.

This occurs if, and only if, the player gets exactly one key from each of 4 locks among the 5 locks available.

First Step: there are C(5,4) = C(5,1) = 5 ways of choosing the locks the player will have exactly one key from.

Second Step: the first step concluded, the player has 2*2*2*2 ways of choosing one key of each pair of keys (pairs of keys chosen in the previous step).

By the multiplicative principle, the "not-wanted" probability is (5*2^4) over C(10,4) = 8/21 ,

because C(10,4) is the total number of equiprobable ways of choosing 4 keys among the 10 available without any restriction.

Our focus is 1 - 8/21 = 13/21.

The above follows the notations and rationale taught in the GMATH method.
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On a game show, there are five boxes, each of which contains a prize.  [#permalink]

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No of ways to select 10 keys = 10C4 = 210

Ways to win 1 prize = 5c1 (no of ways to pick a box) * 2c2 (correct combination that box) * 8c2 - 4 (Way to select remaining keys. We subtract 4 because 4 of the combination is the correct one and it will open the box and we don't want that to happen)
= 5 * 1 *(28-4) = 120

Ways to win 2 prize = 5c2 (any 2 boxes) = 10

Probability = (120+10)/210 = 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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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

To win at least one prize, he must open either one box or both boxes.

Let’s find the probability he can open both boxes first. To open 2 boxes, he must have all 4 keys to the 2 boxes he has chosen. The number of ways of picking 4 keys from 10 is 10C4 = (10 x 9 x 8 x 7)/(4 x 3 x 2) = 10 x 3 x 7 = 210, and only 4C4 = 1 set of 4 keys are the right keys for the 2 boxes he has chosen; thus, the probability of choosing the right set of keys is 1/210. However, there are 5C2 = (5 x 4)/2 = 10 ways to choose 2 boxes from 5, and each of these 10 selections has 1/210 chance of opening the 2 boxes; therefore, the probability of opening 2 boxes is actually 10 x 1/210 = 1/21.

Of course, he can also open only 1 box. To do that, he must have the 2 keys to that box. Since he has 4 keys, the number of ways the 4 keys contain the 2 keys he needs is 4C2 = (4 x 3)/2 = 6, so the probability he can open 1 box is 6/210. However, it’s possible that the 2 keys can open either one of the two boxes he has chosen and there are 10 ways to choose 2 boxes from 5 (as mentioned above), so the probability of opening exactly 1 box is actually 2 x 10 x 6/210 = 120/210 = 12/21.

Therefore, the probability of winning at least one prize is

1/21 + 12/21 = 13/21

Alternate Solution:

We will use the formula:

P(opening at least one box) = 1 - P(opening no boxes)

Thus, we need to calculate the probability that none of the boxes can be opened using the four selected keys.

After the first key is selected, there is a 8/9 probability that the second key is not the matching pair of the first key.

After two keys are selected, there are 2 keys among the remaining 8 keys that form a pair with one of the already selected keys, which means any of the 8 - 2 = 6 keys will not form a pair with one of the previously selected keys. Thus, the probability that the third selected key does not form a pair with the previous two keys is 6/8 = 3/4.

Finally, after three keys are selected, there are 3 keys among the remaining 7 keys that form a pair with one of the already selected keys, which means any of the 7 - 3 = 4 keys will not form a pair with one of the previously selected keys. Thus, the probability that the fourth selected key does not form a pair with the previous three keys is 4/7.

In total, the probability that the four keys do not contain a pair which will open a box is 8/9 x 3/4 x 4/7 = 8/21. Therefore, the probability that the contestant will win at least one prize is 1 - 8/21 = 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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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

FIND 1-P(NOT) WINNING
First: anything = 10/10
Second: anything but previous = 9-1/10-1 = 8/9
Third: anything but the previous two = 8-2/9-1 = 6/8
Fourth: anything but the previous three = 7-3/8-1 = 4/7
P(NOT): 1*8/9*6/8*4/7 = 8/21
1-P(NOT): 1-8/21 = 13/21

Ans (C) Re: On a game show, there are five boxes, each of which contains a prize.   [#permalink] 22 Nov 2019, 05:24
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