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

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
_________________

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 \(\frac{13}{21}\).
_________________

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

Alternatively, we can also divide them into pairs

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

130/210 = 12/21
Ans C

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
_________________

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.
_________________

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

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.

Answer: C
_________________

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)

Hello chetan2u,
Thanks for nice solution! understood the solution you provided. But I am breaking my head with the following issue :
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?
Solution: = 1- (9/9∗6/8∗3/7∗6/6∗2)=9/14.
Here we are multiplying by 2 as this 2nd and 3rd cup can be tasted in 2 ways. However, in your solution of selecting keys we are not multiplying by 3!, although here also the 2nd/3rd/4th keys (not selected) can be arranged in 3! ways. Where I am making the mistake?

Dear IanStewart JeffTargetTestPrep,

I don't understand the highlighted portion above at all. As per my understanding, 4C2 is used to group 2 items from the pool of 4. Why do we need to group here?

Also, why choose 2 boxes from 5 when you are dealing with choosing ONLY 1 boxes?

BTW, I think the two below are both the correct approaches (instead of the solution quoted above)


Could you please share some thoughts sir?

Yes, the line after the one you've highlighted says " it’s possible that the 2 keys can open either one of the two boxes he has chosen", but he is not choosing two boxes. So I don't understand what the explanation has to do with the original problem. You could direct your question instead to the person who wrote that post.

The question itself is so problematically worded I normally wouldn't spend any time on it. Right from the beginning:



the question talks about five boxes, then about "the box". Does the contestant win by opening any of the five boxes, or by opening one particular box? It's completely unclear. Then it says



and again, it's not clear what this means. Is it possible that keys 1 and 2 open the first box, and keys 1 and 3 open the second? Those are "unique" key combinations. They presumably mean the key combinations don't overlap, not simply that they're unique.

Anyway, in probability, you generally have two ways to solve. You can solve the question directly, or you can solve the opposite problem (i.e. find the probability the event you're asked about does not happen) and subtract from 1. Most of the time, solving directly is the best choice; it's conceptually a lot simpler to solve what you're asked, rather than reverse the problem. But there are some questions which are much easier to solve if you do reverse them. This is one of them. When you study questions like this one, the main question you should be asking yourself is "how can I quickly recognize that I should reverse a problem like this?" It's when we get multiple cases solving directly that it can be easier to reverse a problem. When a probability question uses a phrase like "at least" or "at most", that almost always means you can have several cases if you solve directly. It's then worthwhile asking if you'd only have one case if you reverse the problem. That's what happens here: if we want the probability our keys won't open any box, we just need to make sure, after our first selection, that none of our later selections match our earlier ones. The first key can be anything, the second can be 8 of the 9 remaining, the next 6 of the 8 remaining, and the last key 4 of the 7 remaining. So the probability we can't open a box is (8/9)(6/8)(4/7) = 8/21, and the probability we can is thus 1 - (8/21) = 13/21. Solving this problem directly is annoying and tedious, and I wouldn't consider solving this problem that way.

JeffTargetTestPrep chetan2u

Could you please help me figure out the mistake I'm making in my approach below?

We need probability of opening at least 1 box => Prob(open 1 box) + Prob(open 2 boxes)

P(1box) = 5C1 * 10/10 * 1/9 * 8/8 * 6/7
That is,
5C1 = ways to select a box to open
10/10 = Prob of Selecting 1st key (can select any one of 10 keys)
1/9 = Prob of Selecting 2nd Key (must select the pair key of the first selection to open a box)
8/8 = Prob of Selecting 3rd key (can select any one of 8 keys)
6/7 = Prob of Selecting 4th key (can select any one except the pair of the second key)

So, P(1Box) = 5C1 * 10/10 * 1/9 * 8/8 * 6/7 = 30/63

P(2boxes) = 5C2 * 10/10 * 1/9 * 8/8 * 1/7
That is,
5C2 = ways to select 2 boxes to open
10/10 = Prob of Selecting 1st key (can select any one of 10 keys)
1/9 = Prob of Selecting 2nd Key (must select the pair key of the first selection to open a box)
8/8 = Prob of Selecting 3rd key (can select any one of remaining 8)
1/7 = Prob of Selecting 4th key (must select the pair key of the second selection to open 2nd box)

So, P(2Boxes) = 5C2 * 10/10 * 1/9 * 8/8 * 1/7 = 10/63

Therefore,

Prob(AtLeast 1 Box) = P(1Box) + P(2Boxes)
= 30/63 + 10/63 = 40/63

I know I'm missing some calculation here, since 39/63 (= 13/21) is what should be the actual probability, but I cant figure out which piece I'm missing here. Thanks for your help !

If you begin by picking a specific box that we'll open, so if you begin by multiplying by 5C1, then the probability the first key will open that box is 2/10. It's not true that we can just pick any key at all. So you're multiplying by an extra 5 that you shouldn't be multiplying by when your calculation begins with "5C1 * 10/10". But then you're also assuming the very first two keys we pick will match, and that the third and fourth keys will not. We could pick our matching keys in a few ways - maybe the second and fourth keys match, for example, and the first and third do not. It turns out there are 6 different ways in which we could pick the two matching keys, which is why, when you multiplied by the extra 5, you got an answer fairly close to the right one (if you multiplied by 6 instead, it would be perfect). There's a similar issue with the calculation you've done in the second case -- multiplying by 5C2 is not correct.

As I said above, I'd never consider solving this directly, because it's both difficult and error-prone. But this is how you can do it:

- first you can ask, "what is the probability the first two keys open a single box, and the next two keys we pick do not open the same box?" The first key can be anything, the second must be a specific key (1/9 probability), the third key can be anything, and the fourth must not match the third (6/7 probability). So the probability we get this exact sequence is (1/9)(6/7).

- but we might be able to open exactly one box in several ways. Maybe the first two keys match, or maybe the second and the fourth match, for example. There are 4C2 = 6 ways to choose which two of the four keys match, so the probability that we can open precisely one box is (6)(1/9)(6/7)

- then we need to add the probability we can open both boxes. Again, we can imagine the first two keys match, and the last two match. The probability that will happen is (1/9)(1/7), since the first and third keys can be anything.

- But there are other sequences where we can open both boxes - maybe the 1st and 2nd keys match, or maybe the 1st and 3rd, or maybe the 1st and 4th. So there are 3 possible sequences (the first key must match something, so these are all the cases). So the probability we can open both boxes is (3)(1/9)(1/7)

Adding the results from our two non-overlapping cases, the answer is

(6)(1/9)(6/7) + (3)(1/9)(1/7) = 12/21 + 1/21 = 13/21

As IanStewart has also written that it is better to do it in indirect way. But if you want to do it then

Only 1 box
5C1*2/10*1/9*8/8*6/7, but in these we are talking of probability of only 1 box opening and that too in first two keys itself, so multiply by 4C2, as these 2 keys can be anywhere =5*2/10*1/9*6/7*4C2=36/63

Both boxes
5C2*4/10*3/9*2/8*1/7.. here since we take any of the 4 in first pick , so we are not considering order => 10*4/10*3/9*1/4*1/7=3/63

Total 36/63+3/63=39/63=13/21
_________________

Given: 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.

Asked: 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?

Probability that a particular player wins no prize = 10C1*8C1*6C1*4C1/4!/10C4 = (10*8*6*4/4!)/(10*9*8*7/4!) = 6*4/9*7 = 8/21

Probability that a particular player will win at least one prize by drawing four keys = 1 - 8/21 = 13/21

IMO C

1- P(No event)=event

1- [5C4*2C1*2C1*2C1*2C1/10C4]
=1-80/210
=1-8/21
=13/21

chetan2u
A small typo- it should be 4 out of 7..prob=4/7
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________