Each of four different locks has a matching key. The keys

This is first question of Mgmat challenge set of Gmat Club tests.

Though while solving the tests I got it wrong, but later I tried to solve it and got the correct answer. I looked at the explanation it was too long. I have done in a simple way, but I m not 100% if I m correct.

Total ways to assign keys = 4! = 24

Ways to assign keys such that only 2 fit = Choose the 2 that fit = C(4,2) = 6

Note that once you pick the two locks on which the keys fit there is exactly one allocation of keys possible. For eg. If you pick Lock A & Lock B fit, the only allocation possible is [Key A, Key B, Key D, Key C]

So probability = 6/24 = 1/4

Answer (c)

The initial probability of drawing blue ball is 5/8. Without knowing the other results , the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still will be 5/8.

Hope it's clear.

Dear Dhairya275
There are not hugely simple solutions to this. All solutions that occur to me involve using counting techniques. You might take a look at this blog post.
https://magoosh.com/gmat/2013/gmat-proba ... echniques/

Total number of orders for 4 keys = 4! = 4*3*2*1 = 24

Of those 24 possible orders, how many have two keys in the right place and two in the wrong place. Suppose the locks are {a, b, c, d}. If the keys are in the order {A, B, C, D}, then all four are correct. To get two right & two wrong, we would need to select one pair from {A, B, C, D} and switch them. How many different pairs can we select from a set of four?

4C2 = \(\frac{4!}{(2!)(2!)}\)
= \(\frac{4*3*2*1}{(2*1)(2*1)}\)
= \(\frac{4*3}{2}\)
= 6

So, of the 24 sets, 6 of them would have two right & two wrong. P = 6/24 = 1/4

Does this make sense?
Mike

Similar question with all possible scenarios: letter-arrangements-understanding-probability-and-combinats-84912.html?hilit=letter%20arrangements

A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?

The above question is from M14 q1. I solved it using steps by taking different conditions and solved it correctly.

the answer was 5/8.

I was wondering is this true for probability that whatever happens in between, the finally probability will remain same?
if I exclude "Mary will extract one ball at random and keep it. If, after that," still the probability is 5/8.

is this a co-incidence for a particular case?

I know if it had mentioned that mary picked a particular ball, then probability would have changed. I m talking about the case in which no specific thing mentioned. Actually I have seen such question earlier and I though it would be better to confirm it. I think if mary picks 2 balls without mentioning which ones, then the probability will remain 5/8.

A - B - C - D (locks)
a - b - c - d (keys)
a - b - d - c
a - c - b - d
a - c - d - b
a - d - b - c
a - d - c - b

b - a - c - d
b - a - d - c
b - c - a - d
b - c - d - a
b - d - a - c
b - d - c - a

c - a - b - d
c - a - d - b
c - b - a - d
c - b - d - a
c - d - a - b
c - d - b - a

d - a - b - c
d - a - c - b
d - b - a - c
d - b - c - a
d - c - a - b
d - c - b - a

As you can see there are 4!=24 ways to assign keys to locks (arrangement of 4 keys). So, the number of total outcomes is 24.

Next, consider a case when exactly two of the keys fit the locks, say a and b fit and d and c not:
A - B - C - D (locks)
a - b - d - c (keys)

As you can see if a and b fit, d and c not to fit can only be arranged in one way as shown above.

Now, the number of ways to choose which 2 keys fit is \(C^2_4\), so the number of favorable outcomes is \(1*C^2_4=6\).

P = 6/24 = 1/4.

Does this make sense?

Thanks Bunnel I got it....

Your letter arrangement thread is quite good !! I m able to answer probability questions, but these letter arrangement sometimes confuses me. +1

Yes Bunuel, I was thinking the same but you have confirmed it. This concept is quite helpful.

Thanks !!

I'm definitely missing something rather fundamental because I can't make out what connection I'm missing. I'm not able to apply the correct methodology to the correct problem -- ever. The way I tackled this problem(which was obviously wrong was):

There are 4 keys and 4 locks, therefore 16 combinations? Out of 16 possibilities, there are only 2 ways to arrange this so that the locks match - (Lock A, Key A), (LB, KB), (LC,KD) and (LD, KC) and another way is LBKB, LAKA, LCKD, LDKC. Therefore, I calculated 2/16 which is 1/8.

Bunuel, is this valid?
p(2 keys correct)=Match*Match*Miss*Miss=(1/4)*(1/3)*(1/2)*1 = 1/24
P(2 keys correct)=p(2 keys correct)*Total combinations that this event can happen
Total combinations that this event can happen = Number of attempted matches that will be successful = Number of ways we can arrange Ma,Ma,Mi,Mi = C(4,2)
P(2 keys correct)=1/24*C(4,2)=1/24*6=1/4

Hi TooLong150,

Your explanation is ABSOLUTELY CORRECT

Just another Traditional way of solving the same question

\(Probability = \frac{Favourable Outcomes}{Total Outcomes}\)

Total Possible Arrangement of 4 keys to 4 locks = 4! = 24

Favourable Ways of Arrangement of Keys (2 correct and 2 incorrect) = 4C2*1 = 6

4C2 represents the ways to select the lock in which the key is found to be correct
1 is the way to arrange the remaining two keys to remaining two locks such that key doesn't match the lock

i.e. Probability = 6/24 = 1/4

Answer: Option C

The ways of selecting two keys right = 4C2
The total combinations possible = 4!
Thus, probability = 4C2/4! = 4*3/2*4!
= 1/4.

Thus, the correct option is C.

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