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Updated on: 05 Sep 2019, 05:41
Originally posted by tealeaflin on 15 Oct 2006, 14:12.
Last edited by Bunuel on 05 Sep 2019, 05:41, edited 2 times in total.
Last edited by Bunuel on 05 Sep 2019, 05:41, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8
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10 Nov 2009, 10:04
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You can solve this the traditional way:
Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1
1/4 * 2/3 * 1/2 = 1/12
Multiply by 4, representing the four letters in the correct envelope:
1/12 * 4 = 4/12 = 1/3
Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1
1/4 * 2/3 * 1/2 = 1/12
Multiply by 4, representing the four letters in the correct envelope:
1/12 * 4 = 4/12 = 1/3
23 Oct 2006, 09:31
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Hermione
still not getting thru my brain... any other (possibly more basic/ step by step) explanation?OK in that case let me see if I can offer some help -
I hope you are familiar with basic probability fundas -
Let's say you have just ONE letter and TWO envelopes ONE of which is correctly addressed and the other addressed incorrectly.
What's the probability of putting the Letter in the correctly addressed envelope -
To answer this question - we see IN HOW MANY WAYS can the letter be put into the envelope - you could put it (assuming you don't know which envelope is which) in either of the two - so in total you have TWO ways of shoving the letter in. However, there's only ONE way in which it can go into the correctly addressed envelope -
so 1/2 is the prob of putting in correct envelope. This is easy.
Now in our current problem - let's say we have just ONE letter but FOUR envelopes. Only one of these envelopes has the address corresponding to the letter. The remaining three envelopes are incorrectly addressed.
So the probability that you will put the letter correctly is 1/4. Right?
What happens if i ask you the reverse question? what is the prob for putting it in the incorrect envelope. Suddenly you have three envs that are incorrect so you can put the letter incorrectly with a prob of 3/4. Right?
The whole problem can be broken down into Four Events that will fulfill the requirement of the question
Event 1 - E1
We know that prob of putting ONE Letter correctly is 1/4. Now once ONE letter has been put CORRECTLY, what are you LEFT with? You are left with THREE ENVELOPES and the remaining THREE letters. Since the one letter has been put correctly (though technically we have just calculated the PROBABILITY that the first letter goes into the correct envelope) we have the remaining THREE Letters and THREE envelopes.
Event 2 - E2
Let's take letter number 2 now - what is the probability that it LANDS in the INCORRECT envelope. Again by the same logic as above - there are 3 envelopes remaining out of which ONLY ONE has the correct address for LETTER number 2. The remaining 2 have INCORRECT address and LETTER NUMBER 2 could go in either of these 2 to meet our condition.
Thus the probability of this event is 2/3
So till now what we have done is -
we have calculated the prob of shoving Letter number 1 in correct env -- 1/4
we have calculated the prob of shoving Letter number 2 in INcorrect env --- 2/3
Event 3 - E3
Now let's take letter number 3 - again according to question we want to shove this in the WRONG envelope. There are 2 remaining envelopes and hence the prob of shoving this in the wrong env (or equally in the RIght env) is 1/2.
Finally we come to event E4 - the Letter number 4. This has only one way of going in so its probability of being put into the WRONG envelope is 1.
ok so we can see that our grand event is actually a combination of FOUR EVENTS happening - each with a probability of its own. So to calculate the total probability of the Grand Event itself we just multiply the individual probabilities since each event happens INDEPENDENTLY of each other
Egrand = 1/4 * 2/3 * 1/2 * 1/1 = 1/12
However at this point - I must introduce one last element in this question -since there are FOUR Letters - what we saw above was JUST ONE SEQUENCE of events leading to the desired result.
If we arbitrarily call the letters L1 thru L4, and let's say the above was an example in which we started by Picking up Letter L1 and worked thru the remaining letters, we could have equally well started out with letter L2 or L3 or L4.
Thus since each of these events ARE MUTUALLY EXCLUSIVE, meaning THEY CAN NEVER HAPPEN ALL THE SAME TIME BUT ONLY ONE LETTER AT A TIME, to calculate the TOTAL PROBABILITY of we will add the individual probabilities 1/12 + 1/12 + 1/12 + 1/12 which works out to 1/3.
Phew!!
11 Oct 2009, 12:12
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Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?
A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8
Total # of ways of assigning 4 letters to 4 envelopes is \(4!=24\).
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct).
ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8
\(P(C=1)=\frac{8}{24}=\frac{1}{3}\)
Answer: D.
All other possible scenarios: https://gmatclub.com/forum/letter-arrang ... 84912.html
Hope it's clear.
A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8
Total # of ways of assigning 4 letters to 4 envelopes is \(4!=24\).
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct).
ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8
\(P(C=1)=\frac{8}{24}=\frac{1}{3}\)
Answer: D.
All other possible scenarios: https://gmatclub.com/forum/letter-arrang ... 84912.html
Hope it's clear.
