math question - i think it is hard

[montz]

Probability that only 1 letter will be put into the envelope with the correct address is (d) 1/3

Step 1 - No. of ways you can select one letter (out of the four given) = 4
and there is only one way to put this letter into the correct envelope.

Now we need to ensure that none of the other letters (B,C and D) are put into the correct envelopes.

Step 2 - So, the next letter, can be put in an incorrect envelope in 2! ways (because one envelope out of the remaining three envelopes will be the correct one for this letter and hence we cannot use that one)

Step 3 - Now there are two letters remaining and there is only one way to put them into incorrect envelopes.

Required Probability = (4 * 2)/ 4! [Total number of ways is 4!]
= 1/3

Here's an example -
Lets name the letters and envelopes A, B, C and D.

Step 1- We select A and put it into the envelope named A.

Step 2 - we select envelope D for letter B

Step 3 - we now have envelope B and envelope C for letter C and letter D. Letter C cannot go into envelope C..


PS: pls let me know if I have done any mistake, I'll have to revisit my Probability basics

[agps]

I think it's answer A.
1/24

there is only one possibility to get all letters in the correct envelope (letter A in envelope A, letter B in Envelope B, ...)

and you have 4! possibilities on how to arrange the letters (4 possibilities for the 1st letter, 3 for the 2nd, ...)

so 1/24.

what is the OA?

[montz]

Hi,
I do not have any consolidated material on Probability that I can put up here. You can post your doubts or questions on this forum and people (I too if I can ) will defintely help.

Just in case if this helps -

-Understand permutation/combination concepts, probability becomes a lot easier if P&C is clear in your head.
-Most of the problems on probability can be solved with the help of logic and basic formulae (nPr, nCr, repetition allowed, without repetition, odds in favour/odds in against, mutually exclusive and independent events)

[samirpandeyit62]

Hi,
I would like to suggest certains tips that I believe can help solve or at least strategically guess on a Probabilty question.

As you all know that Probabilty of an event is defined as :

Number of favourable outcomes / Total nos of outcomes (Sample Space)

In any Probabilty problem the complexity lies in finding the numerator of this fratcion i.e (Number of favourable outcomes).

However it is relatively easier to find the Sample space by using Permutation/Combinations most often.

One should concentrate on finding the sample space first like in this question it is 4! i.e 24, Once you have this number i.e multiply all the fractions in the answer choice with a number so as the denominator is same as Sample space i.e 24 (this should be done where ever possible)

I have seen certain problems where after this step you may find that the denominator cannot be made equal to Sample space by multiplying it with any number, ELIMINATE THESE FRACTIONS as they cannot be the correct answer.

After this step try to find the Numerator i.e Favourable nos of outcomes, In this problem we can atleast say that 1 letter can be selected out of 4 in 4 ways & it can put into its correct envelope in only 1 way, After this you need to go for the remaining letters, however if you cannot think of the next step then use this DATA of 4 to narrow down your choices even further, JUST ASK your self whether the number of events will be more than this or less than this, Here we know that we have to find nos of ways of arranging the remaining 3 letters along with our first letter hence the nos of ways may only increase, however it cannot be less than 4 i.e probabilty cannot be less than 4/24, Hence we can eliminate first & second choices as they come up to be 1/24 & 3/24, & proceed with reamaining three.

[hopefully]

Probability that only 1 letter will be put into the envelope with the correct address is (d) 1/3

Step 1 - No. of ways you can select one letter (out of the four given) = 4
and there is only one way to put this letter into the correct envelope.

Now we need to ensure that none of the other letters (B,C and D) are put into the correct envelopes.

Step 2 - So, the next letter, can be put in an incorrect envelope in 2! ways (because one envelope out of the remaining three envelopes will be the correct one for this letter and hence we cannot use that one)

Step 3 - Now there are two letters remaining and there is only one way to put them into incorrect envelopes.

Required Probability = (4 * 2)/ 4! [Total number of ways is 4!]
= 1/3

Here's an example -
Lets name the letters and envelopes A, B, C and D.

Step 1- We select A and put it into the envelope named A.

Step 2 - we select envelope D for letter B

Step 3 - we now have envelope B and envelope C for letter C and letter D. Letter C cannot go into envelope C..


PS: pls let me know if I have done any mistake, I'll have to revisit my Probability basics

I have a question to montz here:

Is step 2 above we have 2! ways to pun an envelope in wrong envelopes but we also have 3 envelopes to choose from...

i know i m missin something but some more light on this would be great .

thanks....

[gabriel]

.. @ samirpandeyit62 .. good work .. that was a really good post ..

[montz]

@ hopefully

Look at the explanation given in bracket in step 2 and correlate it with the example given below in that post.

When I say there are 2! ways to put the next letter in an incorrect envelope, I actually mean there are 2! ways to:
i) select one envelope from two incorrect envelopes and
ii) put the letter inside that envelope. This step does not count because this is a combination problem and permutation does not matter...

We can NOT choose from 3 envelopes because one of them is the 'correct' envelope and if that is chosen then there will be 2 letters in correct envelopes instead of exactly one.

Hope this makes it clear. If not then work with the help of the given example and make all possible combinations by naming each of them. Then knock off the ones which do not fit the requirement. I know this can be tiresome but if you are stuck with PnC or probability this is one way to work out the problem. Off course process of elimination helps anytime...if you can eliminate [/quote]

[mayonnai5e]

I learned a slightly different method of solving this type of problem in college. Can anyone verify whether this is the correct math?

P(only 1 correct) = Probability that L1 is correct and others are wrong OR L2 is correct and others are wrong OR L3 is correct and others are wrong OR L4 is correct and others are wrong

so we're looking for P(only L1 right OR only L2 right OR L3 right OR L4 right)

P(only L1 right) = (1/4) * (2/3) * (1/2) * (1) = 1/12
--> 1/4 is prob to get the correct envelope E1
--> having removed 1, there are now 3 total envelopes and 2 of them are incorrect for whichever envelope we choose next (2/3)
--> having removed 2 envelopes, there are now 2 total and 1 of them is incorrect for whichever envelope we choose next (1/2)
--> only 1 choice left (1)

P(only L2 right) = (1/4) * (2/3) * (1/2) * (1) = 1/12
--> same logic as above

P(only L3 right) = (1/4) * (2/3) * (1/2) * (1) = 1/12
P(only L4 right) = (1/4) * (2/3) * (1/2) * (1) = 1/12

since each one of these are independent:
P(only L1 right OR only L2 right OR L3 right OR L4 right) = P(only L1 right) + P(only L2 right) + P(only L3 right) + P(only L4 right)

= 4 * (1/12) = 4/12 = 1/3

Is this math correct?

[samirpandeyit62]

Hi mayonnai5e,
This is the approach I also learned in college, IMO it is completely correct mathematically.

You find the probabilty of each individual event considering each one is mutually exclusive of the other and then OR them together (add).