Probability

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deepak123gmat: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Wed Oct 06, 2010 11:33 pm

by deepak123gmat » Wed Nov 03, 2010 4:02 am

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If 4 letters are to be put into 4 envelope at random, what is the probobility 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

[spoiler]
OA - D[/spoiler]

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Source: Beat The GMAT — Problem Solving | Wed Nov 03, 2010 4:02 am

shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Wed Nov 03, 2010 4:36 am

deepak123gmat wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If 4 letters are to be put into 4 envelope at random, what is the probobility 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

P (Getting one correct) = 1/4
P (Getting one wrong from remaining 3) = 2/3
P (Getting one wrong from remaining 2) = 1/2
P (Getting one wrong from remaining 1) = 1

Total Probability of getting 1 correct and 3 wrong = 1/4 * 2/3 * 1/2 * 1 = 1/12

This combination is CWWW (C is correct and W is wrong).

We can have 4 combinations like this, CWWW, WCWW, WWCW, WWWC.

Thus required probability = 4*(1/12) = 1/3

If the problem is Easy Respect it, if the problem is tough Attack it

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Wed Nov 03, 2010 5:14 am

Solution:
Let the envelopes be E1, E2, E3, E4.
Let the letter be l1, l2, l3 and l4.
Let the right combination be E1l1, E2l2, E3l3, E4l4.
So we can have 4 possiblities.
(a) l1 goes to E1, l2 does not go to E2, l3 does not go to E3 and l4 does not go to E4.
Prob. of this is Â¼*2/3*1/2*1 = 1/12.
(b) l2 goes to E2, l1 does not go to E1, l3 does not go to E3 and l4 does not go to E4.
Prob. of this is Â¼*2/3*1/2*1 = 1/12.
(c) l3 goes to E3, l1 does not go to E1, l2 does not go to E2 and l4 does not go to E4.
Prob. of this is Â¼*2/3*1/2*1 = 1/12.
(d) l4 goes to E4, l1 does not go to E1, l2 does not go to E2 and l3 does not go to E3.
Prob. of this is Â¼*2/3*1/2*1 = 1/12.

Required answer is 1/12 + 1/12 + 1/12 + 1/12 = 1/3.
The correct answer is D.

Rahul Lakhani
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Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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lalitgmat: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Thu Jul 05, 2007 1:28 pm

by lalitgmat » Wed Nov 03, 2010 9:35 am

This problem is using concept called "De-Arrangements". The no of de-arrangement for N distinct stuffs is
N![1/2! - 1/3! + 1/4! .... + (-1)**n 1/N!]

Total ways of arranging 4 letters in 4 places 4! = 24.
Choose one letter 4C1 ways = 4. De-Arrange remaining 3 i.e. 3! [1/2! - 1/3!] = 2.

So Probability is 4 * 2 / 24 i.e 1/3

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pesfunk: Master | Next Rank: 500 Posts; Posts: 286; Joined: Tue Sep 21, 2010 5:36 pmFollowed by: 5 members; Location: Kolkata, India; Thanked: 11 times

by pesfunk » Thu Nov 04, 2010 5:17 am

I will try to explain this in very simple terms:

let us consider there are four letters (L1,L2,L3,L4) with corresponding right addressed envelopes (e1,e2,e3,e4)

Now if we assume - one correct

Total possible combination = 24 (P4)

Consider => L1-e1 is right => then L2 has 2 possible combinations such as e3 or e4 (we consider L2-e3); it cannot take e2 bcoz then there will be a combination with 2 letters in right envelop. Now the obvious choice for L3 to take e4 and not e3 (same logic for L2) and thus L4 is paired bydefault with e2.....Now plz refer below sequence with 2 possibilities

Hence => [L1-e1 | L2-e3 | L3-e4 | L4-e2] OR [L1-e1 | L2-e4 | L3-e2 | L4-e3]

Now this sequence is only for L1-e1 to be the only right and all others are wrong; similarly we have 4 pairs such as (L1-e1 | L2-e2 | L3-e3 | L4-e4 ) with 2 possibilities for each

Hence Ans is 4 x 2 = 8 ways => 8/24 => 1/3

deepak123gmat wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If 4 letters are to be put into 4 envelope at random, what is the probobility 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

[spoiler]
OA - D[/spoiler]

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gmatdriller: Master | Next Rank: 500 Posts; Posts: 418; Joined: Sun Jul 04, 2010 12:48 pmFollowed by: 3 members; Thanked: 6 times

by gmatdriller » Sat Nov 20, 2010 12:30 am

leaning on the explanations given above, I would appreciate further insight
into my question:

how do we know when to multiply events A x B, or when to take them as
arrangements - A Permutation B (A(A-1)(A-2).....into B places)

Examples:
(i) Tanya has 4L and 4E (letters and Envelops respectively).
(ii) Tanya has 4S and 5P (shirts and pair of pants respectively).

According to the explanations above total possible outcome of EL (whether
right or wrong) = 4P4 = 4x3x2x1 = 24 (why not 4x4?)

(ii), according to Lunapower (RonPurewal: Thursdays with Ron on
Manhattangmat.com), total outfits of choosing a shirt AND a pair of pants=4x5
=20..

Can someone please clarify situations regarding when to use permutations and
when to simply multiply the values...

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