Good explanation needed for this one Jane has to pick 3

Oldest

Best Reply

Author

Message

TAGS:

Founder

Status: Traveling...

Affiliations: UA-1K, SPG-G, HH-D

Joined: 04 Dec 2002

Posts: 10408

Location: United States (WA)

GMAT 1: 750 Q49 V42

GPA: 3.5

WE: Information Technology (Hospitality and Tourism)

Followers: 1371

Kudos [?]: 4204 [0], given: 3122

Good explanation needed for this one Jane has to pick 3 [#permalink]

28 Jul 2003, 19:54

Good explanation needed for this one

Jane has to pick 3 subjects out of 6: A, B, C, D, E, or F. If she has already chosen E, what is the probability that she will choose B also?
тАв 0.2
тАв 0.25
тАв 0.4
тАв 0.8
тАв 0.96

GMAT Instructor

Joined: 07 Jul 2003

Posts: 773

Location: New York NY 10024

Schools: Haas, MFE; Anderson, MBA; USC, MSEE

Followers: 5

Kudos [?]: 9 [0], given: 0

Re: Probability 2 [#permalink]

28 Jul 2003, 20:59

bb wrote:

Here are 3 ways of looking at this:
(1)
WE are already given that E is chosen, so the problem is simply what is the prob that she will pick B if she chooses 2 subjects.

This is the probability that she will either choose the subject as the first pick OR the subject as the second pick.

Pr(b 1st) = 1/5
Pr(b 2nd) = Pr(b not 1st AND b 2nd) = 4/5 * 1/4 = 1/5

These 2 are mutually exclusive, hence the probability is 1/5 + 1/5 = 2/5 = 0.4

Method 2:
The probability that she will NOT leave B as one of her 3 unpicked choice is exactly the same problem.
Pr(B is NOT one of 3 unpicked choices) = 4/5 * 3/4 * 3/2 = 2/5 = 0.4

Method 3:
Pr (b picked in 2 choices) = 1 - Pr(B not picked of her two choices)
= 1 - (4/5 * 3/4) = 1 - 3/5 = 2/5 = 0.4.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Manager

Joined: 24 Jun 2003

Posts: 150

Location: India

Followers: 1

Kudos [?]: 1 [0], given: 0

Re: Probability 2 [#permalink]

29 Jul 2003, 06:54

bb wrote:

Another way of looking at this problem:

E is already chosen. Now B can be chosen in 2nd choice or third choice.

Number of ways B can come in second choice = 1C1*4C1 = 4
Number of ways B can come in third choice = 4C1*1C1 = 4

Total number of ways in which two subjects can be chosen, one by one, out of the remaining five = 5C1*4C1 = 20

Therefore, the probability that B will be chosen = 8/20 = 0.4

SVP

Joined: 30 Oct 2003

Posts: 1963

Location: NewJersey USA

Followers: 3

Kudos [?]: 25 [0], given: 0

[#permalink]

23 Jan 2004, 08:19

A different approach. Assume he has chosen E and B then only one other subject has to be chosen in 4C1 ways
Total ways to to choose 2 subjects from 5 is 5C2 = 10 ways

P = 4C1/5C2 = 0.4

[#permalink] 23 Jan 2004, 08:19

	Similar topics	Author	Replies	Last post
Similar Topics:	Need an good explanation on this: Never before had	rigger	3	22 Oct 2005, 13:41
	Need explanation	ashkrs	6	26 Dec 2007, 21:25
	need explanation	gconfused	1	26 Mar 2008, 01:02
	Good One! --------------- Jane: Professor Harper s ideas for	mymba99	6	17 Apr 2008, 13:14
	There are 10 women and 3 men in Room A. One person is picked	HarishV	3	12 Aug 2010, 04:52

Good explanation needed for this one Jane has to pick 3

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

Good explanation needed for this one Jane has to pick 3

Good explanation needed for this one Jane has to pick 3

Main navigation

GMAT Resources

Partners

MBA Resources