probability

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klaud: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Sat Jan 28, 2012 8:22 amFollowed by: 1 member

by klaud » Thu Feb 16, 2012 9:32 am

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Source: Beat The GMAT — Problem Solving | Thu Feb 16, 2012 9:32 am

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Thu Feb 16, 2012 11:02 am

1st quest.
total outcomes 7C2 (select 2 people out of 7 people)
favorable outcomes 4C2 (select 2 women out of 4 women possible)

The required probability=favorable outcomes/total outcomes=4C2/7C2=6/21/=2/7

answer [spoiler]2/7[/spoiler]

2nd quest [spoiler]0.5*0.5=0.25[/spoiler]

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Feb 16, 2012 8:46 pm

klaud wrote:

Explanation to Q1.

Total # of people = 4 + 3 = 7
Number of ways of choosing 2 people from 7 = 7C2 = 7!/[(7 - 2)! * 2! = 7!(5! * 2!) = 21
Number of ways of choosing 2 women from 4 women = 4C2 = 4!/[(4 - 2)! * 2!] = 4!/(2! * 2!) = 6
Required probability = 4C2/7C2 = 6/21 = [spoiler]2/7[/spoiler]

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Feb 16, 2012 8:46 pm

klaud wrote:

Explanation to Q2.

Probability that coin will turn up head when tossed = 0.5
Since the coin is tossed two times, so the probability that both tosses will turn up heads = 0.5 * 0.5 = 0.25

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LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Sat Feb 18, 2012 12:37 pm

1. 4C2/7C2=2/7 OR 4/7*3/6=2/7

2. 0.5*.05=0.25 OR 2C2/2^2=1/4=0.25

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