Jellybeans and probability

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rahulvsd: Master | Next Rank: 500 Posts; Posts: 184; Joined: Tue Sep 07, 2010 9:43 amFollowed by: 1 member; Thanked: 6 times

by rahulvsd » Sun Nov 06, 2011 5:49 am

A jar contains equal amounts of green jellybeans and red jellybeans. Mark selects at random one jellybean at a time from the jar, notes its color, and returns it to the jar. If he repeats this process four times, what is the probability that he selects 3 green jellybeans and 1 red jellybean?

A) 1/4
B) 3/16
C) 1/8
D) 1/16
E) 1/64

[spoiler]OA:A[/spoiler]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Nov 06, 2011 6:07 am

rahulvsd wrote:A jar contains equal amounts of green jellybeans and red jellybeans. Mark selects at random one jellybean at a time from the jar, notes its color, and returns it to the jar. If he repeats this process four times, what is the probability that he selects 3 green jellybeans and 1 red jellybean?

A) 1/4
B) 3/16
C) 1/8
D) 1/16
E) 1/64

[spoiler]OA:A[/spoiler]

P(exactly N times) = P(one way) * total possible ways.

One way to get a good outcome:
The first jellybean is red; the other three are green.
P(RGGG) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16.

Total possible ways to get a good outcome:
Since R could be the 1st, 2nd, 3rd, or 4th jellybean selected -- implying 4 ways to get exactly one R -- the result above must be multiplied by 4.

4 * 1/16 = 1/4.

The correct answer is A.

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Sat Aug 27, 2011 8:50 pm

by [email protected] » Tue Jan 24, 2012 7:12 pm

I did not get the part where we are multiplying 1/16 by 4. How do we know that order is important here ?

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[email protected]: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Mon Dec 21, 2009 6:27 am; Location: Melbourne, Australia; Thanked: 17 times

by [email protected] » Tue Jan 24, 2012 8:18 pm

[email protected] wrote:I did not get the part where we are multiplying 1/16 by 4. How do we know that order is important here ?

We also need to find the number of ways the arrangment RGGG can be done. Since green & red jellybeans can be picked up in any of the following way,
RGGG
GRGG
GGRG
GGGR
if we consider only 1/16 we would be considering one of the above 4 cases, which would be wrong.

Another way to determine the no. of possible arrangement of RGGG is 4!/3!, where 3! denotes arrangment of all G s among themselves/

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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 » Wed Jan 25, 2012 2:48 am

one more good explanation is given by Stuart https://www.beatthegmat.com/coin-flip-qu ... html#75414

since in the q.stem it is said about the selection with replacement, it is obvious, that the chances to get red or green are 1/2.

using this formula nCk/2^n , we can get 4C3/2^4=4/16=1/4

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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Wed Jan 25, 2012 11:53 am

This can happen in 4C3/2^4=1/4

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