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Jellybeans

by dlencz » Tue Apr 17, 2012 3:37 pm
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

OA is A
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by killer1387 » Tue Apr 17, 2012 5:48 pm
dlencz 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

OA is A
probability for picking one green = x/2x = 1/2 = probability for picking one red

number of ways in which 3 green and one red can be chosen is 4!/3!=4
required probability = 4*(1/2)^3*(1/2) =4/16 = 1/4

hence A

Hope this helps..!!
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by aneesh.kg » Tue Apr 17, 2012 9:11 pm
Let the amounts of Green and Red Jellybeaans be X each.

Probability of selecting a Green Jellybean = XC1/2XC1 = X/2X
Probability of selecting a Green Jellybean = XC1/2XC1 = X/2X

Then, the required probability should have been (X/2X).(X/2X).(X/2X).(X/2X) had there been a specific order, say (G,G,G,R) but since the order doesn't matter, we'll have to arrange (G,G,G,R) in 4!/3! ways also.

So, the answer should be = (X/2X).(X/2X).(X/2X).(X/2X).(4!/3!)
= 1/4

(A) is the answer
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