Jellybeans and probability

[rahulvsd]

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]

[GMATGuruNY]

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.

[[email protected]]

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

[[email protected]]

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/

[LalaB]

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

[ArunangsuSahu]

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