Probability

[talaangoshtari]

A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. What is the probability that the second ball drawn will be red if replacement is NOT allowed?

A. 1/36
B. 1/12
C. 7/36
D. 2/9
E. 7/9

[GMATGuruNY]

A bag contains 3 white, 4 black, and 2 red marbles. Two marbles are drawn from the bag. What is the probability that the second ball drawn will be red if replacement is NOT allowed?

A. 1/36
B. 1/12
C. 7/36
D. 2/9
E. 7/9

No math is needed here if we understand the following concept:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(red on the 2nd pick) = P(red on the first pick) = 2/9.

The correct answer is D.

Other problems that test this concept:
https://www.beatthegmat.com/beat-this-pr ... 85719.html
https://www.beatthegmat.com/probablity-ques-t60161.html
https://www.beatthegmat.com/manhattan-pr ... 89481.html (2 posts)
https://www.beatthegmat.com/a-box-contai ... 51368.html

[prachi18oct]

Hi GMATGuruNY,

Will the probability of picking the second ball as red not change if the first ball picked is also red?
I did it as follows:-
P(second ball as red) = Prob of first red ball * prob of second red ball + prob of first non-red ball * prob of second ball as red
=> 2/9 * 1/8 + 7/9 * 2/8 = 16/72 = 2/9

Please advise.

[GMATGuruNY]

I did it as follows:-
P(second ball as red) = Prob of first red ball * prob of second red ball + prob of first non-red ball * prob of second ball as red
=> 2/9 * 1/8 + 7/9 * 2/8 = 16/72 = 2/9

Your solution is perfect.

Will the probability of picking the second ball as red not change if the first ball picked is also red?

Yes.
As your solution illustrates, P(RR) ≠ P(NR):
P(RR) = (2/9)(1/8) = 2/72.
P(NR) = (7/9)(2/8) = 14/72.

However, the SUM of these probabilities is equal to the probability of selecting a red marble on the first pick:
2/72 + 14/72 = 16/72 = 2/9.

As your solution proves, if we account for ALL of the ways to select a red marble on the NTH pick, the result will be equal to the probability of selecting a red marble on the FIRST pick:
P(red marble on the nth pick) = P(red marble on the 1st pick) = 2/9.

[Brent@GMATPrepNow]

This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).

So, someone would hold up n pieces of grass (for n people), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n people had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.

The same applies to the original question here. 2 of the 9 balls are blue, so P(red ball selected second) = 2/9

Cheers,
Brent

[nikhilgmat31]

I did it in long way
selecting a White Ball at first & selecting red in second OR selecting red ball at first & selecting red in second OR selecting BLUE ball at first & selecting red in second

3/9 * 2/8 = 6/72
2/9 * 1/8 = 2/72
4/9 * 2/8 = 8/72

summing all 3 gives = 16/72 2/9

It is same as doing selecting non red & then selecting a red ball OR selecting RED ball and then selecting RED ball.

7/9 * 2/8 + 2/9 * 1/8

16/72 = 2/9