ajluberto wrote:

Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63

B. 1/6

C. 4/9

D. 809/1260

E. cannot be determined

HELP!!!

This is sooooo easy........just kidding,......but here is my try at explaining this more than solving (as two members have already solved the problem), just so people interested can understand the logic behind. This might not be a GMAT Q, but definitely worth understanding, will help in the GMAT:

4 prime and 5 non-prime numbers in the list

1st selection:1 used and 2 dropped

Total selected = 3 of 9 = 9C3

Selecting 1 prime = 4C1 (since only one marble is placed in the cup and we want this one marble to be a prime)

Selecting 2 non prime = 5C2

P(of selecting one prime and 2 non prime from the jar) = \(\frac{(4C1 * 5C2)}{9C3}\)

Now, out of the 3 selected (i.e. one prime and 2 non-prime) only one is placed in the cup and the other 2 are dropped, hence chance of a prime being placed in the cup is 1/3

P(of selecting one prime and 2 non prime from the jar

& placing the prime in the cup) = \(\frac{(4C1 * 5C2)}{9C3} * \frac{1}{3}\)

2nd selection:2 used and 1 dropped

Total selected = 3 of 6 = 6C3

Selecting 2 prime = 3C2 (since only two marbles is placed in the cup and we want these two marbles to be prime)

Selecting 1 non prime = 3C1

P(of selecting two prime and 1 non prime from the jar) = \(\frac{(3C2 * 3C1)}{6C3}\)

Now, out of the 3 selected (i.e. two prime and 1 non-prime) two are placed in the cup and the other is dropped, hence chance of both prime being placed in the cup is 2/3

P(of selecting two prime and 1 non prime from the jar

& placing both prime in the cup) = \(\frac{(3C2 * 3C1)}{6C3} * \frac{2}{3}\)

3rd selection:1 used and 2 dropped

Total selected = 3 of 3 = 3C3 (only 3 remain)

Selecting 1 prime = 1C1 (only 1 prime remains)

Selecting 2 non prime = 2C2 (only 2 non-prime remain)

P(of selecting one prime and two non prime from the jar) = \(\frac{(1C1 * 2C2)}{3C3}\)

Now, out of the 3 selected (i.e. one prime and two non-prime) one is placed in the cup and the other two are dropped, hence chance of prime being placed in the cup is 1/3

P(of selecting one prime and two non prime from the jar

& placing placing prime in the cup) = \(\frac{(1C1 * 2C2)}{3C3} * \frac{1}{3}\)

P(all 4 marbles in the cup are prime) = \(\frac{(4C1 * 5C2)}{9C3} * \frac{1}{3} * \frac{(3C2 * 3C1)}{6C3} * \frac{2}{3} * \frac{(1C1 * 1C2)}{3C3} * \frac{1}{3}\) = \(\frac{10}{21} * \frac{1}{3} * \frac{9}{20} *\frac{2}{3} * 1 * \frac{1}{3}=\frac{1}{63}\)

_________________

If you analyze enough data, you can predict the future.....its calculating probability, nothing more!