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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Probability

by andressa » Wed Feb 02, 2011 6:58 am
Hi, can someone explain to me how to work on this two problems? I tried to work on them but I cant find the right answer.

There are 3 red balls, and 2 black balls in a jar. What is the probability that John gets at least one red ball when he picks two balls at random out of the jar at the same time?
a) 6/25
b) 19/25
c) 1/10
d) 9/10
e) 19/20

In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black.. If John picks 2 balls out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?
a)5/17
b)12/17
c)25/81
d)56/81
e)4/9
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by Geva@EconomistGMAT » Wed Feb 02, 2011 7:15 am
andressa wrote:Hi, can someone explain to me how to work on this two problems? I tried to work on them but I cant find the right answer.

There are 3 red balls, and 2 black balls in a jar. What is the probability that John gets at least one red ball when he picks two balls at random out of the jar at the same time?
a) 6/25
b) 19/25
c) 1/10
d) 9/10
e) 19/20

P(at at least one red) has too many scenarios: red only on first, red only on second, red on both.
On the other end, the only forbidden scenario is "no red" = two blacks. Go for 1-p(forbidden scenario) = 1-p(two blacks) = 1-2/5*1/4 = 1-1/10 = 9/10. Answer is D.

Generally speaking, at least questions lend themselves to the 1-p(forbidden scenario) approach.

In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black.. If John picks 2 balls out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?
a)5/17
b)12/17
c)25/81
d)56/81
e)4/9
Same as above: P(at least one yellow) = 1-p(no yellow)
find forbidden p(not yellow) =this is the same as P(no yellow on first) * p(no yellow on second) = 10/18 * 9/17 = 10/2 * 1/17 = 5/17
So p(at least one yellow) = 1-p(no yellow) = 1-5/17 = 12/17. Answer is B.
Geva
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by andressa » Wed Feb 02, 2011 9:43 am
Thanks a bunch....I was getting confused on the calculation wise!! Argh....sometimes the frustration gets on the way.
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