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A jar contains 3 red and 2 white marbles. Two marbles are picked witho
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01 Mar 2023, 01:05
01 Mar 2023, 01:05
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Question Stats:
37% (01:57) correct
62% (01:22) wrong based on 37 sessions
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A jar contains 3 red and 2 white marbles. Two marbles are picked without replacement.
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability of picking two red marbles |
The probability of picking exactly one red and one white marble |
A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
Re: A jar contains 3 red and 2 white marbles. Two marbles are picked witho
[#permalink]
19 Mar 2023, 11:51
19 Mar 2023, 11:51
Expert Reply
OE
The easiest way to compute the probability in question is through the “1 − x” shortcut. To do so, imagine the opposite of the event of interest, namely, that none of the n throws yields a 6. The probability of a single throw not yielding a 6 is 5/6, and because each throw is independent, the cumulative probability of none of the n throws yielding a 6 is found by multiplication:
P(RR)=3/5*2/4=3/10
Next, consider the probability of picking a red marble followed by a white marble:
P(RW)=3/5*2/4=3/10
However, this is not the only way to pick one red AND one white marble; you could have picked the white one first, followed by the red one:
P(WR)=2/5*3/4=3/10
This event is mutually exclusive from picking a red marble followed by a white marble. Thus, the total probability of picking one red AND one white marble is the sum of the probabilities of RW and WR, yielding an answer of:
P(RW or WR)=3/10+3/10=3/5
A = 3/10
B= 3/5
B>A
The easiest way to compute the probability in question is through the “1 − x” shortcut. To do so, imagine the opposite of the event of interest, namely, that none of the n throws yields a 6. The probability of a single throw not yielding a 6 is 5/6, and because each throw is independent, the cumulative probability of none of the n throws yielding a 6 is found by multiplication:
P(RR)=3/5*2/4=3/10
Next, consider the probability of picking a red marble followed by a white marble:
P(RW)=3/5*2/4=3/10
However, this is not the only way to pick one red AND one white marble; you could have picked the white one first, followed by the red one:
P(WR)=2/5*3/4=3/10
This event is mutually exclusive from picking a red marble followed by a white marble. Thus, the total probability of picking one red AND one white marble is the sum of the probabilities of RW and WR, yielding an answer of:
P(RW or WR)=3/10+3/10=3/5
A = 3/10
B= 3/5
B>A
gmatclubot
Re: A jar contains 3 red and 2 white marbles. Two marbles are picked witho [#permalink]
19 Mar 2023, 11:51
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