A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted marbles?
If a bowl contains only white and black balls, what is the probability of extracting a white ball from the bowl? (1) There are twice as many white balls as black balls in the bowl (2) If two balls are extracted from the bowl, the probability that both of them will be black is 0
probabilty problem
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Total 9 marbles.
1* 6/8 * 3/7 = 18/56
1* 6/8 * 3/7 = 18/56
sukh wrote:A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are extracted from the basket at random, what is the probability that a marble of each color is among the extracted marbles?
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A IMO
If P(2 blacks) = 0, then we know there is only 1 black ball present, we dont know how many whites are present
From (1), we know ratio between Black and White and hence sufficient,
If P(2 blacks) = 0, then we know there is only 1 black ball present, we dont know how many whites are present
From (1), we know ratio between Black and White and hence sufficient,
sukh wrote:
If a bowl contains only white and black balls, what is the probability of extracting a white ball from the bowl? (1) There are twice as many white balls as black balls in the bowl (2) If two balls are extracted from the bowl, the probability that both of them will be black is 0
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You have a total of 9 balls here (3 colors & 3 of each color)
You can first pick any ball from this 9. So that is 9/9 choices
Say you have picked up a blue in the first turn, for the second you don't want to get a blue again, so that leaves you with only 6 choices out of 8 (1 already taken) So thats 6/8 choices
For the 3rd, you don't want to get the colors you have already got in the first 2 turns, so that leaves you with only 3 choices of the remaining 7.
So combining we have (9/9)*(6/8)*(3/7)
You can first pick any ball from this 9. So that is 9/9 choices
Say you have picked up a blue in the first turn, for the second you don't want to get a blue again, so that leaves you with only 6 choices out of 8 (1 already taken) So thats 6/8 choices
For the 3rd, you don't want to get the colors you have already got in the first 2 turns, so that leaves you with only 3 choices of the remaining 7.
So combining we have (9/9)*(6/8)*(3/7)
sukh wrote:plz explain 1st problem in more detail by giving explanation