A bag contains 8 balls from which 5 are red and 3 are white. If a man selects two balls random then what is the probability of getting each color of ball
Answers
Answer:
Here's sort of a non-formula way of looking at this problem.
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I am going to assume that when the first ball is selected, it is not returned to the bag.
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There are two possibilities of success in this problem, success being defined as selecting one ball of each color. If the first ball selected is red, then the second ball selected must be white. Or the second possibility is that the first ball selected is white, and the second ball selected must therefore be red.
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When the first ball is selected, since there are 8 balls total and 5 of them are red, the probability of it ball being red is 5/8. This leaves 7 balls in the bag, 3 of which are white. Therefore, on the second selection there is a 3/7 probability of it being white. Multiply these two probabilities (5/8 times 3/7) and you get a 15/56 possibility of a red ball followed by a white ball.
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But when the first ball is selected, since there are 8 balls total and 3 of them are white, the possibility of 3/8 that a white ball is selected. This again leaves 7 balls in the bag, 5 of which are red. Therefore, there is a 5/7 probability of the second selection being red. Multiply these two probabilities (3/8 times 5/7) and you again get a 15/56 probability of a white ball followed by a red ball.
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Add these two probabilities together any you get a 30/56 probability of two selections resulting in drawing one ball of each colour. (If you divide 56 into 30 and convert to percent, you get a 53.57 percent chance of your two selections resulting in one ball of each colour.)
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What are the chances of selecting two red balls with your two selections? Following the same procedures, 5/8 chance of red on the first selection and 4/7 chance of a red ball on the second selection. Multiply those two probabilities and you get a 20/56 chance of getting two red balls.
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What are the chances of selecting two white balls? Again 3/8 chance of a white ball on the first selection followed by a 2/7 chance of a white ball on the second selection. Multiplying those together you get a 6/56 chance of drawing two white balls. That covers all the possibilities: a red ball followed by a white ball (15/56), a white ball followed by a red ball (15/56), a red ball followed by another red ball (20/56), and a white ball followed by another white ball (6/56). Add all the probabilities together (15/56 + 15/56 + 20/56 + 6/56) and you get 56/56 or 1. This is a good check because it ensures that there is a 100% chance of one of these possibilities (red-white, white-red, red-red, and white-white) happening.
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Hope this helps you to understand the problem a little better. Again, the answer you were looking for is 30/56 or 53.57% is the probability of one white and one red ball on two random selections without replacement.
**13/28**
Number of ways of selecting two balls from 5 + 3 = 8 balls is 8C2 = 8*7/2 = 28
Number of ways of selecting two white balls from 5 white balls = 5C2 = 5*4/2 = 10
Number of ways of selecting two black balls from 3 black balls = 3C2 = 3*2/2 = 3
Favorable ways = 10 + 3 = 13
**Required probability = Favorable ways / Total ways = 13/28**