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An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that

(i) both the balls are red
(ii) one ball is red and the other is black
(iii) one ball is white.

User Aarona
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1 Answer

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Final answer:

The probability of drawing both red balls is 1/35. The probability of drawing one red and one black ball is 1/7. The probability of drawing one white ball is 8/15.

Step-by-step explanation:

(i) Probability that both balls are red:

The probability of drawing the first red ball is 3/15, as there are 3 red balls out of a total of 15 balls. After drawing the first red ball, there are 2 red balls left out of 14 balls in total for the second draw. Therefore, the probability of drawing the second red ball is 2/14.

To find the probability of both balls being red, we multiply the probabilities of each draw: 3/15 * 2/14 = 1/35.

(ii) Probability that one ball is red and the other is black:

The probability of drawing a red ball at the first draw is 3/15. After drawing a red ball, the probability of drawing a black ball on the second draw is 5/14, as there are 5 black balls left out of 14 balls in total. Since the order of drawing does not matter, we multiply this probability by 2 to account for the possibility of drawing a red ball first and then a black ball, or vice versa: 3/15 * 5/14 * 2 = 1/7.

(iii) Probability that one ball is white:

The probability of drawing a white ball on the first draw is 7/15. The probability of drawing a non-white ball on the first draw is 1 - (7/15) = 8/15. Since we are only interested in the probability of drawing one white ball, we multiply this probability by 2 to account for the possibility of drawing a non-white ball first and then a white ball, or vice versa: 8/15 * 7/14 * 2 = 8/15.

User Mario Werner
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