An urn contains three red balls numbered 1, 2, 3, three white balls numbered 4, 5, 6, and five black balls numbered 7, 8, 9, 10, 11. A ball is drawn from the urn. What is the probability that it is red or odd-numbered?

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7/11
BECAUSE:first of all there are three red balls. It doesnt matter if its odd, as long as it is red, it should be included in the numerator.second there is only one odd numbered white ball, which is five. Therefore there is only one white ball that will be included to the numerator.third there are 3 odd numbered black balls, which are 7, 9, and 11. Therefore there are three black balls that will be included in the numerator.so the sum of all the numbers that will be included in the numerator equals 7. (3+1+3=7)then you add all of the balls together, no matter the color or the number.there are three red balls, three white balls, and five black balls. The sum of all those numbers are 11, which will be the denominator therefore the answer would be 7/11

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