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A bag contains 4 red and 5 yellow balls, and 3 are successively drawn out and not replaced. What’s the chance of getting different colours alternatively?
5/42
10/63
35/126
1/126
Total number of balls = 9 Let the first drawn ball is red, so required probability =4/9×5/8×3/7 =5/42 But here we had started with a red ball. When we start with a yellow ball, the required probability =5/9×4/8×4/7 =10/63 Since these two cases are mutually exclusive. Total probability = 5/42 + 10/63 =( 15 + 20 )/126 =35/126
By: bhavesh kumar singh ProfileResourcesReport error
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