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98. 7 white balls and 3 black ball are placed in a rowat random. The probability that no two black ballsare adjacent is |
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Answer» neet balls = 7+3=10let, 2 black balls =1 black balls now, 9 balls er placed in them is : 9! then the probability that two black balls are adjacent is (9!/10!)= (1/10)now, the probability that no two black balls are adjacent is {1-(1/10)}=(9/10) Placing 7 white balls in a row leaves 8 gaps. The 3 black balls can be placed in 8 gaps in 8C3 = 8×7×6/3×2×1= 56 ways. So, the total number of ways of arranging white and black balls such that no two black balls are adjacent = 56*3!*7! Actual number of arrangements possible with 7 white and 3 black balls = (7+3)! = 10! Therefore probability that no two black balls are adjacent = 56∗3!∗7!/10!=7/5∗3=7/15 |
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