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# Three numbers are chosen at random with replacement from {1,2,3....8}. the probability that their minimum is 3 and the maximum is 6,isa) 3/8b) 1/5c) 1/4d)  2/5

Asked by ppavani198 18th December 2018, 9:32 PM
In the question it is given " with replacement ".  This is a standard question in probability but without replacement.

Any how I answered as per the question asked, i.e. "with replacement"

Total number of possibilities of selecting 3 numbers at random from { 1,2,3...8} is 83

possible event for getting 3 = { 111 } ; probaility of getting  3= 1/83

possible events to get  4 = { 112, 121, 211 }; probability of getting 4 = 3/83

possible events to get  5 = { 122, 113, 131, 212, 221,  311 }; probability of getting 5 = 6/83

possible events to get  6 = { 222, 123, 132, 231, 213, 312, 321 }; probability of getting 6 = 7/83

Hence probability of getting minimum 3 and maximum 6 is = (1+3+6+7)/83
Answered by Expert 19th December 2018, 8:31 AM
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Tags: probability