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four tickets 00.01,10,11 respectively are placed ina bag . aticket is drawn at random five times being replaced each time the probability that the sum on the tickets is 22 is
Dear Phani
There are 3 cases to get sum 22
1) 11+11+00+00+00
It can be done with prob of (5!/(3!*2!))*(1/4)^5
2)10+01+10+01+00
It can be done with prob of (5!/(2!*2!*1!))*(1/4)^5
3)11+10 +01+00+00
It can be done with prob of (5!/(2!*1!*1!*1!))*(1/4)^5
So total probability is sum of three which is 25/256
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