SOLUTION: A box contains five tickets, with letters A,B,C,D,E. Seven draws will be made ar random with replacement. Find to the nearest percent, the probability that exactly four out of th

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: A box contains five tickets, with letters A,B,C,D,E. Seven draws will be made ar random with replacement. Find to the nearest percent, the probability that exactly four out of th      Log On

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Question 437781: A box contains five tickets, with letters A,B,C,D,E.
Seven draws will be made ar random with replacement.
Find to the nearest percent, the probability that exactly four out of the seven draws will be the B.

Answer by ewatrrr(10682) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

A box contains 'five' tickets, with letters A,B,C,D,E. 

Seven draws will be made ar random with replacement.

 P(exactly 4 out of seven will be the B) = (.2)^4(.8)^3