Over a very long period of time, it has been noted that on Friday's 25% of the customer at the drive in window at the bank make a deposit. What is the probability that it takes 4 customers at the drive in window before the first one makes a deposit
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As I understand the question, the 1st, 2nd, and 3rd customers do not make a deposit, and 4th one makes a deposit. If it is the fifth customer to make a deposit, then you can change accordingly.
Probability that a customer makes a deposit = 25% = 1/4.
Probability that a customer does not make a deposit = 3/4
We assume that one customer making a deposit is not dependent on another making a deposit.
Probability needed = 3/4 * 3/4 * 3/4 * 1/4 = 27/256
Probability that a customer makes a deposit = 25% = 1/4.
Probability that a customer does not make a deposit = 3/4
We assume that one customer making a deposit is not dependent on another making a deposit.
Probability needed = 3/4 * 3/4 * 3/4 * 1/4 = 27/256
kvnmurty:
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Can u work this using geometric progression
it is probability. it is done using probabilities of events. why use geomtric progression?
Prob (first one makes a deposit) = 1/4. ie., n = 1, P(n=1) = T1 = 1/4. T2= 3/4*1/4 = 3/16; T3 = Prob(3rd customer makes a deposit) = 3/4*3/4*1/4; so it is a GP with ratio 3/4 and first term = 1/4.
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