Worked solution
A specified success position and a remaining tail
Customers at a supermarket pay for their purchases either by cash or credit card. The probability of a randomly chosen customer paying by credit card is 0.3. Find the probability that, on a randomly chosen day, there are more than 6 customers among the first 20 customers who pay by credit card and the fifth customer is the second customer who does so.
[2006/ACJC/II/10(i)]
CJC · Discussion · Question 2
Answer
(3 s.f.).
Split the customers
Let X count credit-card payments among the first 4 customers, so . Let Y count credit-card payments among the last 15 customers, so . For the fifth customer to be the second credit-card payer, the first four must contain exactly one success and the fifth must be a success. There are then two successes among the first five, so more than six in total requires Y>4. Under the independent-payment model, the required probability is , giving 0.0598 (3 s.f.).
Common mistakes
The event is a joint event, not a conditional probability; the fifth customer contributes a separate factor 0.3.