Question

Of all the customers at a gas station, 40% pay by credit card, and the rest use cash. (a) Find the probability that out of 5 random customers, at most 1 uses credit card. 0.375 5 x 0.4 x 0.64 0.69 + 5 x 0.4 x 0.64 0.05184 (b) Knowing that at most 1 out of the 5 customers paid using credit card, what is the conditional probability that all 5 customers used cash payment? Answer: Suppose that the customers leaving the station are interviewed one after the other. (c) What is the probability that the fourth customer was the second who used credit card? (Hint: Look for the number of failures (cash) before the second success (credit card).] Answer: [lf necessary, round your answers to two decimal places.]

Answer 1

Answer a- Option c

pay Cash cell the s cash. Solutions @ In given situation, there are two cases, either 4 pay P(total) = P(4 cash) + P(5 cash) (0.6)"x 10-y)<SCH + 6.6)$ 6.6), ses = 10.6" x 2.6 . option c. (b) PCA) P(Ano) P(B) (0.6) 3. SCS (0.6)"x2.6 0.23 - C) for the fourth th customer to be second one who use credit card, I out of first three should be credit cord user and the fourth one too P : 34,*(0.6)²+(0.4) x 0.4 . = 0.17

B- 0.23

C- 0.17