Chris and John are taking a driving test for the first time. The probability tha

Question

Chris and John are taking a driving test for the first time. The probability that Chris passes the test is 2/3, and the probability that John passes the test is 3/5. Assume that the results for Chris and John are independent. Find:

a. The probability that both pass the test. (3 points)

b. The probability that at least one of them passes the test. (3 points)

c. The probability that none of them pass the test. (3 points)

d. The probability that Chris passes but John fails. (3 points)

e. The probability that only one of them (either Chris or John) passes the test. (3 points)

3. Chris and John are taking a driving test for the first time. The probability that Chris passes the test is 2/3, and the probability that John passes the test is 3/5. Assume that the results for Chris and John are independent. Find: a. The probability that both pass the test. (3 points) b. The probability that at least one of them passes the test. (3 points) c. The probability that none of them pass the test. (3 points) d. The probability that Chris passes but John fails. (3 points) e. The probability that only one of them (either Chris or John) passes the test. (3 points)

Explanation / Answer

we denote the prob of chris to pass as P(C)=2/3 and he falis as P(C')=1/3 (i.e. 1-(2/3))

and same way as John as P(J)=3/5 and P(J')=2/5 (ie 1-3/5).

Here remember an important assumption ie the results for Chris and John are independent

a)The probability that both pass the test = P(C and J) =P(C)*P(J)= 2/3*3/5 =2/5=0.4

b)The probability that atleast one of them passes i.e P(C or J) = P(C)*P(J')+P(J)*P(C')+P(C)*P(J)=2/3*2/5+3/5*1/3+2/3*3/5 =4/15+1/5+2/5 = 13/15= 0.86667 (3 cases 1: only C passes case 2: only J passes case 3: both passes )

c) The probability that none of them pass the test = P(C')*P(J') = 1/3*2/5=2/15=0.13333 (or it can be 1- the prob of above case)

d) The probability that Chris passes but John fails = P(C)*P(J') =2/3*2/5=4/15=0.266667

e)The probability that only one of them (either Chris or John) passes the test = P(C)*P(J')+P(C')*P(J)=2/3*2/5+1/3*3/5 = 0.466667

Please rate the ans if it has really helped you. Good luck!!

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.