Answer:
(a) The probability that a randomly selected parcel arrived late is 0.026.
(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.
(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.
(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.
Step-by-step explanation:
Consider the tree diagram below.
(a)
The law of total probability sates that: [tex]P(A)=P(A|B)P(B)+P(A|B')P(B')[/tex]
Use the law of total probability to determine the probability of a parcel being late.
[tex]P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026[/tex]
Thus, the probability that a randomly selected parcel arrived late is 0.026.
(b)
The conditional probability of an event A provided that another event B has already occurred is:
[tex]P(A|B)=\frac{P(B|A)P(A)}{P(B)}[/tex]
Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:
[tex]P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615[/tex]
Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.
(c)
Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:
[tex]P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192[/tex]
Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.
(d)
Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:
[tex]P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192[/tex]
Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.
