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Alex Ar [27]
3 years ago
10

If A and B are independent events with P(A) = .5 and P(B) = .2, find the following:a) P(A U B)b) P(A^c ? B^c)c) P(A^c U B^c)**No

te A^c or B^c refers to A complement or B complement**
Mathematics
1 answer:
Effectus [21]3 years ago
4 0

If A and B are independent, then P(A\cap B)=P(A)P(B).

a.

P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)

P(A\cup B)=0.5+0.2-0.5\cdot0.2

\boxed{P(A\cup B)=0.6}

b. I'm guessing the ? is supposed to stand for intersection. We can use DeMorgan's law for complements here:

P(A^c\cap B^c)=P(A\cup B)^c=1-P(A\cup B)

P(A^c\cap B^c)=1-0.6

\boxed{P(A^c\cap B^c)=0.4}

c. DeMorgan's law can be used here too:

P(A^c\cup B^c)=P(A\cap B)^c=1-P(A\cap B)=1-P(A)P(B)

P(A^c\cup B^c)=1-0.5\cdot0.2

\boxed{P(A^c\cup B^c)=0.9}

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A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after
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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: P(A)=P(A|B)P(B)+P(A|B')P(B')

Use the law of total probability to determine the probability of a parcel being late.

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

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:

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615

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:

P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192

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:

P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

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