Suppose that A and B are independent events such that P(A)=0.10 and P(\bar{B})=0.20 Find P(A\cap B) and P(A\cup B)

Khaleesi Herbert 2021-05-07 Answered
Suppose that A and B are independent events such that \(P(A)=0.10\) and \(P(\bar{B})=0.20\)
Find \(P(A\cap B)\) and \(P(A\cup B)\)

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Expert Answer

wheezym
Answered 2021-05-08 Author has 23219 answers
Step 1
The value of \(P(A\cap B)\) is obtained below:
From the given information, the events A and B are independent events and the probability values are
\(P(A)=0.10\), and the value of \(P(B)\) is,
\(P(B)=1-P(\bar{B})\)
\(=(1-0.20)\)
\(P(B)=0.80\)
The required probability is,
\(P(A\cap B)=P(A)\times P(B)\)
\(=0.10\times0.80\)
\(=0.08\)
The value of \(P(A\cap B)\) is \(0.08\)
The value of \(P(A\cap B)\) is obtained by taking the product of probability of event A and the probability of event B. It can be expected that \(2\%\) of the event A and B occurs.
Step 2
The value of \(P(A\cup B)\) is obtained as shown below:
The probability is,
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(=0.10+0.80-0.08\)
\(=0.90-0.08\)
\(=0.82\)
The value of \(P(A\cup B)\) is \(0.82\).
The value of \(P(A\cup B)\) is obtained by adding the individual probabilities and then subtracting the probability of A and B to the resulted value. It can be expected that about \(82\%\) of times event A or B happens.
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