This question is solved using probability concepts. We derive the probabilities from the tree given in the exercise, and with this, added to the use of conditional probability, we get the desired probabilities.
The probabilities are:
P(A) = 0.7, P(A and B) = 0.14, P(B) = 0.26, P(A or B) = 0.82, P(not B given A) = 0.8
Conditional probability:
In this problem, conditional probability concepts are used, and for this, we have that:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
P(A)
At the first node, we have that:
P(A) = 0.7
P(A and B):
From the first node, we have that [tex]P(A) = 0.7[/tex]
From A to B, there is 0.2, which means that [tex]P(B|A) = 0.2[/tex]
Thus
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
[tex]0.2 = \frac{P(A \cap B)}{0.7}[/tex]
[tex]P(A \cap B) = 0.7*0.2 = 0.14[/tex]
So
P(A and B) = 0.14
P(B):
P(B) = P(A and B) + P(not A and B).
P(not A) = 0.3, P(B|not A) = 0.4, then:
[tex]P(not A and B) = 0.3*0.4 = 0.12[/tex]
P(B) = 0.14 + 0.12 = 0.26
P(A or B)
We have that:
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
We already have the three of them, so just replace:
[tex]P(A \cup B) = 0.7 + 0.26 - 0.14 = 0.82[/tex]
Then
P(A or B) = 0.82
P(not B given A)
If A happens, either B happens, or it does not. That is:
P(B|A) + P(not B|A) = 1
Since P(B|A) = 0.2
P(not B|A) = 1 - 0.2 = 0.8
Then
P(not B given A) = 0.8
To take another look at conditional probability, you can check https://brainly.com/question/24161830
