Question

Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (c) Are A and B independent? Explain.

Answer 1

Answer:

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

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

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

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$