Question

given p(a∪b)=0.4 and p(a∪b)=0.7. determine p(a)
choose one of the following.
a. p(a)=0.3
b. p(a)=0.7
c. p(a)=0.6
d. p(a)=0.1
e. p(a)=0.4

Explanation:

Step1: Use the formula for the probability of the union

We know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and $P(A\cup B') = P(A)+P(B')-P(A\cap B')$. Also, $P(B') = 1 - P(B)$ and $P(A)=P(A\cap B)+P(A\cap B')$.
We can rewrite $P(A\cup B)$ as $P(A)+P(B)-P(A\cap B)=0.4$ and $P(A\cup B')$ as $P(A)+(1 - P(B))-P(A\cap B')=0.7$.

Step2: Add the two - equations

$(P(A)+P(B)-P(A\cap B))+(P(A)+1 - P(B)-P(A\cap B'))=0.4 + 0.7$.
Simplify the left - hand side: $2P(A)+1-(P(A\cap B)+P(A\cap B'))=1.1$.
Since $P(A)=P(A\cap B)+P(A\cap B')$, we substitute to get $2P(A)+1 - P(A)=1.1$.

Step3: Solve for $P(A)$

Combine like terms: $2P(A)-P(A)=1.1 - 1$.
So, $P(A)=0.1$.

Answer:

D. $P(A)=0.1$