Question

1. (4 points) let a and b be events. suppose that the probability that a does not occur is .6, the probability that b occurs is .5, and the probability that either a or b occurs is .7. (a) find the probability that both a and b occur. (b) find the probability that a occurs and b does not occur. 3) (4 points) (a) is it possible to have two events a and b such that p(a) = .6, p(b) = .8, and p(ab) = .3? explain why. (b) is it possible to have two events a and b with the following assignments of probability: p(a) = .6 and p(ab) = .7? explain why.

Explanation:

Step1: Find $P(A)$

Given $P(\text{not }A)=0.6$, then $P(A)=1 - P(\text{not }A)=1 - 0.6 = 0.4$.

Step2: Use the formula for $P(A\cup B)$

The formula is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. We know $P(A\cup B) = 0.7$, $P(A)=0.4$ and $P(B)=0.5$. Substitute these values into the formula: $0.7=0.4 + 0.5-P(A\cap B)$.

Step3: Solve for $P(A\cap B)$

Rearrange the equation from Step 2: $P(A\cap B)=0.4 + 0.5-0.7=0.2$.

Step4: Find $P(A\cap\text{not }B)$

We know that $P(A)=P(A\cap B)+P(A\cap\text{not }B)$. So $P(A\cap\text{not }B)=P(A)-P(A\cap B)$. Substitute $P(A) = 0.4$ and $P(A\cap B)=0.2$ to get $P(A\cap\text{not }B)=0.4 - 0.2=0.2$.

Step5: For 3(a)

Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Substitute $P(A)=0.6$, $P(B)=0.8$ and $P(A\cap B)=0.3$. Then $P(A\cup B)=0.6 + 0.8-0.3=1.1$. But probability $P(A\cup B)$ must satisfy $0\leq P(A\cup B)\leq1$. Since $1.1>1$, it is not possible.

Step6: For 3(b)

Since $A\cap B\subseteq A$, we have $P(A\cap B)\leq P(A)$. Given $P(A)=0.6$ and $P(A\cap B)=0.7$, and $0.7>0.6$, it is not possible.

Answer:

2(a). $0.2$
2(b). $0.2$
3(a). It is not possible because $P(A\cup B)=1.1>1$.
3(b). It is not possible because $P(A\cap B)=0.7 > P(A)=0.6$.