Question

question 9
determine which two events are mutually exclusive:
a bag contains six yellow jerseys numbered 1 to 6. the bag also contains four purple jerseys numbered 1 to 4. you randomly pick one jersey from the bag.
event a is that your jersey is yellow
event b is that your jersey has number 3
event c is that your jersey has number 5
b and c are mutually exclusive
a and c are mutually exclusive
a and b are mutually exclusive
none of the events are mutually exclusive
all of the events are mutually exclusive
question 10
according to some internet research, 83.8% of adult americans have some form of medical insurance, and 77.4% of adult americans have some form of dental insurance. if 88.2% of adult americans have either medical or dental insurance, then what is the probability that a randomly selected adult american with have both medical and dental insurance?
0.730
0.612
0.233
0.649
0.946
0.818

Explanation:

Question 9

Mutually exclusive events cannot occur at the same time. Event A (yellow jersey) can have number 3 or 5 (since yellow jerseys are numbered 1 - 6). Event B (number 3) and Event C (number 5) can't happen together (a jersey can't have both 3 and 5). Event A and B: a yellow jersey can have number 3 (so not mutually exclusive). Event A and C: a yellow jersey can have number 5 (so not mutually exclusive). So B and C are mutually exclusive.

Step1: Recall the formula for the probability of the union of two events: $P(M \cup D) = P(M) + P(D) - P(M \cap D)$, where $M$ is medical insurance and $D$ is dental insurance.

Step2: Rearrange the formula to solve for $P(M \cap D)$: $P(M \cap D) = P(M) + P(D) - P(M \cup D)$.

Step3: Substitute the given values: $P(M) = 0.838$, $P(D) = 0.774$, $P(M \cup D) = 0.882$. So $P(M \cap D) = 0.838 + 0.774 - 0.882$.

Step4: Calculate the result: $0.838 + 0.774 = 1.612$; $1.612 - 0.882 = 0.73$.

Answer:

B and C are mutually exclusive

Question 10