Answer:
a.
Let N represent Nickel
D represent Dime
Q represent Quarter
H represent Half-Dollar
E1 = (NDQ)
E2 = (NDH)
E3 = (NQH)
E4 = Â (DQH).
Then, S = {E1, E2, E3, E4}
b. 3/4
c. 3/4
Step-by-step explanation:
a. Simple events are those ones that happens at one time and have a single outcome. If the event is denoted by F, then the probability is denoted by P(F), and is usually between 0 and 1. In this light, the simple events in the above are;
Let N represent Nickel
D represent Dime
Q represent Quarter
H represent Half-Dollar
E1 = (NDQ)
E2 = (NDH)
E3 = (NQH)
E4 = Â (DQH).
Then, S = {E1, E2, E3, E4}
b. Probability that the selection will contain the half-dollar = Probability that a nickel, a dime, and a half-dollar is selected + Probability that a nickel, a quarter, and a half-dollar is selected + Probability that a dime, a quarter, and a half-dollar is selected
P(selection will contain the half-dollar) = P(E2) + P(E3) + P(E4) Â
=1/4 + 1/4 + 1/4 = 3/4
c. Probability that the total amount drawn will equal 60¢ or more
Nickel (N) = 5 cents
Dime (D) = 10 cents
Quarter (Q) = 25 cents
Half-Dollar (H) = 50 cents
Hence we look at the simple events and their monetary values:
E1 = NDQ = 40 cents Â
E2 = NDH = 65 cents
E3 = NQH = 80 cents
E4 = DQH = 85 cents
Hence,
P(total amount is 60 cents or more) = P(E2) + P(E3) + P(E4)
= 1/4 + 1/4 + 1/4 = 3/4