A jar contains four coins: a nickel, a dime, a quarter, and a half-dollar. Three coins are randomly selected from the jar.
a. List the simple events in S. b. What is the probability that the selection will contain the half-dollar?c. What is the probability that the total amount drawn will equal 60¢ or more?

a.

Let N represent Nickel

D represent Dime

Q represent Quarter

H represent Half-Dollar

E1 = (NDQ)

E2 = (NDH)

E3 = (NQH)

E4 =  (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 =  (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)  

=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  

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

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation:

no

step-by-step explanation: