Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

John has coins in a jar. Five of the coins are pennies, 8 are dimes, and 3 are
quarters. John randomly selects a coin from the jar and without replacing it
randomly selects another coin. What is the probability that he selects a dime and
then a quarter?


Sagot :

Answer:

Step-by-step explanation:

X = 10 → 2 nickels

P(1st Nickel) = 5/16

P(2nd Nickel) = 4/15

P(X = 10) = P(2 nickels) = (5/16)*(4/15) = 20/240 = 0.08333 (8.333%)

The probability of selecting 2 nickels ($0.10) is 0.08333.

X = 11 → dime and penny

P(Dime) = 3/16

P(Penny) = 8/15

P(X = 11) = P(Dime and Penny) = (3/16)*(8/15) = 24/240 = 0.1 (10%)

You could also choose penny first and then a dime. This will give you the same probability.

P(Penny) = 8/16

P(Dime) = 3/15

P(X = 11) = P(Penny and Dime) = (8/16)*(3/15) = 24/240 = 0.1 (10%)

P(Dime and Penny) + P(Penny and Dime) = 0.2 (20%)

The probability of selecting a dime and a penny no matter the order is 0.2.

Total Value = 8(0.01) + 5(0.05) + 3(0.10) = 0.08 + 0.25 + 0.30 = $0.63

There are a total of 16 coins and you selected 2 coins from the jar. So the probability of selecting 2 coins from the jar is 2/16 or 0.125.

E[X] = ($0.63)*(2/16) = $0.07875

The expected value is about 8 cents.

1/8 or .125 is the answer