Question

3. Anna has a large bag of coins. After counting the coins, she recorded the counts in the table below. She then decided to draw some coins at random, replacing each coin before the next draw.QuartersNickelsDimesPennies22271118(a) What is the probability that Anna obtains a quarter on the first draw? (b) What is the probability that Anna obtains a penny or a dime on the second draw? (c) What is the probability that Anna obtains at most 10 cents worth of money on the third draw? (d) What is the probability that Anna does not get a nickel on the fourth draw? (e) What is the probability that Anna obtains at least 10 cents worth of money on the fifth draw?

Answer 1

(a) P = 0.2821

(b) P = 0.3718

(c) P = 0.7179

(d) P = 0.6538

(e) P = 0.4231

Step-by-step explanation:

There is a total of 78 coins in the bag because:

22 + 27 + 11 + 18 = 78

From those 78 coins, 22 are quarters, so the probability that Anna obtains a quarter on the first draw is:

P = 22/78 = 0.2821

Since Anna replaces each coin before the next draw, the probability for each draw is independent. Therefore, the probability that Anna obtains a penny or a dime on the second draw is calculated as:

P = (18 + 11)/78 = 0.3718

Because there are 18 pennies and 11 dimes in the bag.

In the same way, Anna will obtain at most 10 cents worth of money if she obtains a penny, a dime, or a nickel. So, the probability is equal to:

P = (27 + 11 + 18)/78 = 0.7179

The probability that Anna does not get a nickel on the fourth draw is:

P = (22 + 11 + 18)/78 = 0.6538

Finally, Anna obtains at least 10 cents if he obtains a dime or a quarter, so the probability is:

P = (22 + 11)/78 = 0.4231

Therefore, the answers are:

(a) P = 0.2821

(b) P = 0.3718

(c) P = 0.7179

(d) P = 0.6538

(e) P = 0.4231