Question

each labeled with the numbers 1 to 6. the first column shows the possible outcomes for the first number cube, and the first row shows the possible outcomes of the second number cube. the ordered pair (3,4) indicates a roll of 3 on the first number cube and a roll of 4 on the second number cube. what is the probability of rolling an odd number on the first number cube and an even number on the second number cube? what is the probability that both spinners land on a 1, ( p(k) )? ( p(k)=) evan has 6 quarters, 4 dimes, 3 nickels, and 8 pennies in his pocket. if he randomly removes 3 coins from his pocket and replaces the coin each time, what is the probability of choosing a quarter, then a nickel, then a dime, ( p(c) )? ( p(c)=)

Explanation:

Step1: Calculate the total number of outcomes for the first spinner

The first spinner has 6 sections. So, the total number of outcomes for the first spinner \(n_1 = 6\).

Step2: Calculate the number of favorable outcomes for the first spinner to land on 1

The first spinner has 1 section labeled 1. So, the number of favorable outcomes for the first spinner \(m_1=1\). The probability of the first spinner landing on 1 is \(P_1=\frac{m_1}{n_1}=\frac{1}{6}\).

Step3: Calculate the total number of outcomes for the second spinner

The second spinner has 4 sections. So, the total number of outcomes for the second spinner \(n_2 = 4\).

Step4: Calculate the number of favorable outcomes for the second spinner to land on 1

The second spinner has 1 section labeled 1. So, the number of favorable outcomes for the second spinner \(m_2 = 1\). The probability of the second spinner landing on 1 is \(P_2=\frac{m_2}{n_2}=\frac{1}{4}\).

Step5: Use the multiplication rule for independent events

Since the two spinners are independent events, \(P(K)=P_1\times P_2\). Substitute \(P_1=\frac{1}{6}\) and \(P_2=\frac{1}{4}\) into the formula: \(P(K)=\frac{1}{6}\times\frac{1}{4}=\frac{1\times1}{6\times4}=\frac{1}{24}\).

Step6: Calculate the total number of coins

Evan has \(6 + 4+3 + 8=21\) coins.

Step7: Calculate the probability of choosing a quarter

The probability of choosing a quarter \(P(C_1)=\frac{6}{21}=\frac{2}{7}\) (since there are 6 quarters out of 21 coins).

Step8: Calculate the probability of choosing a nickel

Since the coin is replaced, the total number of coins is still 21. The probability of choosing a nickel \(P(C_2)=\frac{3}{21}=\frac{1}{7}\) (since there are 3 nickels out of 21 coins).

Step9: Calculate the probability of choosing a dime

Since the coin is replaced, the total number of coins is still 21. The probability of choosing a dime \(P(C_3)=\frac{4}{21}\) (since there are 4 dimes out of 21 coins).

Step10: Use the multiplication rule for independent events (with replacement)

Since the events are independent (because of replacement), \(P(C)=P(C_1)\times P(C_2)\times P(C_3)\). Substitute \(P(C_1)=\frac{2}{7}\), \(P(C_2)=\frac{1}{7}\), and \(P(C_3)=\frac{4}{21}\) into the formula: \(P(C)=\frac{2}{7}\times\frac{1}{7}\times\frac{4}{21}=\frac{2\times1\times4}{7\times7\times21}=\frac{8}{1029}\).

Answer:

\(P(K)=\frac{1}{24}\), \(P(C)=\frac{8}{1029}\)