Question

select the correct answer from each drop-down menu. franklin rolls a pair of six - sided fair dice with sides numbered 1 through 6. the probability that the sum of the numbers rolled is either even or a multiple of 5 is dropdown. the probability that the sum of the numbers rolled is either a multiple of 3 or 4 is dropdown.

Explanation:

To solve the problem, we first determine the total number of possible outcomes when rolling two six - sided dice. Each die has 6 possible outcomes, so the total number of outcomes is \(n(S)=6\times6 = 36\).

Part 1: Probability that the sum is either even or a multiple of 5

Let \(A\) be the event that the sum is even, and \(B\) be the event that the sum is a multiple of 5.

Step 1: Find \(n(A)\)

For the sum of two numbers to be even, either both numbers are even or both are odd.

• Number of ways to get two odd numbers: There are 3 odd numbers (\(1,3,5\)) on a die. So, the number of ways to get two odd numbers is \(3\times3=9\).
• Number of ways to get two even numbers: There are 3 even numbers (\(2,4,6\)) on a die. So, the number of ways to get two even numbers is \(3\times3 = 9\).

Thus, \(n(A)=9 + 9=18\).

Step 2: Find \(n(B)\)

The possible sums when rolling two dice range from \(2(1 + 1)\) to \(12(6+6)\). The multiples of 5 in this range are \(5\) and \(10\).

• For sum \(= 5\): The possible pairs \((x,y)\) are \((1,4),(2,3),(3,2),(4,1)\), so \(n_1 = 4\).
• For sum \(= 10\): The possible pairs \((x,y)\) are \((4,6),(5,5),(6,4)\), so \(n_2=3\).

Thus, \(n(B)=4 + 3=7\).

Step 3: Find \(n(A\cap B)\)

We need to find the sums that are both even and a multiple of 5. The even multiples of 5 in the range \(2 - 12\) is \(10\). We already found that the number of ways to get a sum of 10 is 3. So, \(n(A\cap B) = 3\).

Step 4: Use the inclusion - exclusion principle \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\), \(P(B)=\frac{n(B)}{n(S)}=\frac{7}{36}\), \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{3}{36}\)

\(P(A\cup B)=\frac{18 + 7-3}{36}=\frac{22}{36}=\frac{11}{18}\)

Part 2: Probability that the sum is either a multiple of 3 or a multiple of 4

Let \(C\) be the event that the sum is a multiple of 3, and \(D\) be the event that the sum is a multiple of 4.

Step 1: Find \(n(C)\)

The multiples of 3 in the range \(2 - 12\) are \(3,6,9,12\).

• For sum \(= 3\): The pairs are \((1,2),(2,1)\), so \(n_1 = 2\).
• For sum \(= 6\): The pairs are \((1,5),(2,4),(3,3),(4,2),(5,1)\), so \(n_2 = 5\).
• For sum \(= 9\): The pairs are \((3,6),(4,5),(5,4),(6,3)\), so \(n_3 = 4\).
• For sum \(= 12\): The pair is \((6,6)\), so \(n_4 = 1\).

Thus, \(n(C)=2 + 5+4 + 1=12\).

Step 2: Find \(n(D)\)

The multiples of 4 in the range \(2 - 12\) are \(4,8,12\).

• For sum \(= 4\): The pairs are \((1,3),(2,2),(3,1)\), so \(n_1 = 3\).
• For sum \(= 8\): The pairs are \((2,6),(3,5),(4,4),(5,3),(6,2)\), so \(n_2 = 5\).
• For sum \(= 12\): The pair is \((6,6)\), so \(n_3 = 1\).

Thus, \(n(D)=3 + 5+1=9\).

Step 3: Find \(n(C\cap D)\)

The numbers that are multiples of both 3 and 4 (i.e., multiples of 12) in the range \(2 - 12\) is \(12\). The number of ways to get a sum of 12 is 1. So, \(n(C\cap D)=1\).

Step 4: Use the inclusion - exclusion principle \(P(C\cup D)=P(C)+P(D)-P(C\cap D)\)

\(P(C)=\frac{n(C)}{n(S)}=\frac{12}{36}\), \(P(D)=\frac{n(D)}{n(S)}=\frac{9}{36}\), \(P(C\cap D)=\frac{n(C\cap D)}{n(S)}=\frac{1}{36}\)

\(P(C\cup D)=\frac{12 + 9-1}{36}=\frac{20}{36}=\frac{5}{9}\)

Final Answers

• The probability that the sum of the numbers rolled is either even or a multiple of 5 is \(\boldsymbol{\frac{11}{18}}\).
• The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is \(\boldsymbol{\frac{5}{9}}\).

Answer:

To solve the problem, we first determine the total number of possible outcomes when rolling two six - sided dice. Each die has 6 possible outcomes, so the total number of outcomes is \(n(S)=6\times6 = 36\).

Part 1: Probability that the sum is either even or a multiple of 5

Let \(A\) be the event that the sum is even, and \(B\) be the event that the sum is a multiple of 5.

Step 1: Find \(n(A)\)

For the sum of two numbers to be even, either both numbers are even or both are odd.

• Number of ways to get two odd numbers: There are 3 odd numbers (\(1,3,5\)) on a die. So, the number of ways to get two odd numbers is \(3\times3=9\).
• Number of ways to get two even numbers: There are 3 even numbers (\(2,4,6\)) on a die. So, the number of ways to get two even numbers is \(3\times3 = 9\).

Thus, \(n(A)=9 + 9=18\).

Step 2: Find \(n(B)\)

The possible sums when rolling two dice range from \(2(1 + 1)\) to \(12(6+6)\). The multiples of 5 in this range are \(5\) and \(10\).

• For sum \(= 5\): The possible pairs \((x,y)\) are \((1,4),(2,3),(3,2),(4,1)\), so \(n_1 = 4\).
• For sum \(= 10\): The possible pairs \((x,y)\) are \((4,6),(5,5),(6,4)\), so \(n_2=3\).

Thus, \(n(B)=4 + 3=7\).

Step 3: Find \(n(A\cap B)\)

We need to find the sums that are both even and a multiple of 5. The even multiples of 5 in the range \(2 - 12\) is \(10\). We already found that the number of ways to get a sum of 10 is 3. So, \(n(A\cap B) = 3\).

Step 4: Use the inclusion - exclusion principle \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\), \(P(B)=\frac{n(B)}{n(S)}=\frac{7}{36}\), \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{3}{36}\)

\(P(A\cup B)=\frac{18 + 7-3}{36}=\frac{22}{36}=\frac{11}{18}\)

Part 2: Probability that the sum is either a multiple of 3 or a multiple of 4

Let \(C\) be the event that the sum is a multiple of 3, and \(D\) be the event that the sum is a multiple of 4.

Step 1: Find \(n(C)\)

The multiples of 3 in the range \(2 - 12\) are \(3,6,9,12\).

• For sum \(= 3\): The pairs are \((1,2),(2,1)\), so \(n_1 = 2\).
• For sum \(= 6\): The pairs are \((1,5),(2,4),(3,3),(4,2),(5,1)\), so \(n_2 = 5\).
• For sum \(= 9\): The pairs are \((3,6),(4,5),(5,4),(6,3)\), so \(n_3 = 4\).
• For sum \(= 12\): The pair is \((6,6)\), so \(n_4 = 1\).

Thus, \(n(C)=2 + 5+4 + 1=12\).

Step 2: Find \(n(D)\)

The multiples of 4 in the range \(2 - 12\) are \(4,8,12\).

• For sum \(= 4\): The pairs are \((1,3),(2,2),(3,1)\), so \(n_1 = 3\).
• For sum \(= 8\): The pairs are \((2,6),(3,5),(4,4),(5,3),(6,2)\), so \(n_2 = 5\).
• For sum \(= 12\): The pair is \((6,6)\), so \(n_3 = 1\).

Thus, \(n(D)=3 + 5+1=9\).

Step 3: Find \(n(C\cap D)\)

The numbers that are multiples of both 3 and 4 (i.e., multiples of 12) in the range \(2 - 12\) is \(12\). The number of ways to get a sum of 12 is 1. So, \(n(C\cap D)=1\).

Step 4: Use the inclusion - exclusion principle \(P(C\cup D)=P(C)+P(D)-P(C\cap D)\)

\(P(C)=\frac{n(C)}{n(S)}=\frac{12}{36}\), \(P(D)=\frac{n(D)}{n(S)}=\frac{9}{36}\), \(P(C\cap D)=\frac{n(C\cap D)}{n(S)}=\frac{1}{36}\)

\(P(C\cup D)=\frac{12 + 9-1}{36}=\frac{20}{36}=\frac{5}{9}\)

Final Answers

• The probability that the sum of the numbers rolled is either even or a multiple of 5 is \(\boldsymbol{\frac{11}{18}}\).
• The probability that the sum of the numbers rolled is either a multiple of 3 or 4 is \(\boldsymbol{\frac{5}{9}}\).