Question

(e) four on one die or on both dice
when rolling two dice, the probability of rolling a four on one die or on both dice is

part 6 of 8
(f) sum that is odd
when rolling two dice, the probability of rolling a sum that is odd is

Explanation:

Step1: Calculate total number of outcomes

When rolling two dice, each die has 6 possible outcomes. So the total number of outcomes when rolling two dice is \(n(S)=6\times6 = 36\)

Step2: Calculate the number of outcomes for part (e)

Let \(A\) be the event of getting a 4 on the first - die and non - 4 on the second die, \(B\) be the event of getting a 4 on the second - die and non - 4 on the first die, and \(C\) be the event of getting a 4 on both dice.

• For event \(A\): The first die shows 4 (1 way) and the second die shows non - 4 (5 ways). So \(n(A)=1\times5 = 5\)
• For event \(B\): The first die shows non - 4 (5 ways) and the second die shows 4 (1 way). So \(n(B)=5\times1 = 5\)
• For event \(C\): The first die shows 4 (1 way) and the second die shows 4 (1 way). So \(n(C)=1\times1 = 1\)

Using the formula \(n(A\cup B\cup C)=n(A)+n(B)+n(C)\) (since \(A\), \(B\), and \(C\) are mutually - exclusive events), we have \(n = 5 + 5+1=11\)

The probability \(P=\frac{n}{n(S)}\), so \(P=\frac{11}{36}\)

Step3: Calculate the number of outcomes for part (f)

Let the first die show \(x\) and the second die show \(y\). The sum \(z=x + y\). For \(z\) to be odd, one of \(x\) and \(y\) must be even and the other must be odd.

• If the first die (\(x\)) is even (3 possibilities: 2, 4, 6) and the second die (\(y\)) is odd (3 possibilities: 1, 3, 5), the number of outcomes \(n_1=3\times3 = 9\)
• If the first die (\(x\)) is odd (3 possibilities: 1, 3, 5) and the second die (\(y\)) is even (3 possibilities: 2, 4, 6), the number of outcomes \(n_2=3\times3 = 9\)

The number of favorable outcomes \(n=n_1 + n_2=9 + 9=18\)

The probability \(P=\frac{n}{n(S)}\), so \(P=\frac{18}{36}=\frac{1}{2}\)

Answer:

(e) \(\frac{11}{36}\)
(f) \(\frac{1}{2}\)