Question

(b) sum of 8 or 9
when rolling two dice, the probability of rolling a sum of 8 or 9 is \\( \frac { 1 } { 4 } \\).
part 3 of 8
(c) sum greater than or equal to 8
when rolling two dice, the probability of rolling a sum greater than or equal to 8 is \\( \frac { 5 } { 12 } \\).

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: Find number of outcomes for sum of 8

The pairs \((x,y)\) such that \(x + y=8\) are \((2,6)\), \((3,5)\), \((4,4)\), \((5,3)\), \((6,2)\). So \(n(8)=5\)

Step3: Find number of outcomes for sum of 9

The pairs \((x,y)\) such that \(x + y = 9\) are \((3,6)\), \((4,5)\), \((5,4)\), \((6,3)\). So \(n(9)=4\)

Step4: Calculate probability for sum of 8 or 9

Using the formula \(P(A\cup B)=P(A)+P(B)\) (since sum of 8 and sum of 9 are mutually - exclusive events, \(A\cap B=\varnothing\)), \(P(\text{sum}=8\text{ or }9)=\frac{n(8)+n(9)}{n(S)}=\frac{5 + 4}{36}=\frac{9}{36}=\frac{1}{4}\)

Step5: Find number of outcomes for sum greater than or equal to 8

Sum of 8: \(n(8) = 5\); sum of 9: \(n(9)=4\); sum of 10: \((4,6)\), \((5,5)\), \((6,4)\) so \(n(10)=3\); sum of 11: \((5,6)\), \((6,5)\) so \(n(11)=2\); sum of 12: \((6,6)\) so \(n(12)=1\)
\(n(\text{sum}\geq8)=5 + 4+3+2+1=15\)

Step6: Calculate probability for sum greater than or equal to 8

Using the formula \(P=\frac{n(\text{sum}\geq8)}{n(S)}\), \(P(\text{sum}\geq8)=\frac{15}{36}=\frac{5}{12}\)

Answer:

For part (b): \(\frac{1}{4}\)
For part (c): \(\frac{5}{12}\)