Question

1. rolling two dice if two dice are rolled one time, find the probability of getting these results:

a. a sum of 5
b. a sum of 9 or 10
c. doubles

Explanation:

Step1: Calculate total number of outcomes

When two dice are rolled, each die has 6 possible outcomes. By the fundamental counting principle, the total number of outcomes \(n(S)=6\times6 = 36\)

Step2: Find number of outcomes for sum of 5

The pairs \((x,y)\) such that \(x + y=5\) are \((1,4)\), \((2,3)\), \((3,2)\), \((4,1)\). So \(n(A)=4\). The probability \(P(A)=\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

Step3: Find number of outcomes for sum of 9 or 10

For sum of 9: \((3,6)\), \((4,5)\), \((5,4)\), \((6,3)\) (\(n_1 = 4\))
For sum of 10: \((4,6)\), \((5,5)\), \((6,4)\) (\(n_2=3\))
\(n(B)=n_1 + n_2=4 + 3=7\). The probability \(P(B)=\frac{n(B)}{n(S)}=\frac{7}{36}\)

Step4: Find number of outcomes for doubles

The doubles are \((1,1)\), \((2,2)\), \((3,3)\), \((4,4)\), \((5,5)\), \((6,6)\). So \(n(C)=6\). The probability \(P(C)=\frac{n(C)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

Answer:

a. \(\frac{1}{9}\)
b. \(\frac{7}{36}\)
c. \(\frac{1}{6}\)