Question

1. game of craps in a game of craps, a player wins on the first roll if the player rolls a sum of 7 or 11, and the player loses if the player rolls a 2, 3, or 12. find the probability that the game will last only one roll.

Explanation:

Step1: Determine the total number of outcomes

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

Step2: Find the number of favorable outcomes (sum is 2,3,7,11,12)

• Sum of 2: \((1,1)\), so \(n(2)=1\)
• Sum of 3: \((1,2),(2,1)\), so \(n(3) = 2\)
• Sum of 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\), so \(n(7)=6\)
• Sum of 11: \((5,6),(6,5)\), so \(n(11)=2\)
• Sum of 12: \((6,6)\), so \(n(12)=1\)

The number of favorable outcomes \(n(A)=n(2)+n(3)+n(7)+n(11)+n(12)=1 + 2+6+2 + 1=12\)

Step3: Calculate the probability

The probability formula is \(P(A)=\frac{n(A)}{n(S)}\). Substituting \(n(A) = 12\) and \(n(S)=36\), we get \(P(A)=\frac{12}{36}=\frac{1}{3}\)

Answer:

The probability that the game will last only one roll is \(\frac{1}{3}\)