Question

solve the following problems. (see example 3)

1. you roll two six - sided dice. find the probability that:

(a) the sum is not 4 and
(b) the sum is greater than 5

Explanation:

Part (a)

Step 1: Find total outcomes

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

Step 2: Find number of outcomes with sum 4

We find the pairs \((x,y)\) where \(x + y=4\) and \(x,y\in\{1,2,3,4,5,6\}\). The pairs are \((1,3)\), \((2,2)\), \((3,1)\). So the number of outcomes with sum 4, \(n(A) = 3\).

Step 3: Find number of outcomes with sum not 4

Using the formula for the complement of an event, if \(A\) is the event that the sum is 4, then the event that the sum is not 4 is \(A'\). We know that \(n(A')=n(S)-n(A)\). Substituting the values, we get \(n(A') = 36 - 3=33\).

Step 4: Calculate probability of sum not 4

The probability \(P(A')=\frac{n(A')}{n(S)}=\frac{33}{36}=\frac{11}{12}\).

Part (b)

Step 1: Find total outcomes

As before, when rolling two six - sided dice, the total number of outcomes \(n(S)=6\times6 = 36\).

Step 2: Find number of outcomes with sum \(\leq5\)

We find the pairs \((x,y)\) where \(x + y\leq5\):

• For \(x + y = 2\): \((1,1)\) (1 pair)
• For \(x + y=3\): \((1,2)\), \((2,1)\) (2 pairs)
• For \(x + y = 4\): \((1,3)\), \((2,2)\), \((3,1)\) (3 pairs)
• For \(x + y=5\): \((1,4)\), \((2,3)\), \((3,2)\), \((4,1)\) (4 pairs)

The total number of outcomes with sum \(\leq5\) is \(1 + 2+3 + 4=10\).

Step 3: Find number of outcomes with sum \(>5\)

Let \(B\) be the event that the sum is greater than 5. The complement of \(B\) is the event that the sum is less than or equal to 5. So \(n(B)=n(S)-n(B')\), where \(n(B') = 10\). Then \(n(B)=36 - 10 = 26\).

Step 4: Calculate probability of sum greater than 5

The probability \(P(B)=\frac{n(B)}{n(S)}=\frac{26}{36}=\frac{13}{18}\).

Answer:

s:
(a) \(\frac{11}{12}\)

(b) \(\frac{13}{18}\)