Question

video lesson 5.1 example 3 - finding probabilities of complements. find the probability of each event when two six - sided dice are rolled. 3. the sum is greater than 8. 4. the sum is not 2.

Explanation:

Problem 3: The sum is greater than 8.

Step1: Determine 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 two dice are rolled is \(n(S)=6\times6 = 36\).

Step2: Find favorable outcomes

We need to find the number of outcomes where the sum of the two dice is greater than 8. Let the outcomes of the first die be \(x\) and the second die be \(y\), so we need \(x + y>8\), where \(x,y\in\{1,2,3,4,5,6\}\).

• If \(x = 3\), \(y\) must be 6 (\(3 + 6=9>8\))
• If \(x = 4\), \(y\) can be 5, 6 (\(4 + 5 = 9>8\), \(4+6 = 10>8\))
• If \(x = 5\), \(y\) can be 4, 5, 6 (\(5 + 4=9>8\), \(5 + 5 = 10>8\), \(5+6 = 11>8\))
• If \(x = 6\), \(y\) can be 3, 4, 5, 6 (\(6+3 = 9>8\), \(6 + 4=10>8\), \(6+5 = 11>8\), \(6 + 6=12>8\))
• If \(x=1\) or \(x = 2\), there are no values of \(y\) such that \(x + y>8\)

Now, count the number of favorable outcomes:
For \(x = 3\): 1 outcome
For \(x = 4\): 2 outcomes
For \(x = 5\): 3 outcomes
For \(x = 6\): 4 outcomes

Total favorable outcomes \(n(E)=1 + 2+3 + 4=10\)? Wait, no, let's list all pairs:
\((3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)\). Wait, that's 10? Wait, no:
Wait, when \(x = 3\), \(y = 6\) (1)
\(x = 4\), \(y = 5,6\) (2)
\(x = 5\), \(y = 4,5,6\) (3)
\(x = 6\), \(y = 3,4,5,6\) (4)
Wait, actually, the correct pairs are:
Sum = 9: \((3,6),(4,5),(5,4),(6,3)\) (4 outcomes)
Sum = 10: \((4,6),(5,5),(6,4)\) (3 outcomes)
Sum = 11: \((5,6),(6,5)\) (2 outcomes)
Sum = 12: \((6,6)\) (1 outcome)

Total number of favorable outcomes \(n(E)=4 + 3+2 + 1=10\)? Wait, no, \(4 + 3+2+1=10\)? Wait, \(4\) (sum 9)+\(3\) (sum 10)+\(2\) (sum 11)+\(1\) (sum 12) \(= 10\)? Wait, no, \(4+3 = 7\), \(7 + 2=9\), \(9+1 = 10\). Yes.

Step3: Calculate probability

The probability \(P(E)=\frac{n(E)}{n(S)}=\frac{10}{36}=\frac{5}{18}\)

Problem 4: The sum is not 2.

Step1: Determine total outcomes

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

Step2: Find the number of outcomes where the sum is 2

The only outcome where the sum of two dice is 2 is when \(x = 1\) and \(y = 1\) (i.e., \((1,1)\)), so \(n(\text{sum}=2)=1\).

Step3: Use the complement rule

The complement of the event "the sum is 2" is "the sum is not 2". Let \(A\) be the event that the sum is 2. Then \(P(A)=\frac{n(A)}{n(S)}=\frac{1}{36}\). By the complement rule \(P(\text{not }A)=1 - P(A)\).
So \(P(\text{sum is not }2)=1-\frac{1}{36}=\frac{36 - 1}{36}=\frac{35}{36}\)

Answer:

for Problem 3: \(\frac{5}{18}\)