Question

for the experiment of rolling an ordinary pair of dice, find the probability that the sum will be odd or a multiple of 3. (you may want to use a table showing the sum for each of the 36 equally likely outcomes.) the probability that the sum of the pair of dice is odd or a multiple of 3 is

Explanation:

Step1: Calculate the number of odd sums

When rolling two dice, the sum is odd if one die shows an odd number and the other shows an even number.
There are \(3\) odd numbers (\(1,3,5\)) and \(3\) even numbers (\(2,4,6\)) on a die.
The number of ways to get an odd sum is \(3\times3 + 3\times3=18\)

Step2: Calculate the number of sums that are multiples of \(3\)

The possible sums that are multiples of \(3\) are \(3,6,9,12\)
For sum \(3\): \((1,2),(2,1)\) (2 ways)
For sum \(6\): \((1,5),(2,4),(3,3),(4,2),(5,1)\) (5 ways)
For sum \(9\): \((3,6),(4,5),(5,4),(6,3)\) (4 ways)
For sum \(12\): \((6,6)\) (1 way)
Total number of sums that are multiples of \(3\) is \(2 + 5+4 + 1=12\)

Step3: Calculate the number of sums that are both odd and multiples of \(3\)

The sums that are both odd and multiples of \(3\) are \(3,9\)
For sum \(3\): 2 ways; for sum \(9\): 4 ways. Total \(2 + 4=6\)

Step4: Use the inclusion - exclusion principle

The formula for \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
Let \(A\) be the event that the sum is odd and \(B\) be the event that the sum is a multiple of \(3\)
\(n(A) = 18\), \(n(B)=12\), \(n(A\cap B)=6\)
\(n(A\cup B)=n(A)+n(B)-n(A\cap B)=18 + 12-6=24\)

Since the total number of outcomes \(n = 36\)
The probability \(P=\frac{n(A\cup B)}{n}=\frac{24}{36}=\frac{2}{3}\)

Answer:

\(\frac{2}{3}\)