Question

video lesson 5.4 example 1 – finding the probability of disjoint events

1. two six - sided dice are rolled. what is the probability that the sum of the numbers rolled is equal to a multiple of 4 or equal to 5?

event a: 4, 8, or 12 event b: sum is 5
the sample space is provided below:
sample space image of dice combinations
hint: count the number of elements in event a, event b, and the sample space and use the probability of disjoint events formula.

Explanation:

Step1: Determine Sample Space Size

When rolling two six - sided dice, the total number of possible outcomes (sample space) is calculated by the multiplication principle. Each die has 6 faces, so for two dice, the number of outcomes \(n(S)=6\times6 = 36\).

Step2: Count Outcomes in Event A

Event A is the sum being 4, 8, or 12.

• For sum = 4: The possible pairs \((x,y)\) where \(x\) is the result of the first die and \(y\) is the result of the second die are \((1,3)\), \((2,2)\), \((3,1)\). So there are 3 outcomes.
• For sum = 8: The possible pairs are \((2,6)\), \((3,5)\), \((4,4)\), \((5,3)\), \((6,2)\). So there are 5 outcomes.
• For sum = 12: The only pair is \((6,6)\). So there is 1 outcome.

The number of outcomes in Event A, \(n(A)=3 + 5+1=9\).

Step3: Count Outcomes in Event B

Event B is the sum being 5. The possible pairs are \((1,4)\), \((2,3)\), \((3,2)\), \((4,1)\). So the number of outcomes in Event B, \(n(B) = 4\).

Step4: Check for Disjoint Events

Since a sum cannot be both a multiple of 4 (or 12) and 5 at the same time, Events A and B are disjoint. For disjoint events, the probability of \(A\) or \(B\) is given by \(P(A\cup B)=P(A)+P(B)\).

The formula for probability of an event \(E\) is \(P(E)=\frac{n(E)}{n(S)}\).

So \(P(A)=\frac{n(A)}{n(S)}=\frac{9}{36}\) and \(P(B)=\frac{n(B)}{n(S)}=\frac{4}{36}\).

Then \(P(A\cup B)=\frac{9 + 4}{36}=\frac{13}{36}\).

Answer:

\(\frac{13}{36}\)