Question

the table shows the total possible outcomes of rolling two number cubes, each labeled with the numbers 1 to 6. the first column shows the possible outcomes for the first number cube, and the first row shows the possible outcomes of the second number cube. the ordered pair (3,4) indicates a roll of 3 on the first number cube and a roll of 4 on the second number cube. what is the probability of rolling the two number cubes and rolling two 3s? study the worked example and then answer each question. consider the example. calculate the probability of rolling an odd number on the first number cube, ( p(o) ), and an even number on the second number cube, ( p(e) ). ( p(o)= ) the probability of rolling an odd number on the first number cube. ( p(e)= ) the probability of rolling an even number on the second number cube. ( p(o \text{ and } e)= ) the probability of rolling an odd number on the first number cube and an even number on the second number cube. the probability of two independent events occurring is the product of the probabilities of the events. thus, if ( a ) and ( b ) are independent events, ( p(a \text{ and } b)=p(a)\times p(b) )

Explanation:

Step1: Calculate the total number of outcomes

When rolling two number - cubes, each cube has 6 possible outcomes. By the fundamental counting principle, the total number of outcomes \(n(S)=6\times6 = 36\).

Step2: Find the number of favorable outcomes for rolling two 3s

The favorable outcome for rolling two 3s is \((3,3)\), so \(n(A)=1\).
The probability \(P(A)=\frac{n(A)}{n(S)}=\frac{1}{36}\).

Step3: Calculate \(P(O)\)

The odd numbers on a number cube are 1, 3, 5. So for the first cube, \(n(O)=3\). Then \(P(O)=\frac{3}{6}=\frac{1}{2}\).

Step4: Calculate \(P(E)\)

The even numbers on a number cube are 2, 4, 6. So for the second cube, \(n(E)=3\). Then \(P(E)=\frac{3}{6}=\frac{1}{2}\).

Step5: Calculate \(P(O\ and\ E)\)

Since the events of rolling the first cube and the second cube are independent, \(P(O\ and\ E)=P(O)\times P(E)\).
Substitute \(P(O)=\frac{1}{2}\) and \(P(E)=\frac{1}{2}\) into the formula: \(P(O\ and\ E)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\).

Answer:

The probability of rolling two 3s is \(\frac{1}{36}\). \(P(O)=\frac{1}{2}\), \(P(E)=\frac{1}{2}\), \(P(O\ and\ E)=\frac{1}{4}\)