Question

leave your answer as a fraction. c) find the probability of getting a 4 on the first die. leave your answer as a fraction. d) find the probability of getting a sum of 8 and a 4 on the first die. leave your answer as a fraction. e) find the probability of getting a sum of 8 or a 4 on the first die. leave your answer as a fraction. f) find the probability of getting doubles. leave your answer as a fraction. g) find the probability of getting both dice to be even numbers. leave your answer as a fraction. h) find the probability of getting doubles and both dice to be even numbers. leave your answer as a fraction. i) find the probability of getting doubles or both dice to be even numbers. leave your answer as a fraction.

Explanation:

Step1: Determine sample - space size

When two dice are rolled, the sample - space size \(n(S)=6\times6 = 36\) since each die has 6 possible outcomes.

Step2: Calculate probability for g

The event of getting both dice as even numbers: The first die can be 2, 4, or 6 (3 possibilities) and the second die can be 2, 4, or 6 (3 possibilities). So \(n(E)=\ 3\times3=9\). The probability \(P=\frac{n(E)}{n(S)}=\frac{9}{36}=\frac{1}{4}\).

Step3: Calculate probability for h

The event of getting doubles and both dice as even numbers: The possible outcomes are \((2,2)\), \((4,4)\), \((6,6)\), so \(n(E) = 3\). The probability \(P=\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12}\).

Step4: Calculate probability for i

Use the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Let \(A\) be the event of getting doubles (\(n(A) = 6\), \(P(A)=\frac{6}{36}\)) and \(B\) be the event of getting both dice as even numbers (\(P(B)=\frac{9}{36}\)), and \(A\cap B\) be the event of getting doubles and both dice as even numbers (\(P(A\cap B)=\frac{3}{36}\)). Then \(P(A\cup B)=\frac{6 + 9-3}{36}=\frac{12}{36}=\frac{1}{3}\).

Answer:

g) \(\frac{1}{4}\)
h) \(\frac{1}{12}\)
i) \(\frac{1}{3}\)