First of all, note that there are 36 possible outcomes (all the numbers from 1 to 6 from the first die, combined with all the numbers from 1 to 6 from the second die).
a. Out of these 36 outcomes, the only ways to get 8 as sum are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). There are 5 favourable cases over 36 possible cases, so the probability is 5/36.
b. We want to get 8 as sum, and the first die must be a 2. Out of the five pairs we got before, only the first satisfies this request, so the probability is 1/36
c. Knowing that the first die was a 2, we have 6 possible outcomes:
The second die rolls a 1, for a total of 3.
The second die rolls a 2, for a total of 4.
The second die rolls a 3, for a total of 5.
The second die rolls a 4, for a total of 6.
The second die rolls a 5, for a total of 7.
The second die rolls a 6, for a total of 8.
So, there is 1 favourable outcome out of 6 possible outcomes, so the probability is 1/6
d. The die is fair, so every number occurs with probability 1/6.
The right answers will be "[tex]\frac{5}{36}[/tex]", "[tex]\frac{1}{36}[/tex]", "[tex]\frac{1}{6}[/tex]" and "[tex]\frac{1}{6}[/tex]". A further solution is provided below.
According to the question,
P(E) = [tex]\frac{1}{6}[/tex]
For event F, the Sample space is:
→ [tex]F = { (2,6),(3,5),(4,4),(6,2),(5,3) }[/tex] (First no. of each pair denotes first dice and second one denoted the second dice).
(a)
→ P(F) = [tex]\frac{1}{6}\times \frac{1}{6} +\frac{1}{6}\times \frac{1}{6} +\frac{1}{6}\times \frac{1}{6} +\frac{1}{6}\times \frac{1}{6} +\frac{1}{6}\times \frac{1}{6}[/tex]
= [tex]\frac{5}{36}[/tex]
(b)
→ P(E ∩ F) = [tex]\frac{1}{6}\times \frac{1}{6}[/tex]
= [tex]\frac{1}{36}[/tex]
(c)
→ P(F | E) = [tex]\frac{P(F \cap E)}{P(E)}[/tex]
= [tex]\frac{\frac{1}{36} }{\frac{1}{6} }[/tex]
= [tex]\frac{1}{6}[/tex]
(d)
→ P(E) = [tex]\frac{1}{6}[/tex]
Learn more about P(E ∩ F) here,
https://brainly.com/question/17173629
