About A red and a blue die are thrown. Both dice are fair. The events A, B, and C are defined as follows: A: The sum on the two dice is even B: The sum on the two dice is at least 10 C: The red die comes up 5 (a) Calculate the probability of each individual event. That is, calculate p(A), p(B), and p(C). (b) What is p(A|C)?

[tex]p(A) = \frac{1}{2}[/tex][tex]p(B) = \frac{1}{6}[/tex][tex]p(C) = \frac{1}{6}[/tex][tex]P(A | C)=\frac{1}{2}[/tex]Solution:The probability of an event is given as:[tex]\text { probability of an event }=\frac{\text { number of favorable outcomes }}{\text { total number of outcomes }}[/tex]In throwing one die, the total number of outcomes = 6 { 1, 2, 3, 4, 5 , 6}First let us calculate p(A):The event is defined as: The sum on the two dice is evenSum on two dice is even if and only if either both dice turn up odd or both even.The odd outcomes in thowing a single die = 3 {1, 3, 5}The even outcomes in throwing a single die = 3 {2, 4, 6} The probability that both turn up odd is: [tex]\text { probability of both die turing up odd }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}[/tex]Similarly, the probability that both turn up even is:[tex]\text { probability of both die turing up even }=\frac{3}{6} \times \frac{3}{6}=\frac{1}{4}[/tex]probability that the sum on two dice is even = probability that both turn up odd  + probability that both turn up even[tex]\text { probability of sum on two dice is even }=\mathrm{p}(\mathrm{A})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}[/tex]Thus [tex]p(A) = \frac{1}{2}[/tex]Let us calculate p(B):The event B is defined as: The sum on the two dice is at least 10The total possible outcomes of two die is given as:{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }Since each individual die can turn up any of the numbers 1, 2, 3, 4, 5, 6 the event "sum of the two dice will be at least 10" is :atleast 10 means that sum can be 10 or greater than 10 {(4,6), (6,4), (5,5), (5,6), (6,5), (6,6)} Here favourable outcomes = 6Total number of outcomes = 36 Hence, the probability that the sum of the two dice will be at least 10 is:[tex]\text { probability that the sum of the two dice will be at least } 10=\frac{6}{36}=\frac{1}{6}[/tex]Thus [tex]p(B) = \frac{1}{6}[/tex]Let us calculate p(C):The event C is defined as: The red die comes up 5Favourable outcomes = {(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)}[tex]\text { probability of red die comes up } 5 \text { is the event }=\frac{6}{36}=\frac{1}{6}[/tex]Thus [tex]p(C) = \frac{1}{6}[/tex]B) What is p(A l C)[tex]P(A | C)=\frac{p(A \cap C)}{P(C)}[/tex][tex]\mathrm{A} \cap \mathrm{C}=\{(1,5),(3,5),(5,5)\}[/tex][tex]p(A \cap C)=\frac{3}{36}=\frac{1}{12}[/tex][tex]P(A | C)=\frac{p(A \cap C)}{P(C)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}[/tex]Thus [tex]P(A | C)=\frac{1}{2}[/tex]