Die A has 4 red faces and 2 black faces. Die B has 2 red faces and 4 black faces. A coin is flipped once. If it were heads, only die A is used (die B is discarded). If it were tails, only die B is used (die A is discarded). We do not get to know which die was chosen. a) The first roll of the chosen die gives red. What is the probability that the second roll (with the same die) will be red?
b) The first two rolls of the die were both red. What is the probability that the third roll (with the same die) will be red?

a) 5/9

b) 1/3

Step-by-step explanation:

a) P(Die A or Die B) = P(red and red with Die A) + P(red and red with Die B)

                               = 4/6 × 4/6 +  2/6×2/6

                               = 5/9

b)  P(Die A or Die B) = P(red and red and red with Die A) + P(red and red and red with die B)

                           = 4/6×4/6×4/6 + 2/6×2/6×2/6

                          = 1/3

i just took the test, the correct answer is .. 31.5 in.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.