Chapter 13 25 If possible can you do as many as possible It

Chapter 13 25>>>

If possible can you do as many as possible. It would be greatly appreciated. Have a great day

Solution

25)E)11/36

because

the probability of rolling two six-sided dice and getting at least one six is=>

1/6 (for the given first die) + 1/6 (for the given second die) - 1/36 (as the given possibility of both was included twice, and you only want it once)

=>1/6 + 1/6 - 1/36 = 6/36 + 6/36 - 1/36 = (6 + 6 - 1)/36 = 11/36.

26)A)3/4

P(A)=probability of getting even number on dice A

P(B)=probability of getting even number on dice B

We want to find P(AUB) =>
P(AUB)=P(A)+P(B)-P(AnB) [where P(AnB) is when the the probability that both A and B occur - called A intersection B]

This imlies=>
P(AnB)=P(A)xP(B)
Now, we must assume that they are fair dice. Then P(A)=P(B)=3/6=1/2
so, P(AnB)=(1/2)x(1/2)=1/4
so P(AUB)=1/2 + 1/2 - 1/4
=3/4

27)C)1/2 as one would be P and another would be p=>Pp(50%)

28)C)1/2 as one would be P and another would be p=>Pp(50%)

27)