probability?

Question

Either terry, chris, or kim will attend a party. The probability terry attends is 0.31 and the probability chris attends is 0.5. What is the probability that kim will attend?

a. 0.33
b. 0.5
c. 0.81
d. 0.19

Answer 1

terry attends 31 out of 100 (31/100)
chris attends 50 out of 100 (50/100)

so then 31 + 50 = 81
between them they attend 81 out of 100 parties (81/100)

the balane therefor must be attended by kim

100- 81= 19

kim attends 19 out of 100 parties (19/100)

19 divided by 100 equals 0.19 or 19%

does this help?

Answer 2

Sience only one of Terry, Chris and Kim attend the party. The three outcomes are mutually exclusive. therefore probability that kim attend the party is 1- 0.31-0.5= 0.19.

Answer 3

1.0 - 0.31 - 0.5 = 0.19

The problem says that EITHER Terry OR Chris OR Kim will attend. This means that if one attends, the others won't; i.e. there's no chance Terry and Chris both showing up, for example.

Since at least one will attend, the sum of the probabilities of each must be 1.0, or 100%.

Therefore, the probability of Kim attending is 1 - the probability of Terry attending - the probability of Chris attending.

Answer 4

prob that one of them will attend party = 1
prob kim attending party = 1 -(0.31+0.5) = 0.19

Answer 5

Let the probability of Kim be x
So,
X+0.31+0.5=1
X= 1-0.31-0.5
X=0.19

Answer 6

The answer is D. You add Terry and Chris's probabilities and subtract that from 1 so you get .19 I think.

Answer 7

d

Because one and only one of them will attend the probabilities should add up to 1.

1 - (0.31+0.5) = 0.19

Answer 8

0.5 + 0.31 = .81

1.00 - .81 = 0.19

The answer is d. 0.19

Answer 9

.19, 1-.5-.31 = .19

Answer 10

1-(0.5+0.31)=0.19
So d??????