# Venn Diagram: prob's P(1)=0.2, P(2)=0.25, P(3)=0.35, P(4)=0.05, P(5)=0.1, P(6)=0.05.

#### megarust

##### New member
A sample space contains six sample points and events A, B, and C as shown in the Venn diagram. The probablities of the sample points are P(1)=0.2, P(2)=0.25, P(3)=0.35, P(4)=0.05, P(5)=0.1, P(6)=0.05. Use the Venn diagram and the probabilities of the sample points to find:
P(C)= 0.15
_ _
P(B|C)=
P(A|C)= 0
I figured out two of the probabilities, but I'm having trouble with the third one. For the first one I added all portions of C to get 0.15. I wasn't able to answer the second one.

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#### tkhunny

##### Moderator
Staff member
You'll have to provide the picture. Please also demonstrate how you arrived at your answers t the first two.

#### megarust

##### New member

You'll have to provide the picture. Please also demonstrate how you arrived at your answers t the first two.
I uploaded the picture.

#### tkhunny

##### Moderator
Staff member
P(B|C)
1) P(5) is what?
2) P(5)/P(C) is what?
Assuming that you are in C, what is the probability that you are ALSO in B?

P(A|C)
Assuming that you are in C, what is the probability that you are ALSO in A?

#### j-astron

##### New member
[Not the OP] Can I ask a dumb question about this problem? Maybe I just don't understand the Venn diagram representation of the sample space of possible outcomes, but what is the difference between point 1 and point 2? It seems to be that both represent the outcome that 'A has occurred, while neither B nor C has occurred."

#### JeffM

##### New member
[Not the OP] Can I ask a dumb question about this problem? Maybe I just don't understand the Venn diagram representation of the sample space of possible outcomes, but what is the difference between point 1 and point 2? It seems to be that both represent the outcome that 'A has occurred, while neither B nor C has occurred."
The probability that event A but neither event B nor event C has occurred is indeed the sum of the probabilities that 1 or 2 has occurred. This does not make them the same event. If A is the event that "the animal seen was a mammal," 1 is the event that "the animal seen was a whale," and 2 is the event that " the animal seen was a bat," we do not conclude that bats are whales because both are mammals. Ultimately, Venn diagrams are about categorization and the fact that the same "atomic" event may fit into more than one category of sets of events.

Does that help?

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#### stapel

##### Super Moderator
Staff member
A sample space contains six sample points and events A, B, and C as shown in the Venn diagram.

View attachment 9109

The probablities of the sample points are P(1)=0.2, P(2)=0.25, P(3)=0.35, P(4)=0.05, P(5)=0.1, P(6)=0.05. Use the Venn diagram and the probabilities of the sample points to find:

i. P(C)= 0.15
_ _
ii. P(B|C)=

iii. P(A|C)= 0
I figured out two of the probabilities, but I'm having trouble with the third one. For the first one I added all portions of C to get 0.15. I wasn't able to answer the second one.
You say that you "figured out two of the probabilities", but that you are "having trouble with the third one" and weren't "able to answer the second one"...? Does this mean that actually you've only completed what I've labelled as (i), and still need help with (ii) and (iii)? Also, between the first probability (P(C)) and the second (P(B|C)) in your post, there is a line with two underscore characters. What is the significance of this?

Thank you!

#### j-astron

##### New member
The probability that event A but neither event B nor event C has occurred is indeed the sum of the probabilities that 1 or 2 has occurred. This does not make them the same event. If A is the event that "the animal seen was a mammal," 1 is the event that "the animal seen was a whale," and 2 is the event that " the animal seen was a bat," we do not conclude that bats are whales because both are mammals. Ultimately, Venn diagrams are about categorization and the fact that the same "atomic" event may fit into more than one category of sets of events.

Does that help?
Ah yeah. I should have realized it was just "two possible events that both fall into the category of A", but I wasn't sure. Thanks.