How do you find P(A' U B)

If P(A) = 0.3, P(B) = 0.2, and P(A and B) = 0.1

To find the value you can find conditional probabilities

P(A|B) = P(A and B) / P(B)

= 1/10 / 2/10

= 1/2

P(A|B) = 1/2

Now find P(A'|B)

P(A'|B) = P(A' and B) / P(B)

= P(A' and B) / 2/10

Problem: We don't know what P(A' and B) is.

Solution: We know that P(A|B) + P(A'|B) = 1, these are complement events and this property can be shown by a Venn diagram

So we have

P(A|B) + P(A'|B) = 1

1/2 + P(A' and B) / (2/10) = 1

P(A' and B) / (2/10) = 1/2

10/2 * P(A' and B) = 1/2

10 P(A' and B) = 1

P(A' and B) = 1/10

Now that we know P(A' and B) we can use the formula

P(A' U B) = P(A') + P(B) - P(A' and B)

= 7/10 + 2/10 - 1/10

= 8/10

Answer: P(A' U B) = 8/10




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From drawing a Venn diagram, I get P(A' U B) = 8/10

Here is how I did the diagram.

1.) Draw a box

2.) Draw 2 overlapping circles (label one circle A and labe the other circle B)

3.) I put 1/10 in the overlapping part of the 2 circles

4.) I put x in the non-overlapping part of the B circle, and y in the non-overlapping part of the A circle

5.) Since P(B) = 2/10 we have that x + 1/10 = 2/10, which mean that x = 1/10 and so the nonoverlapping part of circle B gets the value 1/10

6.) Since P(A) = 3/10 we have that y + 1/10 = 3/10, which means that y = 2/10, and so the nonoverlapping part of the A circle gets the value of 2/10.

7.) I labeled the rest of the rectagular box around the circle with the letter z and this part has the value 6/10 (since the values in the two circles minus the intersection equals 4/10).

8.) P(A' U B) is the probability of everything in B and everything in not A, so I added all values up (except for the value in the nonoverlapping part of circle A) and I got 8/10 or 0.80.

P(A' U B) = P(A') + P(B) - P(A' n B)

P(A) = 1 - P(A')

so P(A') = 1 - P(A)

P(A') = 1 - 0.3 = 0.7

so we have one peice have to find the intersection

not sure how to find the intersection