Need help solving this please. If you can explain how you got the answer, that w
ID: 2927878 • Letter: N
Question
Need help solving this please. If you can explain how you got the answer, that would be very helpful. Thanks in Advance.
1) Given P(A) = .10, P(B) = .12, P(C) = .21, P(AC) = .05, and P(BC) = .03, solve the following. (Round your answers to 2 decimal places.)
a. P(AC) =
b. P(BC) =
c. If A and B are mutually exclusive, P(AB) =
2) Use the values in the joint probability table to solve the equations given.
*(Round your answers to 2 decimal places.)
**(Round your answer to 3 decimal places.)
a. P ( A U F) = ____________*
b. P ( E U B) = ____________*
c. P ( B U C) = ____________*
d. P ( E U F) = ____________**
E F A .10 .03 B .04 .12 C .27 .06 D .31 .07Explanation / Answer
1) (Key symbols are missing. What does P(AC) mean? P(A|C) or P(A C)?)
P(A) = .10, P(B) = .12, P(C) = .21, P(A|C) = .05, and P(B|C) = .03
a) P(A C) = P(A|C) P(C) = 0.05 * 0.21 = 0.015
b) P(B C) = P(B|C) P(C) = 0.03 * 0.12 = 0.0036
c) If A and B are mutually exclusive, P(A B) = P(A|B) = 0.
2) Note, all the probabilities add to 1.
a) P(A U F) = 0.10 + 0.03 + 0.12 + 0.06 + 0.07 = 0.38
b) P(E U B) = 0.10 + 0.04 + 0.27 + 0.31 + 0.12 = 0.84
c) P(B U C) = 0.04 + 0.12 + 0.27 + 0.06 = 0.49
d) P(E U F) = 1
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