Probability
Question:
Given 3 random events A, B, C, and P(A) = P(B) = P(C) = 1/4, P(AB)=P(BC) = 0, P(AC) = 1/8, find the probability that at least one event has occured.
Solution: We are to find P(A∪B∪C).
Note that P(A∪B∪C) = P(A)+P(B)+P(C)-P(AB)-P(BC)-P(AC)+P(ABC).
Since P(ABC) >= 0 and P(ABC) < P(AB) = 0, we have P(ABC) = 0
Now we have 1/4 x 3 - 0 - 0 - 1/8 + 0 = 5/8.
Another question:
Denote a random event A such that A + A* = 1.
Now we have another random event B such that A and B are independent.
Given P(at least one event occuts) = 8/9 and P(A B*) = P(A* B).
Compute P(A).
Another question:
Denote a random event A such that A + A* = 1.
Now we have another random event B such that A and B are independent.
Given P(at least one event occuts) = 8/9 and P(A B*) = P(A* B).
Compute P(A).
Given P(at least one event occurs)
Solution:
P(A B*) = P(A) - P(AB)
P(A* B) = P(B) - P(AB)
Since P(A B*) = P(A* B),
we have P(A) = P(B)
Now since P (A∪B) = 8/9
We have P(A) + P(B) - P(AB) = 8/9
Also, P(AB) = P(A)P(B)
Then P(A) + P(B) - P(A)P(B) = 8/9
2P(A) - P^2(A) = 8/9
Denote P(A) = a, we have
2a - a^2 = 8/9
a^2 - 2a + 8/9 = 0
a = 2/3 or 4/3 (rejected)
i.e. P(A) = 2/3
QED
Note:
When P(AB) = P(A)P(B), P(AC) = P(A)P(C), P(BC) = P(B)P(C), P(ABC) = P(A)P(B)P(C), we say that A, B, C are mutually exclusive.
So we have another question:
When A and B are mutually exclusive and P(C) = 0,
are A*, B*, C* also mutually exclusive?
Note:
When P(AB) = P(A)P(B), P(AC) = P(A)P(C), P(BC) = P(B)P(C), P(ABC) = P(A)P(B)P(C), we say that A, B, C are mutually exclusive.
So we have another question:
When A and B are mutually exclusive and P(C) = 0,
are A*, B*, C* also mutually exclusive?
Solution:
Given P(AB) = P(A)P(B), and:
0 <= P(AC) <= P(C)
0 <= P(BC) <= P(C)
Now P(AC) = P(BC) = 0
Also,
0 <= P(ABC) <= P(C)
Then P(ABC) = 0
Now P(A) P(B) P(C) = P(A) P(B) x 0 = 0
Then P(ABC) = 0 = P(A)P(B)P(C)
Since A, B, C are mutually exclusive,
A*. B*, C* are also mututally exclusive.
So the answer is Yes.
A*. B*, C* are also mutually exclusive.
Note:
When P(AB) = P(A)P(B), P(AC) = P(A)P(C), P(BC) = P(B)P(C), P(ABC) = P(A)P(B)P(C), we say that A, B, C are mutually exclusive.
So we have another question:
When A and B are mutually exclusive and P(C) = 0,
are A*, B*, C* also mutually exclusive?
Solution:
Given P(AB) = P(A)P(B), and:
0 <= P(AC) <= P(C)
0 <= P(BC) <= P(C)
Now P(AC) = P(BC) = 0
Also,
0 <= P(ABC) <= P(C)
Then P(ABC) = 0
Now P(A) P(B) P(C) = P(A) P(B) x 0 = 0
Then P(ABC) = 0 = P(A)P(B)P(C)
Since A, B, C are mutually exclusive,
A*. B*, C* are also mututally exclusive.
So the answer is Yes.
Errata:
mutually exclusive independent
Question (MC):
When P(A) = 0 and P(B) = 1, we have:
(A) A = Ø, B = Ω
(B) A⊂ B
(C) A and B are mutually exclusive.
(D) A and B are independent.
Solution:
Note that P(A) = 0 does not imply impossibility and P(B) = 1 does not imply certainty. So option (A) is false.
Recall 0 <= P(AB) <= P(A) = 0
So P(AB) = 0
Also, P(A)P(B) = 0 x P(B) = 0
Then P(AB) = 0 = P(A)P(B)
So the correct option is (D).
Solution:
Note that P(A) = 0 does not imply impossibility and P(B) = 1 does not imply certainty. So option (A) is false.
Recall 0 <= P(AB) <= P(A) = 0
So P(AB) = 0
Also, P(A)P(B) = 0 x P(B) = 0
Then P(AB) = 0 = P(A)P(B)
So the correct option is (D).
Note that option (C) is valid for AB = Ø only.