Probability

Academic
#1 X
26/05/24 01:12

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.

#2 X
26/05/24 01:16

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.

#3 X
26/05/24 01:19

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).

#4 X
26/05/24 01:20

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)

#5 X
26/05/24 01:25

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

#6 X
26/05/24 01:29

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?

#7 X
26/05/24 01:34

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.

#8 X
26/05/24 01:35

A*. B*, C* are also mutually exclusive.

#9 X
26/05/24 01:39

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

#10 X
26/05/24 01:42

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.

#11 X
26/05/24 01:46

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).

#12 X
26/05/24 01:47

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.

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