What is the Probability of Independent Events with Given Probabilities?

[DAKIK]

Homework Statement

The two events A and B have probabilities 0.2 and 0.4. Also P (A n B)=0:08.
(a)Are the two events A and B independent? Explain.
(b) Find the probability that either A or B or both occur.
(c) Find the probability that neither A nor B occurs.
(d) Find the probability that exactly one of A or B occurs.

Homework Equations

P(A n B) = P(A)*P(B)

The Attempt at a Solution

My working out:

(a) For 2 events to be independent they must satisfy:
P(A n B) = P(A)*P(B)
so 0.08 = 0.2x0.4
which makes them independent .. Correct ?

(b) hmm not sure about this one! need some helpp

(c) P(neither A or B) = 0.8*0.6 Correct?

(d) hmm maybe P = 0.2*0.6 + 0.4*0.8 = 0.44 ?

 

[eumyang]

(A) is right.

For (B), you need a formula for P(A \cup B). But note that the problem ends with "or both."

For (C), what would be the compliment of "neither A nor B"?

(D) is wrong. You again need to find P(A \cup B), but note the difference between (D) and (B).69

 

[DAKIK]

(A) is right.

For (B), you need a formula for P(A \cup B). But note that the problem ends with "or both."

For (C), what would be the compliment of "neither A nor B"?

(D) is wrong. You again need to find P(A \cup B), but note the difference between (D) and (B).


69

so for (B) i should use: P(A \cup B) = P(A) + P(B) - P(A n B)

for (C) use P(A^c n B^c) = 1 - P(A \cup B) ??

for (D) use: P(A \cup B) = P(A) + P(B)

Need help asap

thankss

 

[eumyang]

so for (B) i should use: P(A \cup B) = P(A) + P(B) - P(A n B)

for (D) use: P(A \cup B) = P(A) + P(B)

You sure about these? Again, the end of (B) says "or both". P(A \cap B) would represent the "both," would it not? And in (D), we want "exactly one," so we cannot include both A and B.69

 

[DAKIK]

You sure about these? Again, the end of (B) says "or both". P(A \cap B) would represent the "both," would it not? And in (D), we want "exactly one," so we cannot include both A and B.


69

ok so would it look something like that

so for (B) i should use: P(A \cup B) = P(A) + P(B) + P(A n B)

for (D) use: P(A \cup B) = P(A) + P(B) - P(A n B)

 

[HallsofIvy]

so for (B) i should use: P(A \cup B) = P(A) + P(B) - P(A n B)

for (C) use P(A^c n B^c) = 1 - P(A \cup B) ??

for (D) use: P(A \cup B) = P(A) + P(B)

Need help asap

thankss

You sure about these? Again, the end of (B) says "or both". P(A \cap B) would represent the "both," would it not?

For (B), he is correct. "P(A\cap B)" is already included in both P(A) and
P(B). In order to count it only once we must subtract off one: P(A\cup B)= P(A)+ P(B)- P(A\cap B).

And in (D), we want "exactly one," so we cannot include both A and B.


69

So for D, you subtract off P(A\cap B) completely- twice.

 

[DAKIK]

For (B), he is correct. "P(A\cap B)" is already included in both P(A) and
P(B). In order to count it only once we must subtract off one: P(A\cup B)= P(A)+ P(B)- P(A\cap B).


So for D, you subtract off P(A\cap B) completely- twice.

what do u mean by subract off P(A\cap B) completely ?
so - P(A\cap B) twice ?

 

[eumyang]

For (B), he is correct. "P(A\cap B)" is already included in both P(A) and
P(B). In order to count it only once we must subtract off one: P(A\cup B)= P(A)+ P(B)- P(A\cap B).

That's it, no more posting early in the morning. >.<


69

 

[DAKIK]

Thanks for the help guys
Appreciate it