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).


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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. >.<


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[DAKIK]

Thanks for the help guys
Appreciate it