Probability?

Question

The tow events A and B have probabilities .3 and .7.
i. If A and B are independent, find P(A U B).
ii. If A and B are mutually exclusice, find the prob that neither A nor B occurs

I totally dont understand part ii. How can neither occur! one of thm has to occur dsnt it?

Answer 1

Note that all of the answers posted above are wrong except Joe Mkt's.

Well if you solve the problem then you will know if one has to occur or not. Since both probabilities add up to 1, one will have to occur as long as they are mutually exclusive. Mutually exclusive means they can't happen at the same time so they must add up to one.

For independent events you can multiply the events together when you have a intersection (an upside down U). I will call it an intersection here since I can't write upside down U's.

i. P(A U B) = P[A] + P[B] - P[A intersection B] which equals:
.3 + .7 - (.3 * .7) = 1 - .21 = .79 Note that normally we can't multiply A and B together but we can in this problem because they told us they were independent.

ii. If they are mutually exclusive then we can add the unions without subtracting the intersection of the two events because we don't have to worry about them intersecting. The reason for this is mutually exclusive events can't happen at the same time. For example if they are mutually exclusive then P[AUB] = P[A] + [PB] because P[A intersection B] = 0 so we don't need to subtract it.

The probability that neither A or B occurs is written as
P[A' intersection B'] P[ (AUB)' ] Anytime we have mutually exclusive events we most likely want the probability in union form so we can add the two events together so thats what we will do.By De Morgans law P[A' intersection B'] = P[ (AUB)' ]. This in turn equals 1 - P[AUB]. We know P[AUB] = 1 since we can just add P[A] and P[B] together since they are mutually exclusive. Therefore we get 1 - 1 = 0

I hope this helped and good luck with probability in the future. It is an exciting subject.

Answer 2

If they are independent events (i.e., the occurrence of one has no bearing on the occurrence of the other), then it's quite possible that neither will occur. You can expect that to happen 21% of the time, since the probability that neither A nor B occurs is (1 - 0.3)(1 - 0.7) = 0.7 * 0.3 = 0.21

Answer 3

P(A)=0.3
P(B)=0.7

i) If A and B are independent that means that
P(A U B)=P(B)*P(B)
P(A U B)=0.3*0.7 = 0.21

ii)A and B mutually exclusive means that
P(A n B) = impossible
The probability that neither A or B occurs is
P(A(compliment) U B(compliment))
= ( 1-P(A) ) + ( 1-P(B) ) - P(A(compliment) n B(compliment)) .
=(1-0.3 )+( 1-0.7) - (1 - P(A U B))
=1-(1-0.21)
=0.21

(draw a Venn-diagram if you are having trouble with part ii)

Answer 4

i. P(A U B) = P(A) + P(B) - P(A)P(B) #A and B are independent
P(A U B) = .3 + .7 - .21 = .79
ii. P(A U B) = P(A) + P(B) = 1
P(A U B)' = 1 - P(A U B) = 1 - 1 = 0

Answer 5

Leave the calculation part of it .take an example .Boy comes to see a girl with parents .Her parents also are there,Friends and relatives are also there .fOR A DECISION TO BE TAKEN ,SEE HOW MANY RESULTS WE CAN CONCLUDE .BOY OR GIRL MAY NOT EVEN AGREE OR DIS AGREE .BOTH OF THE ALSO KEEP THEIR SECRETS WITH THEM .SO CAN YOU BLAME ANY ONE IF 100% POSSIBILITY OF THEY TWO WILL MARRY..IS IT NOT COMING DOWN ALSO 0%
LINK YOUR QUESTION AND DOUTS U HAVE .NEIGHTHER OCCURS .BOY AGREES .GIRL DOES NT AGREE.SO ONE SAYING HAS A POINT THAT IS WHY WE CALL PROBALITY WHICH IS JUST TO SAY BUT NOT TO HAPPEN