Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability help (2 Viewers)

[blackops23]

Hi guys, I need help with understanding an crucial concept regarding P(AB), i.e. the intersection of events A and B, in relation to non-mutually exclusive events, and other probability questions.


It's known for non-mutually exclusive events,
P(A and B) = P(A) + P(B) - P(AB)

All this time I thought P(AB) = P(A) * P(B), i.e. the product of the probabilities of A and B.
For some questions, finding P(AB) by doing P(A) * P(B) ACTUALLY WORKED.

e.G. for this question:

Q1. A pack of cards is numbered 1 to 10, one card is drawn at random, What is the probability that the card drawn is: less than 5 or divisible by 2?

Now its obviously an extremely simply question, but it's the theory I'm having trouble with. So in the question:

P(A) + P(B) - P(AB) = 4/10 + 5/10 - 2/10 = 7/10
Now in this question P(AB) = P(A) * P(B), as (4/10)*(5/10)=2/10

Also in this question:
Q2. A set of 30 discs are numbered 1 to 30, one disc is selected at random. What's the probability that it is a MULTIPLE OF EITHER 5 OR 3?

Answer: Let A = event of multiple of 5
Let B = event of multiple of 3
Now P(A) + P(B) - P(AB) = 1/5 + 1/3 - 1/15
Once again, P(AB) = P(A)*P(B)

------------------------------------------------------------------------------

But for other questions, P(AB) = P(A)*P(B) does not work.
E.g. in this one:

Q3. A set of 30 discs are numbered 1 to 30, one disc is selected at random. What's the probability that it is a MULTIPLE OF EITHER 3 or 7?

Answer:
P(A) + P(B) - P(AB) = 10/30 + 4/30 - 1/30
= 1/3 + 2/15 - 1/30
= 13/30

Note in this question P(AB) does not equal P(A) * P(B).

So what I want to know is:

WHY does P(AB) = P(A)*P(B) in some questions, but not all. Is drawing a Venn Diagram to find the intersection values in A and B the ONLY way to calculate P(AB)??

Thanks guys

 

[Bored Of Fail 2]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

for the first one you are only making one choice

P(AB) = P(A) x P(B) is used when you have two events following, one after the other

 

[Bored Of Fail 2]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

similarly for the second question, the events are not following on from each other

 

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

1) for mutually exclusive (also called disjoint) events A & B:

P(A + B) = P(a) + P(B)


In general: P(A + B) = P(A) + P(B) - P(A x B)

But for mutually exclusive events:

P(A x B) = 0 [i.e. prob(event A & event B occuring together) = 0]

That's why for mutually exclusive events. the P(A x B) term simply drop out.

2) P(A + B) is often written and P(A x B) or P(AB) as

3) P(A + B) simply means: the probability of A or B occuring whereas P(A x B) means: the probability of the 2 events occuring together (i.e. jointly)

4) In Q1: event A: "less than 5" i.e. you draw 1, 2, 3 or 4
. . . . . . . event B: "divisible by 2" i.e. you draw a 2,4, 6,8 or 10
. . . . . . . joint event AxB: "divisible by 2 as well as less than 5" i.e. you draw a 2 or 4

Here A and B are not mutually exclusive; so P(A + B) = P(A) + P(B) - P(AxB) = 4/10 + 5/10 - 2/10 = 7/10

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

similarly for the second question, the events are not following on from each other

Looking at Q2 and Q3, you can see they are the same, except for what it is asking of you, i.e. Q2 asks for multiples of 3 and 5, and Q3 asks for 3 and 7.

So why does P(AB)=P(A)*P(B) work for Q2 but not Q3?

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

1) for mutually exclusive (also called disjoint) events A & B:

P(A + B) = P(a) + P(B)


In general: P(A + B) = P(A) + P(B) - P(A x B)

But for mutually exclusive events: P(A x B) = 0 [i.e. prob(event A & event B occuring together) = 0

That's why for mutually exclusive events. the P(A x B) term simply drop out.

No, I was asking for non-mutually exclusive events, e.g. look at Q3.

In Q3, P(A) = 10/30, P(B)= 4/30
Therefore P(A) * P(B) = 2/45
BUT, P(AB) = 1/30 because 21 is the only number is the set{1to30} that is a multiple of 3 and 7.

So here, a non-mutually exclusive event, P(AB) doesn't equal P(A) * P(B). I'd really like to know why?

thanks.

 

[deterministic]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

P(A and B)=P(A)*P(B) has nothing to do with non-mutually exclusive events. Rather the result implies that A and B are statistically independent events. Intuitively, independent events are those where probability of event B occurring is not affected by the fact that event A occurred, and vice versa. This involves the concept of conditional probability. In your examples:

1) P(x3)=1/3, P(x5)=1/5. Suppose A has occurred ie. I know the number is a multiple of 3. That means I have 10 numbers to choose from, and 2 of them are multiple of 5. So:
P(choosing a multiple of 5 given a multiple of 3 has been chosen)=1/5
Thus, we can see that choosing a multiple of 3 has no effect on the probability of choosing the multiple of 5, so hence these are independent. So P(A)P(B)=P(AB) holds.

2) P(x3)=1/3, P(x7)=2/15. Suppose A has occurred ie. I know the number is a multiple of 3. That means I have 10 numbers to choose from, and 1 of them are multiple of 7. So:
P(choosing a multiple of 7 given a multiple of 3 has been chosen)=1/10
Thus, we can see that choosing a multiple of 3 has changed the probability of choosing the multiple of 7, so hence these are not independent, so P(A)*P(B)=P(AB) does not hold.

For these types questions, it is advisable to use the venn diagram instead of relying on P(A)P(B)=P(AB), as independence may not be intuitive or immediately implied by the question. For questions like:

"What is the probability of rolling a 6 and tossing a head?"

we can easily see that rolling a 6 should have no physical effect on the toss of the coin, so they are physically independent, which implies statistical independence. For your examples, it is not apparent the events will be independent.

 

[deterministic]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

In general
P(A and B) = P(A)*P(B GIVEN A has occured) = P(B)*P(A GIVEN B has occured)
where P(A GIVEN B has occured) is a conditional probability (the condition is that B has occured)
Only when P(B)=P(B GIVEN A has occured) and P(A)=P(A GIVEN B has occured) are events A and B are independent, so then we can use P(A)*P(B)=P(AB)

"Consider a deck of cards. I choose 2 cards. What is the probability of choosing an ace of spades first and then the ace of diamonds?"

For the first pick, I can choose ace of spades out of 52 cards.
Let A=ace of spades
B=ace of diamonds
P(A)=1/52
For the second pick, I can choose ace of diamonds out of 51 cards.
P(B GIVEN A [ie.I chose the ace of spades first])=1/51
So P(event)=P(A)*P(B GIVEN A)=1/52*1/51
Note that P(B) (non conditional)=1/52 so P(B)=/= P(B GIVEN A)

"Consider a coin and a fair die. Find the probability of throwing a head and rolling a 6.

Let A=head
B=roll a 6
P(A)=1/2
P(B GIVEN A [ie.I threw a head])=1/6
Note that P(B) =1/6 so P(B)= P(B GIVEN A) so A and B are independent.

 

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

It's known for non-mutually exclusive events,
P(A or B) = P(A) + P(B) - P(AB)

Note in this question P(AB) does not equal P(A) * P(B).




Thanks guys

P(A x B) = P(A) x P(B) if and only if A & B are INDEPENDENT events.

As deterministic has correctly pointed out: P(AxB) = P(A) x P(B|A) also = P(B) x P(A|B)

where P(A|B) is the conditional probability of the event A given that the event B has occured. If events A and B are independent, then P(A|B) simply = P(A); it's like event A saying "who cares; event B has nothing to do with me; i.e. whether B occurs or not makes no difference to whether or not I occur"

Note also: it is not correct to write P(A and B) when you mean P(A or B)

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

P(A x B) = P(A) x P(B) if and only if A & B are INDEPENDENT events.

As deterministic has correctly pointed out: P(AxB) = P(A) x P(B|A) also = P(B) x P(A|B)

where P(A|B) is the conditional probability of the event A given that the event B has occured. If events A and B are independent, then P(A|B) simply = P(A); it's like event A saying "who cares; event B has nothing to do with me; i.e. whether B occurs or not makes no difference to whether or not I occur"

Note also: it is not correct to write P(A and B) when you mean P(A or B)

Correct me if I'm wrong, but isn't that the same as mutually exclusive events, that is, occurrence of one event automatically excludes all others. If I'm wrong, what is the difference between this condition probability and mutually exclusive events?

 

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Correct me if I'm wrong, but isn't that the same as mutually exclusive events, I mean the occurrence of one event automatically excludes all others. If I'm wrong, what is the difference between this condition probability and mutually exclusive events?

events A & B are mutually exclusive means: if A occurs then B cannot occur and vice versa

events A & B are independent means whatever happens to event A has no effect on event B and vice versa; they can both occur together.

One simple example: Throw a coin twice.

Let A be the occurence of a head in the 1st throw

and B the occurence of, say, a tail in the 2nd throw.

Then the occurence of B does not depend on the occurence of A. In fact whatever the outcome of the 1st throw, the probability of getting a tail in the 2nd throw remains the same; event B is indifferent to the outcome of the 1st throw.

It's interesting that if we have tossed a coin 10 times and got all heads, we tend to think that the next one is more likely to be a tail; but this is not true. The coin has no memory of what has happened previously.

On the other hand consider this situation. An urn has 3 red and 5 blue balls. Say you randomly pick the 1st ball and do not put it back. What is the prob of getting a red in the 2nd draw? Well this 2nd outcome depends on the 1st outcome. If the 1st drawn was a red, then P(2nd red|1st red) = 2/7. If the 1st was blue, then P(2nd red|1st blue) = 3/7.

Note: P(1st red & 2nd red) = P(1st red) x P(2nd red|1st red) = 3/8 x 2/7 = 3/28

and P(1st blue & 2nd red) = P(1st blue) x P(2nd red|1st blue) = 5/8 x 3/7 = 15/56

 

[Bored Of Fail 2]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

The coin has no memory of what has happened previously.

how do you know

 

[Bored Of Fail 2]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

thats just like something I was reading about pokie machines. Supposely a pokie machine works by randomly generating tons of combinations every millisecond or whatever, and the outcome that you get all depends on the exact combination ( out of like how every many hundred thousand it generates each second ) that was generated at the time you pressed the "bet" button.

But if the pokie machine has no memory then how do they enforce the minium payout laws, there is some law that says that a pokie machine must payout a certain percentage of money that is put into it.

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

events A & B are mutually exclusive means: if A occurs then B cannot occur and vice versa

events A & B are independent means whatever happens to event A has no effect on event B and vice versa; they can both occur together.

One simple example: Throw a coin twice.

Let A be the occurence of a head in the 1st throw

and B the occurence of, say, a tail in the 2nd throw.

Then the occurence of B does not depend on the occurence of A. In fact whatever the outcome of the 1st throw, the probability of getting a tail in the 2nd throw remains the same; event B is indifferent to the outcome of the 1st throw.

It's interesting that if we have tossed a coin 10 times and got all heads, we tend to think that the next one is more likely to be a tail; but this is not true. The coin has no memory of what has happened previously.

On the other hand consider this situation. An urn has 3 red and 5 blue balls. Say you randomly pick the 1st ball and do not put it back. What is the prob of getting a red in the 2nd draw? Well this 2nd outcome depends on the 1st outcome. If the 1st drawn was a red, then P(2nd red|1st red) = 2/7. If the 1st was blue, then P(2nd red|1st blue) = 3/7.

Note: P(1st red & 2nd red) = P(1st red) x P(2nd red|1st red) = 3/8 x 2/7 = 3/28

and P(1st blue & 2nd red) = P(1st blue) x P(2nd red|1st blue) = 5/8 x 3/7 = 15/56

Ok so I guess there is conditional probability in questions involving item not being replaced?? Could you please relate the conditional probability to Q2 and Q3 in my 1st post, because for Q2, I don't understand why P(x3) doesn't affect P(x5) yet in Q3 P(x3) affects P(x7)??

Thanks

 

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

how do you know

The coin is inanimate; it has no brain; it has no memory. Say we have a not-unbiased coin with P(head) = 0.4 and P(tail) = 0.6

Each throw the coin will obligingly land head with a prob of 0.4 and a tail with a prob of 0.6. After a toss of the coin, it does not remember (nor does it care) how it landed, in the nest toss.

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

could someone please look at my above post? Thanks

 

[Drongoski]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Ok so I guess there is conditional probability in questions involving item not being replaced?? Could you please relate the conditional probability to Q2 and Q3 in my 1st post, because for Q2, I don't understand why P(x3) doesn't affect P(x5) yet in Q3 P(x3) affects P(x7)??

Thanks

Q2: P(A) = P(x5) = P(5 or 15 or . . . 30) =6/30

and P(B) = P(x3) = P(3 or 6 or . . . 30) = 10/30

P(A&B) = P(x15) = P(15 or 30) = 2/30 [events A abd B are not mutually exclusive]

Here P(A&B) = 2/30 = 6/30 x 10/30 = P(A) x P(B) ==> A and B are independent


Q3: P(A) = P(x7) = P(a 7 or 14 or 21 or 28) = 4/30

and P(B) = P(x3) = 10/30

P(A&B) = P(a multiple of 3x7) = P(a 21) = 1/30 [again not mutually exclusive, since there is an outcome '21']

Now P(A&B) = 1/30 =/= 4/30 x 10/30 = P(A) x P(B) ==> events A & B not statistically independent.

 

[cutemouse]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Conditional probability isn't in the course. Don't worry about it until uni.

 

[blackops23]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

Q2: P(A) = P(x5) = P(5 or 15 or . . . 30) =6/30

and P(B) = P(x3) = P(3 or 6 or . . . 30) = 10/30

P(A&B) = P(x15) = P(15 or 30) = 2/30 [events A abd B are not mutually exclusive]

Here P(A&B) = 2/30 = 6/30 x 10/30 = P(A) x P(B) ==> A and B are independent


Q3: P(A) = P(x7) = P(a 7 or 14 or 21 or 28) = 4/30

and P(B) = P(x3) = 10/30

P(A&B) = P(a multiple of 3x7) = P(a 21) = 1/30 [again not mutually exclusive, since there is an outcome '21']

Now P(A&B) = 1/30 =/= 4/30 x 10/30 = P(A) x P(B) ==> events A & B not statistically independent.

Could I just ask why in Q2, A and B are statistically independant, yet it in Q3, A and B, they aren't statistically independant?

 

[deterministic]

Re: Don't understand concept of P(AB) in non-mutually Exclusive Events - Probability

purely by coincidence