This video is provided as supplementary

material for courses taught at Howard Community College and this video is going to be about the

probability of mutually exclusive and non-mutually exclusive events, and this topic is a the whole lot easier

than it sounds. So let’s get started. I’ve got a die. It’s a cube, so it has six sides. One side has a ‘1’ on it, one dot, then there’s two dots, three dots, four, five and six. And I’ve got two events. Event A is throwing either a ‘1’ or a ‘3’ and event B is throwing either a ‘2’ or

a ‘4’. Now I can figure out the probability of each of these events. The probability of event A is the number of ways that it could

happen, and since there are two different ways it could happen, either a ‘1’ or a ‘3’, that will be 2 divided by the number of different outcomes, which

is 6. So the probability of A is 2 over 6, or 1/3. I can do the same thing for the

probability of B. It’s also got two different ways of happening — either a ‘2’ or a ‘4’ — so the probability of B is also going to be 2 over 6. or 1/3. If I want to find the probability of either A or B happening, then I can think of that as the

probability of A union B. Now let’s just work this out logically,

and then come up with a formula for it. So the probability of A union B means I’ve got to get either a ‘1’ or a ‘3’ or a ‘2’ or a ‘4’. Now there are 4 different ways that this

could happen – ‘1’, ‘3’, ‘2’, and ‘4’ — so the probability would be 4 over 6, the total number of outcomes. I could reduce that to 2/3. And realize that’s just the same as the probability of A plus the probability of B. So if I want a general formula, I can say the probability the union of A and B, the probability of A happening

and the probability of B happening is going to equal the probability of A plus the probability of B. Now if this works because A and B are what we call mutually

exclusive events. I’ll write that… mutually exclusive events. Mutually exclusive events are events that don’t have anything in

common. I can think of this is a Venn diagram.

If I have Venn diagram, I would have two circles which don’t

overlap. One would be the circle for event A, with a ‘1’ and a ‘3’ in it. The other would be the circle for

event B, with a ‘2’ and a ‘4’. So if we have mutually exclusive events, we can just add up to their individual

probabilities and get the probability of their union. Now let’s compare that to a different situation. So now I’ve got event A, and event A

is a ‘2’, a ‘4’ or ‘6’, and event B would be throwing a ‘4’, a ‘5’,

or a ‘6’. Now I’ve got some overlap here. Both of these events have a ‘4’ in them

and a ‘6’ in them. In terms of a Venn diagram, — I’ll draw a Venn diagram down here — I would have two overlapping circles. Event A is going to have a ‘2’, and then the overlapping area

will have a ‘4’ and a ‘6’, because event B is ‘4’ ‘5’ and ‘6’. So I’ve got a ‘4’ and a ‘6’ in both of those area. Now let’s look at their individual

probabilities. and then try finding the probability of A or B. So the probability of A is going to be the number of ways A

could happen. There are three different ways A could happen. So that’s three over six, or 1/2. The probability of B is going to be the number of ways

B could happen. There are three different ways

B could happen, so that’s also three over six or 1/2. Now if I use the formula for

mutually exclusive events, I would say that the probability of A union B — A or B — equals the probability of B plus the probability of B. Now if I applied that formula, what that

means is I would be adding the probability of A, which is 1/2, and the probability of B, which is 1/2… I’ve be adding 1/2 and 1/2 and that equals one, or 100 percent, and that’s like saying that there’s a 100 percent certainty that either A or B will happen. But there isn’t, because I might throw a ‘1’, which is not in event A or event B. Or I might throw a ‘3’, which is not in event A or event B. So there can’t be a 100 percent

probability. Here’s what the problem is. Because I have that overlap, I’m actually including the probability of a ‘4’ or a ‘6’ two times, one with event A and one with event B. So what I want to do is get rid of one of

those times when I have the probability of ‘4’ or ‘6’. So what I’m going to do is take this formula,

the probability of A union B equals the probability of A plus the

probability of B, and I’m going to subtract the probability of that intersection the ‘4’ and the ‘6’.

So I’m subtracting the probability of A intersect B. Let’s work that out with the numbers we have. So the probability of A was 1/2.

I’m gonna use 3/6, that will make it easier to do this. And the probability of B is also 3/6. Let’s find the probability of A intersect B.

Well, A intersect B is ‘4’ and ‘6’. So that’s two different ways that A intersect B could happen. So that means I’m subtracting 2 divided by 6. When I do that, I’m going to get 3/6 plus 3/6 minus 2/6. That works out to 3 plus 3 which is 6, minus 2, which is 4 over 6. And that reduces down to 2/3. Let’s look at this a different way and make

sure it works. If I want the probability of either A

or B, that means I’m looking for the

probability of a ‘2’ or a ‘4’ or a ‘5’ or a ‘6’. Those are really the numbers I’m dealing with.

There are 4 ways that could happen. There are 6 different total outcomes. So the probability of A or B of ‘2’, ‘4’, ‘5’ or ‘6’ happening is just 4 over 6, or 2/3, which is what I get with this formula. So here’s what that means in terms of

how much you might have to memorize. If I have mutually exclusive events, there is no

overlap. Since there is no overlap, the

intersection of those two events is zero. So I can use this formula. I can say that the probability of those

mutually exclusive events of A or B, is the probability of A plus the probability of B minus the probability of their

intersection, which is zero. iIf I have non-mutually exclusive events, like the ones i have here, with a ‘2’, ‘4’ and ‘6’ and then ‘4’, ‘5’ and ‘6’, once again I’ll use this formula — the

probability of A or B is equal to the probability of A plus

the probability of B minus the probability of the intersection of A and B. So that’s basically how this works. Run through this a couple of times.

It seems like it’s confusing until you just work through it and

realize how logically it works out with actual

numbers. So give it a try. Take care. I’ll see you next time.

Hello, how are you coming up with the 6 for the bottom number? Can you help me with a problem? the possibility of space consists of the integers from 1 to 20 exclusively. A is the event the number is a multiple of 3. B is the event the number is a multiple of 4. An integer is selected at random. Find P(A) and P(B).

Also, what do you do when you find the person to look for in finding the median is a half or quarter person? (The cumulative frequency plus one divided by 2) ?

this video really helped me..thank you a lot for doin this

This video really helped me on that minus 2/6 part as my Stat textbook doesn't explain it well. Thank you!

Thank you 🙂 This helped a lot!

Cleared my doubt of difference between the 2.Thanks a lot.

nice vid

Very good video. Made things very simple.

THANK YOU VERY MUCH, VERY CLEAR EXPLANATION!

Isnt P(A andB) = ¼ ??? I thought P(A and B) is P(A) * P(B)?

great teaching, very clear.

what we would do if we are not even given sample space?

e.g find P(C and D)=?

when P(C)=19/30, P(D)=2/5, P(CUD)=4/5, P(C and D)=?

I've been trying to understand it but my homework is different from what you are doing. For example, you spin the spinner and flip a coin. Find the probability of the events.

1. spinning a 2 and flipping tails

2. not spinning a 4 and flipping tails

Please explain why

how do you get 2/6 at 1:35 please?

this was very helpful. thank you very much. you have no idea how much i appreciated this.

Thank you so much for this video! It helped me understand what this complicated book was saying! （＾ｖ＾）

Mutually Exclusive=No common elements.

Non Mutually Exclusive=Some or All Common Elements.

Thanx for the video i just now know how to deal with probablity problems.

Awesome!!

Thanks a lot for video! Now I got what that mean)))

Thanks

Thank you, very helpful!

Thanks

thanks !!!!It helped me a lot !!!Doubts are completely cleared!!!

thanks so much

Thank you for the video. I appreciate it!

Hi guys, I have only one question, the JOINT and NON-DISJOINT events are the same thing or not? please help me with this. and if this possible, explain if NO then what is this…

god bless you sir! take care always too :))

I get why it makes sense to subtract the 2/6 at the end but mathematically if P(A) = 1/2 and P(B) = 1/2 then P(AnB) = 1/8. What am I missing?

thamks

THANK YOU SO MUCH I'M CRYIN' YOU ARE A BLESSINg

Nice learned it in one night! Thank you sooo much sir!!!

Thank you very much sir!

screenshot… now i have some more notes lol. Thanks, great vid

WOW .. THANK U SO MUCH FOR THIS

Thank you so much! I actually understand it more when you discuss it compared to my prof.

really clears things up

You sound like Patrick Dempsey!

could you please tell me, how did you got 6 in the denominator in the first problem? I thought it was 4

NICE

Where did you get 6?

This is so obvious

Wtf

Fuck your formula

This helped me sooo much thank you so so so so much you are the best

tank you

niece video sir everything were understand easilu

the real question is when to add or when to multiply, when i think about it i go nuts, let god bring his wrath on the guy who found this miserable piece of shit probability.

I got it. Thankyouu so much

OMG , I FINALLY UNDERSTAND THIS, THANK YOU SO MUCH, BEST EXPLANATION EVER. GOD BLESS U 🙂

THANK YOU!!!

What a great guy

Hello,

I'm using the formula P(A) + P(B)-P(A and B) for the second part and can't get the same answer.

I re-wrote the formula to be P(A)+P(B)-P(A)P(B|A) using the conditional probability rule P(B|A) = P(BA)/P(A)

wouldn't the two P(A) cancel each other out leaving you with P(BA) = 1/4?

Great video! Too bad it is wrong… Consider remaking it before people take this to classes, lectures or exams. The p(AuB)=p(A)*p(B), whilst the p(AnB)=p(A)+p(B). Exactly the other way around

This is actually so crappy

Still confused

Good

Thank you SO much for these videos. I learned more in the past 3 hours watching them, than I have in my entire Statistics class combined this semester.

I just wanted to let you know that your videos are awesome. You have the best way of explaining things. I'm studying for the GRE and I've watched several videos on probability, and yours is by far the best. It's been 10+ years since I graduated from college with a BA and I'm wanting to go to grad school. I'm having to learn all my math all over again and it's been such a frustrating journey. You've made it so much easier. Thank you so much. You should be a GRE tutor.

why are you placing every probability under the same denominator 6…….i dont understand

I have exam tomorrow

Thank you Teather

thank you for uploading this vid! I think this vid just save my grades for midterms HAHAHAHA

THANK U A LOT

PERFECT EXPLANATION thank u so much

Bless your soul

I love it

Good luck to me for reporting tommorrow

Ohhh thank you sir

My confusion is gone now bcoz of u

this is wonderful job .keep up .you educate students

so you use the last number for the bottom number

thanks

Thank you for the great video.

fck we hav quiz tommorow…

Thank you for this!!!

You earn a sub bro it help me in our basic statistic subject thank you man.

kumakapit sa youtube kay bobo ag maestro rawrrr

you should be a teacher, you're so good !!!

Thank you so much for the help!

This was wonderful! Thank you!

Amazing, I experienced a clarity in your voice, it was so logical… thank you so much

😍

But how do find if it is inclusive? How do you know which formula to use?

I am really confused on that!!🤨😥😟

At 7:28 where did you get 2/6 from 4/6???