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## Statistics and probability

# Probability with Venn diagrams

AP.STATS:

VAR‑4 (EU)

, VAR‑4.E (LO)

, VAR‑4.E.4 (EK)

CCSS.Math: , Probability of compound events. The Addition Rule. Created by Sal Khan.

## Want to join the conversation?

- Maybe this is explored later in the play list but is it just coincidence that P(J and H) = P(J)*P(H) or can we always use this? I know he goes on to show the formula for P(A or B) but was wondering about the formula for P(A and B).(21 votes)
- It is not a coincidence. It is the same thing.

For P(J and H) you can simply count within the deck the number of Jacks that are also Hearts, like in the video, and you'll get 1/52. Or you can look at the chances of picking a Jacks, and out of those jacks, what are the chances of choosing one that is a Hearts.

If I look at the chances of grabbing a Jacks, P(J), we get that there are 4 Jacks in the deck, so 4/52 = 1/13. And now, out of those four Jacks, how many are Hearts? Only one, so we get a probability P(H) of 1/4 to pick one card that is Hearts. This means that there is a 1/4 chance within the 1/13 chance to get a Hearts that is also a Jacks. This is represented by multiplying both probabilities (1/13)*(1/4) or P(J)*P(H) like you stated. You're essentially applying a probability to another probability. :)(28 votes)

- In the first example the answer is 1/13. Isn't that means that if you randomly draw 13 cards from the deck then at least 1 card is a Jack?(15 votes)
- No, This is not necessary. Probability of 1/13 means that if we keep drawing out 13 cards at random, then on an average we should have one jack per lot of 13 drawn out. The whole point of Probability is that there is one chance out of every 13 chances that the card drawn out is a jack. In fact, in the worst case scenario, its very much possible that out of 52 cards, we draw out 48 cards one by one, and not one of them is a jack!

But then we'll compulsorily have to draw jacks on the next 4 turns, so finally the probability turns out to be 1/13 only.(21 votes)

- A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart ?(8 votes)
- there 1 "queen of club" and 1 "king of heart" so

the probability a queen of club is 1/52 because there is only one in the neck.

and the probability a king of heart is 1/52 because there is only one in the neck.

1/52 + 1/52 = 2/52 = 1/26

there for you have a probability of 1/26 to get a queen of club or a king of heart.(16 votes)

- 1. In a standard deck of cards, find the probability of selecting a club or an ace on a single draw.(5 votes)
- In a 52 card deck there 13 cards of clubs and 4 aces. So, on a single draw the probability of selecting a club is 13/52 and the probability of selecting an ace is 4/52.

If you want the probability of selecting a club or an ace you should be carefull not count the ace of clubs twice. The probability is then 16/52 (13 clubs + 4 aces - ace of clubs)

Sometimes 40 card decks are used instead (10 cards of clubs and 4 aces). In that case the probability of selecting a club is 10/40 and the probability of selecting an ace is 4/40. The probability of selecting a club or an ace is then 13/40 (10 clubs + 4 aces - ace of clubs)(10 votes)

- HI, I thought P(J or H) would be = P(J) + P(H). In this case, 1/4 + 1/13. Why not?(6 votes)
- P[J or H]=P[J]+P[H]-P[J and H],because P[J and H] repeat.(2 votes)

- how many jacks,hearts,queens in playing cards(4 votes)
- There are 52 cards in a deck with 4 suits, so there are 13 cards in each suit. There are 13 hearts. 4 jacks and 4 queens because there is one in each suit(2 votes)

- what does mutually exclusive mean ??(2 votes)
- Two events are mutually exclusive if they cannot occur together.(4 votes)

- Around9:40ish where we add the two sets {4/52}A{13/52} ie; (4+13)/(52) and then proceed to subtract the shared set, I believe we should be subtracting the fractional {1/52} instead of 1 - this would fit our answer more appropriate to convention. Mathematically speaking, subtracting a unit of one from a fraction is the same as negating the fraction by a unit whole based on its denominator.

So in this case, our equation would be :

=>P(j) + P(h) - P(j + h) = P(j or h)

=>4/52 + 13/52 - 1 ~ 4/52 + 13/52 - 52/52

=>17/52 - 52/52 => -(35/52)

.:.P(j or h) = -35/52

which doesn't make sense. So in order for this to be accurate, we need to be subtracting 1/52, which would give us (17/52) - (1/52) = 16/52 = 4/13(1 vote)- Yes, the answer is 16/52, Sal says and writes out this fact. The "-1" is in the numerator, he just didn't extend the fraction bar all the way.(6 votes)

- why isn't the joker card being used ?(2 votes)
- Just so it's consistent no matter the deck. A deck may have more than 1 joker card, or none at all. All decks have the base 52 though.(3 votes)

- If there are cards 1, 2, 3, 4, 5 and 6, what is the probability of selecting a 3 first?(0 votes)
- 1/6 because there are 6 choices and only one 3 so 1/6 is the answer(6 votes)

## Video transcript

Let's do a little bit of
probability with playing cards. And for the sake of
this video, we're going to assume that our
deck has no jokers in it. You could do the same
problems with the joker, you'll just get slightly
different numbers. So with that out of
the way, let's first just think about
how many cards we have in a standard playing deck. So you have four
suits, and the suits are the spades, the diamonds,
the clubs, and the hearts. You have four suits and
then in each of those suits you have 13 different
types of cards-- and sometimes it's
called the rank. You have the ace, then you have
the two, the three, the four, the five, the six,
seven, eight, nine, ten, and then you have the Jack,
the King, and the Queen. And that is 13 cards. So for each suit
you can have any of these-- you can
have any of the suits. So you could have a Jack of
diamonds, a Jack of clubs, a Jack of spades,
or a Jack of hearts. So if you just multiply
these two things-- you could take a deck of playing
cards, take out the jokers and count them-- but
if you just multiply this you have four suits, each
of those suits have 13 types. So you're going to
have 4 times 13 cards, or you're going to have 52 cards
in a standard playing deck. Another way you could
have said, look, there's 13 of these
ranks, or types, and each of those come in four
different suits-- 13 times 4. Once again, you would
have gotten 52 cards. Now, with that of
the way, let's think about the probabilities
of different events. So let's say I
shuffle that deck. I shuffle it really,
really well and then I randomly pick a
card from that deck. And I want to think about
what is the probability that I pick a Jack. Well, how many equally
likely events are there? Well, I could pick any
one of those 52 cards. So there's 52 possibilities
for when I pick that card. And how many of those 52
possibilities are Jacks? Well you have the Jack of
spades, the Jack of diamonds, the Jack of clubs, and
the Jack of hearts. There's four Jacks in that deck. So it is 4 over 52-- these
are both divisible by 4-- 4 divided by 4 is 1, 52
divided by 4 is 13. Now, let's think
about the probability. So I'll start over. I'm going to put that
Jack back and I'm going to reshuffle the deck. So once again, I
still have 52 cards. So what's the probability
that I get a hearts? What's the probability
that I just randomly pick a card from a shuffled
deck and it is a heart? Well, once again,
there's 52 possible cards I could pick from. 52 possible, equally likely
events that we're dealing with. And how many of those
have our hearts? Well, essentially 13
of them are hearts. For each of those suits
you have 13 types. So there are 13
hearts in that deck. There are 13 diamonds
in that deck. There are 13 spades
in that deck. There are 13 clubs in that deck. So 13 of the 52 would result
in hearts, and both of these are divisible by 13. This is the same thing as 1/4. One in four times
I will pick it out, or I have a one in four
probability of getting a hearts when I randomly pick a card
from that shuffled deck. Now, let's do something that's
a little bit more interesting, or maybe it's a little obvious. What's the probability
that I pick something that is a Jack-- I'll just
write J-- and it is a hearts? Well, if you are reasonably
familiar with cards you'll know that
there's actually only one card that is
both a Jack and a heart. It is literally
the Jack of hearts. So we're saying, what
is the probability that we pick the exact
card, the Jack of hearts? Well, there's only
one event, one card, that meets this criteria
right over here, and there's 52 possible cards. So there's a one
in 52 chance that I pick the Jack of hearts--
something that is both a Jack and it's a heart. Now, let's do something a
little bit more interesting. What is the
probability-- you might want to pause this and think
about this a little bit before I give you the answer. What is the probability
of-- so I once again, I have a deck of 52
cards, I shuffled it, randomly pick a card from that
deck-- what is the probability that that card that I pick from
that deck is a Jack or a heart? So it could be the
Jack of hearts, or it could be the
Jack of diamonds, or it could be the
Jack of spades, or it could be the
Queen of hearts, or it could be
the two of hearts. So what is the
probability of this? And this is a little bit
more of an interesting thing, because we know, first
of all, that there are 52 possibilities. But how many of
those possibilities meet these conditions that
it is a Jack or a heart. And to understand that,
I'll draw a Venn diagram. Sounds kind of fancy,
but nothing fancy here. So imagine that this
rectangle I'm drawing here represents all of the outcomes. So if you want, you could
imagine it has an area of 52. So this is 52 possible outcomes. Now, how many of those
outcomes result in a Jack? So we already learned, one out
of 13 of those outcomes result in a Jack. So I could draw a
little circle here, where that area-- and I'm
approximating-- represents the probability of a Jack. So it should be
roughly 1/13, or 4/52, of this area right over here. So I'll just draw it like this. So this right over here is
the probability of a Jack. There's four possible
cards out of the 52. So that is 4/52,
or one out of 13. Now, what's the probability
of getting a hearts? Well, I'll draw another
little circle here that represents that. 13 out of 52 cards
represent a heart. And actually, one of those
represents both a heart and a Jack. So I'm actually going
to overlap them, and hopefully this will
make sense in a second. So there's actually 13
cards that are a heart. So this is the number of hearts. And actually, let me write this
top thing that way as well. It makes it a little bit
clearer that we're actually looking at the number of Jacks. And of course,
this overlap right here is the number of Jacks
and hearts-- the number of items out of this 52 that
are both a Jack and a heart-- it is in both sets here. It is in this green circle and
it is in this orange circle. So this right over here--
let me do that in yellow since I did that problem in
yellow-- this right over here is a number of Jacks and hearts. So let me draw a
little arrow there. It's getting a little
cluttered, maybe I should draw a little
bit bigger number. And that's an
overlap over there. So what is the probability
of getting a Jack or a heart? So if you think about
it, the probability is going to be the
number of events that meet these conditions,
over the total number events. We already know the total
number of events are 52. But how many meet
these conditions? So it's going to be the
number-- you could say, well, look at the green
circle right there says the number that gives us a Jack,
and the orange circle tells us the number that
gives us a heart. So you might want to say,
well, why don't we add up the green and the
orange, but if you did that, you would
be double counting, Because if you add
it up-- if you just did four plus 13--
what are we saying? We're saying that
there are four Jacks and we're saying that
there are 13 hearts. But in both of these, when we
do it this way, in both cases we are counting
the Jack of hearts. We're putting the
Jack of hearts here and we're putting the
Jack of hearts here. So we're counting
the Jack of hearts twice, even though there's
only one card there. So you would have to subtract
out where they're common. You would have to
subtract out the item that is both a Jack and a heart. So you would subtract out a 1. Another way to
think about it is, you really want to figure
out the total area here. And let me zoom in-- and I'll
generalize it a little bit. So if you have one
circle like that, and then you have another
overlapping circle like that, and you wanted to figure
out the total area of both of these circles
combined, you would look at the area of this circle. And then you could add it
to the area of this circle. But when you do that, you'll
see that when you add the two areas, you're counting
this area twice. So in order to only
count that area once, you have to subtract
that area from the sum. So if this area has
A, this area is B, and the intersection
where they overlap is C, the combined area is
going to be A plus B-- -- minus where they
overlap-- minus C. So that's the same
thing over here, we're counting all
the Jacks, and that includes the Jack of hearts. We're counting all
the hearts, and that includes the Jack of hearts. So we counted the
Jack of hearts twice, so we have to subtract
1 out of that. This is going to be
4 plus 13 minus 1, or this is going to be 16/52. And both of these things
are divisible by 4. So this is going to be the
same thing as, divide 16 by 4, you get 4. 52 divided by 4 is 13. So there's a 4/13 chance that
you'd get a Jack or a hearts.