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### Course: Statistics and probability > Unit 7

Lesson 8: Multiplication rule for dependent events# Dependent probability example

It's important to practice these probability problems as they get more complex eventually. Take a stab on this one...with our help, of course. Created by Sal Khan.

## Want to join the conversation?

- 0:21What if you pick a new coin for every flip instead of keeping it? It seems to me it would be mindbogglingly difficult to calculate it, is this true?(13 votes)
- No, it is not difficult.

If you were to pick a new coin for every flip, then you would be evaluating "what are the chances of getting a coin out of the bag, flipping it, getting a Heads, and doing it all over again four times in a row?". You would first need to determine what the chances are for getting a coin and getting a Heads, and then multiplying this value by itself 4 times.

The chances for getting a coin and getting a Heads, it would be the addition of the chances of getting a Fair coin and getting a Heads, plus the chances of getting an Unfair coin and getting a Heads. So, (1/4)*0.5 + (3/4)*0.55 = 53.75%. This is the probability of getting a coin, any coin, and getting a Heads.

To determine the chances of getting four of these situations in a row, simply multiply 0.5375 times itself four times. So, (0.5375)^4 = 8.35%(36 votes)

- Why do you add the prob of the unfair coin to the prob of the fair coin? why not multiply their probabilities?(8 votes)
- When the probability is about A AND B, then you multiply. For example, to find the probability of getting fair coin AND 4 heads you need to multiply.

When the probability is A OR B, you add. To find the probability of getting fair coin OR unfair coin, you added their probabilities.(30 votes)

- sorry, i have a question about the exercise,the question is if haven't already been hit,

captain emily have 1/2 chance of hitting pirate ashley,if captain emily is hit,then she

always miss,if haven't already been hit,pirate ashley have 1/7chance of hitting captain

emily,is she is hit,she always miss,the question is :if emily fires first,what is the probability of emily hits,ashley missed.my understanding is emily has 1/2chance times (1-1/7)=6/7,i know it's wrong,i checked the hints.but the hints says the pirate is already hit,so 1/2*(1-emily missed)=1/2*(1-0)=1/2,but it didn't state in the question that pirate has

already been hit,can someone help me understand this?thank you in advance.(5 votes)- I believe that the question mentions that Emily shoot the pirate FIRST which has 1/2 chance of a accurate shot and if it does hit, the pirate would always miss so it's 1/2 indeed.(3 votes)

- at4:57, why are the probabilities added and not multiplied? when should probabilities be added and when multiplied??(1 vote)
- Probabilities are generally multiplied when something is done over and over in a row ( ex. flipping a coin 4 times and getting heads 4 times in a row is .5^4 or .5 * .5 * .5 *.5 ).

Probabilities are generally added together when you are trying to find a total probability between two independent probabilities on a probability tree ( ex. the whole video shows a probability chart. )(9 votes)

- Why is the following calculation incorrect?

And if so, what is the difference in the situation with my calculation and the one on the video?

If I calculate the odds of gettin one heads in one throw -> (3/4*55% + 1/4*50%)) = 53,75%

Now the odds of getting 4 in a row would be 53,75^4=8,346...

And the result in the video was 8,425...

EDIT: I guess this was answered already by Jean Rambo... But the interesting point is that there is only 0,079% advantage in picking a coin and using it 4 times than picking a coin 4 times and then flipping it...(2 votes)- I think the difference can be better understood by setting both scenarios in this way:

When the probability gives you 0.08425 is under the circumstance where you are asked to pick any coin from the bag and then flip it 4 times, hoping to get heads (as the problem states).

When the probability gives you 0.08347 is like you were asked to pick any coin, flip it just one time hoping to get heads, put it back in the bag, and repeat this whole process 4 times.

Now we can conclude that the slight difference in these probabilities is due to the persistent possibility (in the 2nd case) of getting a fair coin in each of the 4 flips (which offers a lower probability of getting heads than an unfair one), while in the first case that possibility just occurs once.(7 votes)

- so if the probabilities are A and B you have to multiply, and if they are A or B you have to add? just trying to make sure I understand(5 votes)
- I believe that is correct. According to sal, (around4:30)you do add a and b if it is talking about 'either way'. And you also multiply when the question is asking a and b.(2 votes)

- Still unclear about this: This is called dependent probability, but we're multiplying prob's together as if they were independent.

How does this fit with what he was saying earlier about you can multiply prob's together if they are independent?(3 votes)- If they are independent, then P(A and B) = P(A)^P(B) which is just the probability of A times the probability of B.

If they are dependent, then P(A and B) = P(A)*P(B|A) which is the probability of A times the probability of "B happening if A has occurred," which is different than the "Probability of B if A has not occurred."(4 votes)

- 3:20That calculator looks so real. It's scary.(4 votes)
- what i don't understand is why he doesn't lay out the procedure for finding P(A&B) from P(A|B) while incorporating P(B). this is the layout of one of my quiz questions. can anyone explain this?(4 votes)
- 2.02 Why is it 0.55 TIMES o.55 instead of 4 times 0.55? I don't see the reason to the repeated multiplication.(2 votes)
- 4·0.55 = 2.2, that would mean that there's a 220% probability of this event occurring, and that doesn't make much sense since the probability can't exceed a 100%.

Imagine if you were flipping not one, but some huge number of coins, let's say, a million of them. You would expect, that given 55% chance that each coin will land heads, after flipping all of them, you would end up with roughly 550 thousands heads, or 1,000,000·0.55. If you then take the coins that landed heads, and flip them all a second time, out of 550 thousands of them, roughly 302,500 or 550,000·0.55 will land heads once again. So the expected amount of coins out of million that will land heads on both flips is 1,000,000·0.55·0.55 or 1,000,000·0.55². The expected amount of coins that will land heads on each of three flips would be about 1,000,000·0.55³ and so on.

With that in mind, if you pick one of the million initial coins and ask what is the probability that this would be one of the coins that lands heads three times in a row, then that would be: (1,000,000·0.55³)/1,000,000 = 0.55³(4 votes)

## Video transcript

Let's do another one of these
dependent probability problems. You have 4 coins in a bag. 3 of them are
unfair in that they have a 45% chance of coming
up tails when flipped. The rest are fair. So for the rest of them, you
have a 50% chance of tails or a 50% chance of heads. You randomly choose
one coin from the bag and flip it 4 times. What is the percent
probability of getting 4 heads? So let's think about it. When we put our hand in the bag
and we take one of the coins out, there's some probability
that we get an unfair coin. And 3 of the 4 coins are unfair. So there's a 3/4 probability
that we get an unfair coin. And then there is only 1 out
of the 4 coins that's fair. So there was a 1/4 probability
that I get a fair coin. So unfair, let's
remind ourselves-- an unfair coin has a 45%
chance of coming up tails. So this means that I have a
45% chance of tails, which also means-- and we have
to be careful here because they're asking us
about heads-- if I have a 45% chance of
getting tails, that means I have a 55%
chance of getting heads. I have a 100% chance of
getting one of these two. If it's 45% for tails, 100
minus 45 is 55 for heads. For the fair coin, I have
a 50% chance of tails and a 50% chance of heads. 50% heads. Fair enough. Now I want to know, in
either of these scenarios, what is the percent probability
of getting four heads? So given I've got
the unfair coin, the probability of
getting four heads is going to be 55% for
each of those flips. So the probability of
getting exactly four heads is going to be 0.55 times
0.55 times 0.55 times 0.55. And so the probability
of picking an unfair coin and getting four
heads in a row is going to be equal to 3/4 times
all of this business over here. So that's 3/4 times--
and this is 0.55 times itself four times, so I
could write it as 0.55 to the fourth power. And we'll get the
calculator out in a second to actually calculate
what this is. Now let's do the same
thing for the fair coin. If I did pick a fair
coin, the probability of getting heads four times in
a row is going to be 0.5 times 0.5 times 0.5 times 0.5. Or the probability of
getting the fair coin, which is 1/4 chance, times
the probability-- and getting four heads in a
row is going to be 1/4 times all of this. So it's going to be 1/4 times--
this is just 0.5 times itself four times, so that's
0.5 to the fourth power. So let's get the
calculator out to calculate either one of these. So we get 3 divided
by 4 times-- and it knows that when I do
the multiplication, it's not in the
denominator here. So it's 3/4 times-- and I'll
just do it in parentheses, which I don't have to
do it in parentheses, because it knows
order of operations-- so 0.55 to the fourth
power is equal to 0.-- let me write it down. Let me take it off the screen
so I can write it down properly. Actually, let me just do
both of these calculations. So this probability is
that one right over there. And then this one down here
is 1 divided by 4 times 0.5 to the fourth power. So it's equal to that
right over there. And so let's be clear. The probability of
picking the unfair coin and then getting four heads
in a row is this top number. It's like roughly 6.9 chance
that you get the unfair coin and then get four
heads in a row. The probability that
you get the fair coin and then get four heads
in the row is even lower. It's only a 1.6% chance. Now the probability of getting
four heads in a row either way is going to be the
sum of this and this, or the sum of that
and that, which is going to be-- let me
keep my calculator out-- so it's going to be
equal to-- I can just take the previous answer. Let me just retype it
so I don't confuse you. So 0.015625 plus 0.0686296875. I'm going to round it anyway,
so it won't matter too much. So if I take the sum-- let
me take this off screen so I can still see it
and then let me write it. So what I got here,
this one is 0.068629. And I'll round it, 7. This down here was 0.015625
and when you add these two up-- because we just care about
getting four heads either way. There's a probability
of getting it this way with the unfair coin. This is the probability of
getting it with a fair coin. We want it either way. So let's add the two, which we
already did on our calculator. So if you add that
number to that number. You get 0.08425
and it keeps going, but I'm just going to round it. This is equal to
8.425%, or if I want to round it a little
bit more, 8.43% chance of getting
four heads in a row. And once again, that's
a slightly higher number than if all of the
coins were fair. Because there's a
3/4 chance that I get a coin that has a better
than even chance of getting heads. So that's why this number is
going to be a little bit higher than the probability if I had
a fair coin, of just getting four heads in a row.