Statistics and probability
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.
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- 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?(12 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%(34 votes)
- Why do you add the prob of the unfair coin to the prob of the fair coin? why not multiply their probabilities?(6 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.(28 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)
- 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. )(8 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.(4 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.(2 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(4 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.(1 vote)
- 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?(2 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."(3 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?(3 votes)
- How would you determine that a coin in the real world is an "unfair" coin?(2 votes)
- You cannot tell absolutely for certain, so the next best thing is to toss the coin some large number of times to find a sample proportion of tosses that are heads. Then perform a hypothesis test with null hypothesis p = 0.5, alternative hypothesis p unequal to 0.5, and a significance level of your choice (5% is commonly chosen for this level), where p is the theoretical probability (population proportion) of heads.(2 votes)
- Can't this be done like that,
For 4 coins avg probability of getting heads is (.55+.55+.55+.5)/4 which is = 0.5375 now for each flip this is the avg probability of each coin of getting heads thus multiply it 4 times itself for each flip event which is = 0.08346682129?
There is a difference between my answer and Sal's of about minor 0.93% only but i feel this method is right one, isn't it?(2 votes)
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.