In this example we are figuring out the probability of randomly picking a non-blue marble from a bag. Again, we'll have to think about the possible outcomes first. Created by Sal Khan.
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Why do you always say "a fair coin"...and then begin the question?
What defines a fair coin? Does this refer to a coin that is not weighed down on one side to make it land a certain way, or does this refer to something else?(186 votes)
- A fair coin is flipped 4 times. What is the probability of getting more than 2 heads? I found the answer using the hints, which told me to write out all the possible outcomes. Needless to say, it is far from convenient as methods go. Isn't there a simpler way for finding the answer?(93 votes)
- This problem can be solved using binimial distribution.
n = 4, Probabilitiy of getting a head = 1/2 probability of not getting a head = 1/2.
The binomial probability formula is,
P(x) = nCx px qn-x
p(x>2) = P(X=3) + P( X = 4)
= 4C3 (1/2)^3 (1/2)^1 + 4C4 (1/2)4 (1/2)^0
= 4(1/2)^4 + 1 (1/2)4
= (1/2)^4 (4+1)
= 1/16 (5)
= 5/16(80 votes)
- Hi Sal - We considered the area of outer circle but only circumference of the inner circle, should we not consider areas of both the circles ? in that case answer would be : 64 pi / 324 pi and not 16pi / 324 pi(10 votes)
- Circumference of bigger circle and area of smaller circle are given in the question. First, we had to find the radius. Because we have to find the bigger circle's area. Then we put the radius to area formula to find the bigger circle's area.(4 votes)
- I have a small question
Probability should always be in the form of fractions right ? then why do some books or solutions have their answers in the decimal form?or is it the other way around?(8 votes)
- At around5:54he says "infinite number of points" what doe s that mean?(8 votes)
- By "infinite number of points" Sal means that you can always add points on the circle between points that you have on it, to infinity.
It's easier to see this on a one-dimensional line. If you have points (0,0) and (1,0), you can add (0.5, 0) which is half-way. Then you can add half the way between (0,0) and (0.5, 0), which is (0.25, 0) and so on. Because a line of any length has infinite points on it, a circle (in which we can draw line segments) does as well.
Because of that, you can't use the number of points to determine the probability that the point lies in the smaller circle. You have to use the area of the circles instead.
I hope this helps.
--Phi φ(0 votes)
- is permutation and combination a higher level of probability?(3 votes)
- Permutations and combinations are operations to count. They are used to find probabilities, but are not probabilities. Remember that a probability is just a ratio of all desirable outcomes over the sample space. We can use permutations and combinations to find the number of desirable outcomes and the size of the sample space.(11 votes)
- still a bit confused about that circle one... does anyone have any clarification?(5 votes)
- first you would have to calculate the area of both circles, one of which is already given (16 x pi(50)), for the larger circle it involves finding the area which is pi x 18 squared (1,017) then I assume you would just do 50/1,017 x 100 = approx. 5 (percentage of selecting a point within both circles is 5 %)(5 votes)
- is there another way to figure out how to find the space around the smaller circle
- 70% of adults in a European city say that they enjoy using public transportation. If a random sample of 8 adults is selected from the city, find the probability that exactly 3 adults respond that they enjoy using public transportation.(0 votes)
- probability that an adult do not like PT = 100-70 = 30% = 0.3
no. of adults who should not like PT among random 8 = 5. => 5/8
therefore, the probability is 0.3*5/8= 15/80 [ANSWER]
PLEASE CORRECT ME IF I AM WRONG
if i solve the question other way, I am getting answer 21/80.
PLEASE LET ME KNOW THE ANSWER, SO THAT I CAN THINK UPON WHICH OF THE METHOD IS CORRECT AND WHY?
though i presume that the first method and answer is correct, because of some weird logic that will be difficult to explain, though this weird logic may be the fundamental one!!!
THANKS! :)(9 votes)
- Would you do this problem? A student plans to simultaneously toss a fair number cube, with faces numbered 1 through 6, and a fair coin. What is the probability that the cube will land with the face numbered 4 up and the coin will land heads up?(2 votes)
- I will not tell you the answer, but I will tell you the solution.
Since each of rolling a 4 or flipping a heads are both independent, that means the probability of having both is simply multiplied. The chances of rolling a 4 on a fair 6-sided cube is 1/6, and the chances of flipping a fair coin and landing heads is 1/2. Just multiply the chances.(5 votes)
Let's do a couple of exercises from our probability one module. So we have a bag with 9 red marbles, 2 blue marbles, and 3 green marbles in it. What is the probability of randomly selecting a non-blue marble from of the bag? So let's draw this bag here. So that's my bag, and we're going to assume that it's a transparent bag, so it looks like a vase. But we have 9 red marbles, so let me draw 9 red marbles. 1, 2, 3, 4, 5, 6, 7, 8, 9 red marbles. They're kind of orange-ish, but it does the job. 2 blue marbles, so we have 1 blue marble, 2 blue marbles. And then we have 3 green marbles, let me draw those 3, so 1, 2, 3. What is the probability of randomly selecting a non-blue marble from the bag? So maybe we mix them all up, and we have an equal probability of selecting any one of these. And the way you just think about it is what fraction of all of the possible events meet our constraint? So let's just think about all of the possible events first. How many different possible marbles can we take out? Well that's just the total number of marbles there are. So are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 possible marbles. So this is the number of possibilities. And then we just have to think what fraction of those possibilities meet our constraints. And the other way you could have gotten 14 is just taking 9 plus 2 plus 3. So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble from the bag. Another way to think about it is a red or green marble, because the only other two colors we have are red and green. So how many non-blue marbles are there? Well, there's a couple ways to think about it. You could say there's 14 total marbles. 2 are blue. So there are going to be 14 minus 2, which is 12 non-blue marbles. Or you could just count them. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 non-blue marbles. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to-- this isn't in simplified form right here, since both 12 and 14 are divisible by 2. So let's divide both the numerator and the denominator by 2, and you get 6 over 7. So we have a 6/7 chance of selecting a non-blue marble from the bag. Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of 5? So once again, we want to find the fraction of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple of 5, that's 6 times 5. 55 is 11 times 5. Not 56, not 28. This is clearly 5 times 10, this is 8 times 5, this is the same number again, also 8 times 5. So all of these are multiples of 5. 45, that's 9 times 5. 3 is not a multiple of 5. 25, clearly 5 times 5. So I've circled all the multiples of 5. So of all the possibilities, the ones that meet our constraint of being a multiple of 5, there are 1, 2, 3, 4, 5, 6, 7 possibilities. So 7 meet our constraint. So in this example, the probability of a selecting a number that is a multiple of 5 is 7/12. Let's do another one. the circumference of a circle is 36 pi. Let's draw this circle. The circumference of a circle is 36 pi, so let's say the circle looks-- I can draw a neater circle than that. So let's say the circle looks something like that. And its circumference-- we have to be careful here, they're giving us interesting-- the circumference is 36 pi. Then they tell us that contained in that circle is a smaller circle with area 16 pi. So inside the bigger circle, we have a smaller circle that has an area of 16 pi. A point is selected at random from inside the larger circle, so we're going to randomly select some point in this larger circle. What is the probability that the point also lies in this smaller circle? So here's a little bit interesting, because you actually have an infinite number of points in both of these circles, because it's not kind of a separate balls or marbles, like we saw in the first example, or separate numbers. There's actually an infinite number of points you could pick here. And so, when we talk about the probability that the point also lies in the smaller circle, we're really thinking about the percentage of the points in the larger circle that are also in the smaller circle. Or another way to think about it is the probability that if we pick a point from this larger circle, the probably that it's also in the smaller circle is really just going to be the percentage of the larger circle that is the smaller circle. I know that might sound confusing, but we're really just have to figure out the areas for both of them, and it's really just going to be the ratios so let's think about that. So there's a temptation to just use this 36 pi up here, but we have to remember, this was the circumference, and we need to figure out the area of both of these circles. And so for area, we need to know the radius, because area is pi r squared. So we can figure out the radius from the circumference by saying, well, circumference is equal to 2 times pi times the radius of the circle. Or if you say 36 pi, which we were told is the circumference, is equal to 2 times pi times the radius, we can divide both sides by 2 pi, and on the left hand side, 36 divided by 2 is 18 the pi's cancel out, we get our radius as being equal to 18 for this larger circle. So if we want to know its area, its area is going to be pi r squared, which is equal to pi times 18 squared. And let's figure out what 18 squared is. 18 times 18, 8 times 8 is 64, eight times 1 is 8 plus 6 is 14, and then we put that 0 there because we're now in the tens place, 1 times 8 is 8, 1 times 1 is 1. And really, this is a 10 times the 10, and that's why it gives us 100. Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12, and then 1 plus 1 plus 1 is a 3, so it's 324. So the area here is equal to pi times 324, or we could say 324 pi. So the area of the entire larger circle, the part that I shaded in yellow, including what's kind of under this orange circle, if you want to view it that way, this area right over here is equal to 324 pi. So the probability that a point that we select from this larger circle is also in the smaller circle is really just a percentage of the larger circle that is the smaller circle. So our probability-- I'll just write it like this-- the probability that the point also lies in the smaller circle-- so all of that stuff I'll put in it. The probability of that is going to be equal to the percentage of this larger circle that is this smaller one, and that's going to be-- or we could say the fraction of the larger circle's area that is the smaller circle's area. So it's going to be 16 pi over 324 pi. And the pi's cancel out, and it looks like both of them are divisible by 4. If we divide the numerator by 4, we get 4, if we divide the denominator by 4, what do we get? 4 goes into 320 80 times, it goes into 4 once, so we get an 81. So a probability-- I didn't even draw this to scale, this area is actually much smaller when you do it to scale-- the probability that if you were to randomly select a point from the larger circle, that it also lies in the smaller one is the ratio of their areas, the ratio of the smaller circle to the larger one. And that is 4/81, I guess is the best way to say it.