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Add uniform conditional

In this video, we will determine a conditional probability of a uniform distribution. For the background on the distribution, it's based upon the 55 smiling times in seconds of an eight-week-old baby shown here. We assume the smiling times, in seconds, follow a uniform distribution between 0 and 23 seconds inclusive. This means that any smiling time from 0 to and including 23 seconds is equally likely. We let the random variable x equal the length in seconds of an eight-week-old baby's smile, and we use a notation here for the uniform distribution where the least x-value is 0 and the maximum x-value is 23. And therefore the probability density function is equal to f of x equals 1/23 where x takes on all the values from 0 to 23, including both end points. And now going to our problem, we will find the probability two ways. In the first method, we will find a new probability density function. We're asked to find the probability that a random eight-week-old baby smiles more than 12 seconds knowing that the baby smiles more than eight seconds. So we know for the given probability density function f of x equals 1/23, the area under the curve from 0 to 23 is equal to 1 or 100%. We know this area is equal to the base times the height where the base is 23 and the height is 1/23. But in this case, we already know the baby smiles more than eight seconds. And therefore we're not considering the entire interval from 0 to 23, only from 8 to 23, again, because we know the baby smiles more than eight seconds. So we're only considering the interval from x equals 8, to x equals 23 represent 100%. So we know the area under this curve must be equal to 1, but now notice how the base is not 23 units. And therefore this is going to change the probability density function f of x. So to find this conditional probability, we will find the new probability density function using these x-values from 8 to 23. And therefore the new probability function is equal to 1 divided by the quantity 23 minus 8, which is equal to 1/15. So we know this is f of x equals 1/15, which gives us the height of this rectangle. And now let's read the probability again. We're asked to find the probability that a random eight-week-old baby smiles more than 12 seconds knowing that the baby smiles more than 8 seconds. So using the new probability density function, the smiling time of 12 seconds is here. So the probability that a random eight-week-old baby smiles more than 12 seconds knowing that baby smiles more than 8 seconds is equal to the area of the red rectangle to the right of 12, this area here. So let's go ahead and set this up. We have the probability that x is greater than 12 given x is greater than 8, and this is equal to the area of the red rectangle where the base is equal to 23 minus 12, and the height is equal to the new probably density function value of 1/15. Simplifying, 23 minus 12 is equal to 11. 11 times 1/15 is 11/15. Let's also get a decimal approximation and a percent. 11 divided by 15 is equal to 0.73 with the three repeating, or to four decimal places, approximately 0.7333, which is equal to 73.33%. Let's also find this probability a second way using the conditional probability formula, which is the probability of A given B is equal to the probability of A and B divided by the probability of B. So again, we're looking for the probability that x is greater than 12 given x is greater than 8, which is equal to the probability that x is greater than 12 and x is greater than 8 divided by the probability x is greater than 8. Let's focus on the numerator first. The probability that x is greater than 12 and x is greater than 8 is just equal to the probability that x is greater than 12, which gives us a probability x is greater than 12 divided by the probability that x is greater than 8. So let's first find the probability that x is greater than 12, which should be the area under the curve to the right of 12, which is the area of this red rectangle here. The length of the base is equal to 23 minus 12 times the height, which is the original probably density function value of 1/23 divided by the probability that x is greater than 8, which should be the area to the right of 8, between 8 and 23. Let's go ahead and shade that in green, where the base is equal to 23 minus 8 times, again, the height of 1/23. There's a couple of ways to simplify this, but notice that 1/23 divided by itself simplifies to 1. Notice now the numerator is 23 minus 12, which is 11, and 23 minus 8 is equal to 15, giving us the same probability of 11/15, which we already saw as approximately 0.7333 or 73.33%. I hope you found this helpful.