in this video we'll be reviewing probability the binomial distribution and the sampling distribution of the sample proportion we've already discussed these topics in more detail in my previous videos so if you need more review please watch those videos before you watch this one this video is just a way to put all of these Concepts together using practice questions so you can understand them better because sometimes it can be confusing so let's start with the video suppose we have a jar that contains 200 green Marbles and 300 blue marbles if our marble is drawn three times with replacement what is the probability of drawing at least two green marbles to solve this question let's first write down what we know the probability of success is equal to the number of successful outcomes divided by the total number of outcomes since a success in this case is drawing a green marble the probability will be equal to 200 / 500 because there are 200 green Marbles and there are a total of 500 marbles Al together 200 / 500 gives us an answer of 0.4 and the probability of failure or drawing a Blue Marble is equal to 300 / 500 which is equal to 0.6 there are many different ways of drawing three marbles for example one outcome could be drawing two green marbles in a row followed by one one Blue Marble this can be written as ggb for short another outcome could be drawing a Blue Marble then a green marble and then another Blue Marble this can be written as bgb for short the entire list of possible outcomes is called the sample space to calculate the probability for each outcome all we have to do is multiply the probability of each independent event together for example the probability of drawing a green marble followed by another green marble followed by a Blue Marble is equal to 0.4 * 0.4 * 0.6 which is equal to 0.096 and the probability of drawing three blue marbles in a row is equal to 0.6 * 0.6 * 0.6 which is equal to 0.216 and so on since the question asks for the probability of drawing at least two green marbles this means we need to take account for the probability of drawing more than two green marbles as well therefore the probability of drawing at least two green marbles is equal to the probability of getting exactly two green marbles plus the probability of getting exactly three green marbles as you can see from the sample space there are three different ways we can draw exactly two green marbles to find the probability of getting exactly two green marbles we add these probabilities together so 0.096 plus 0.096 plus 0.096 gives us an answer of 0.288 and the probability of drawing exactly three green marbles is equal to 0.064 and finally the probability of drawing at least two green marbles is equal to the probability of drawing exactly two green marbles plus the probability of drawing exactly three green marbles which is equal to 0.288 Plus 0.064 which gives us a final answer of 0.352 let's change the question a little bit what happens if five marbles are drawn instead of three so if a marble is drawn five times with replacement what is the probability of drawing at least two green marbles to answer this question we can extend the sample space and calculate the probabilities for each outcome however this may take time and may not be feasible so instead we can use the binomial formula to solve this problem the binomial formula calculates for the probability of the exact number of successes this is represented by the letter k n is equal to the number of Trials and Little P is equal to the probability of success so to answer the question the probability of drawing at least two green marbles is equal to the probability of drawing exactly two green marbles plus the probability of drawing exactly three green marbles plus the probability of drawing exactly four green marbles plus the probability of drawing exactly five green marbles the binomial formula will be used to calculate each number of success we will use it a total of four times one for calculating exactly two successes one for calculating exactly three successes one for calculating exactly four successes and one for calculating exactly five successes let's do the calculation for exactly two successes we know that K is equal to 2 because we are calculating for exactly two successes we also know that P is equal to 0.4 because that's the probability of success which is just the probability of drawing a green marble and we know that n is equal to 5 because we are drawing a total of five marbles if we plug all of these numbers into the formula we will get an answer of 03456 we will do the same process for calculating the rest of the probabilities after we make thisal calculation we will add these values together to get the answer which is equal to 0. 6634 let's change the question again if a marble is drawn 100 times with replacement what is the approximate probability of drawing at least 35 green marbles there are three ways we can solve this question we can calculate for the answer by extending the sample space like we did in the first example but we know that won't be feasible since it requires a lot of work we can also calculate for this by using the binomial formula but we know that won't be feasible either because we have to calculate the probability of success 65 different times since we'd have to calculate a success for drawing 35 green marbles 36 green marbles 37 green Marbles and so on until we get to 100 successes the other way we can solve this question is by using the sampling distribution of the sample proportion this is what we talked about in the pre previous video but first we need to check if the central limit theorem can be applied for a sampling distribution of the sample proportion it needs to follow two conditions for the central limit theorem to work the first condition is that n * P must be greater than or equal to 10 and the second condition is that n * 1 - P must also be greater than or equal to 10 in this example n is equal to 100 because a marble is being drawn a total of 100 * let's see if the first condition is met we will have n * P which is equal to 100 * 0.4 which is equal to 40 let's see if the second condition is met we will have n * 1 - P which is equal to 100 * 1 - 0.4 which is equal to 60 60 is greater than 10 so the second condition has also been met since these two conditions are satisfied the central limit theorem can be applied and if the central limit theorem can be applied the zcore table can be used by using the appropriate standardization formula for proportions the standardization formula is equal to the sample proportion P hat minus the population proportion P / Sigma P hat which is just equal to theare < TK of P * 1 - P / n in this case P hat is equal to 0.35 as that is the minimum proportion of green marbles that we are trying to draw out of 100 and P is equal to 0.4 because that is the probability of drawing a green marble and N is equal to 100 because that is the amount of trials that we have plugging all of these numbers into the formula gives us a zcore of - 1.02 a z score of 1.02 corresponds to an area of 01539 to the left of this value when we look at this visually on a bell curve this means that a value of 35 corresponds to a value of - 1.02 and the area to the left of this value is equal to 01539 since we use the standardization formula this means that our value to the left of 35 is also equal to 01539 however since we are looking for the approximate probability of drawing at least 35 green marbles we need to include the probability of drawing 44 green marbles 48 green Mar marbles 54 green marbles 60 green Marbles and so on so we need to find the area Associated to the right of this value since the bell curve is a density curve this means that the total area is equal to one or 100% so we will do 1 minus 01539 and that is equal to 0.846 1 which is equal to the area to the right of 35 as a result the approximate probability of drawing at least 35 green marbles is equal to 0.846 one or 84.61% it's important to know that we have calculated for an approximate probability not an exact probability to calculate for the exact probability you would have to use either the sample space method that we showed you in the first example or by using the binomial formula that we showed you in the second example it's important to remember that whenever you are applying the central limit theorem you are calculating for an approximate probability not an exact probab ability so you would be very close to the real answer and normally that's good enough and why they teach you this for introductory statistics if you found this video helpful consider supporting us on patreon to help us make more videos you can also visit our website at simple learning proo to get access to many study guides and practice questions thanks for watching