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Sal graphs the results of using the binomial distribution to find the probabilities of making different numbers of free throws.

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• In the 'Binomial distribution' video, the probability was calculated by finding the total number of events and then using the combinatorics formula to find the chance of X occurring however many times and dividing that by the total number of possibilities to get the probability. On the other hand in the 'Probability of making 2 shots in 6 attempts' the value derived from using combinatorics is multiplied with the probabilities to get the final answer. Why is there a difference in the two approaches ? Sorry my question is a little confusing
• He just skipped the multiplication part since both outcomes are equaly likely. If you multiply by the probabilities raised to the correct exponant, like he does here, it comes to the same result.
• Is it true that even binomial distributions that don't have 50-50 probability of failure-success approaches normal distribution as the number of trials goes to infinity? And if so, do binomial distributions that don't have 50-50 approaches normal slower than the 50-50 ones?
• Yes, according to the Central Limit Theorem and the De Moivre-Laplace theorem, binomial distributions tend to approach a normal distribution as the number of trials n increases, even if the probability of success p is not 0.5. However, the speed at which they approach a normal distribution can indeed vary. Distributions with p closer to 0.5 tend to "normalize" quicker than those with p skewed heavily towards 0 or 1. This is because the variance is maximized at p = 0.5, making the distribution's shape more symmetric and thus more closely approximating a normal distribution even with a lower number of trials.
• at , you said "size". Did you mean to?
• At , Sal said "6 free throws". The caption is incorrect it says "size".
• when you did the calculation for P(X=i) why didnt you show that you mutipy by the "6 choose i" , when u just mutipy the way you did you dont get the correct percentages
• When calculating P(X = i) for a binomial distribution, the correct approach indeed involves multiplying by (6Ci), where i represents the number of successful outcomes (made free throws). The formula for the probability of exactly i successes in n trials is P(X = i) = (nCi)p^i(1 − p)^n−i, where (nCi) is the binomial coefficient representing the number of ways to choose i successes out of n attempts, p is the probability of success on a single trial, and (1 − p) is the probability of failure. If the multiplication by (6Ci) wasn't explicitly shown or mentioned in the calculation process, it's a crucial step that must be included to obtain the correct probabilities.
(1 vote)
• This a histogram ? I thought histograms requires ranges.