1 throw 1 success or 3 throw 2 success brain teaser

It isn't so much that we care about the order, but the probability of getting 2 successful scores in 6 attempts depends on how many ways you can get those 2 successful scores.

If you calculated the probability for one 'score' in six attempts and subtract it from 1, you would get the probability for two or more 'scores' in six attempts. Adding these up, I get a proability of making two or more 'scores' in six attempts to be 0.989, or approximately 99%. (If you think about the fact that this shooter makes seventy percent of his free throws, this probability makes sense.) The probability of five 'scores' and six 'scores' is 0.303 and 0.118, respectively. The probability of four 'scores' in six attempts is 0.324. I calculated the probability of three 'scores' in six attempts to be 0.185. We already have the probability of two 'scores' in six attempts at 0.059. So, for three 'scores' in six attempts, the 0.7 would be raised to the third power and the 0.3 would be raised to the third power. For example, when calculating the probability for three 'scores' in six attempts, the exponents in the factors for 0.7 and 0.3 should add up to six. Since the probabilities for 'score' and 'miss' are not equal, we need to calculate the probabilities the same way Sal did in this video. If we wanted to calculate the probability of two or more 'scores,' then we need to calculate each individual probability for the other events (X>=2,3,4,5,6) and add them together. If combinations are found by the total number of permutations divided by #of ways to arrange (Like 6 ways to arrange ABC), what would the permutations and # of ways to arrange Successes and Misses be? I get this formula from the first Combinations video. where each attempt is a new thing?Īm I even remotely on the right track here? I've watched the videos on permutations and combinations and understand them quite well. I see it as SSMMMM, SMSMMM, SMMSMM, SMMMSM, SMMMMS all being the same thing, which I would consider a permutation and not a combination. Is this because we only have two options, Success and Miss? Or is each attempt at a free throw a unique thing and should be counted independently? So ABC is the same as BAC (CAB, CBA, BCA, CBA) but not the same as ABD.Īre not all Successes the same and all Misses the same? This sounds to me like a variation of permutation. Why do you use combinations here? As I understand it, combinations do not consider order, but rather unique ways things are arranged.

1 throw 1 success or 3 throw 2 success brain teaser