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Jun 13, 2017 at 15:50 vote accept Amelius
Jun 13, 2017 at 15:46 comment added Wintermute You just using the binomial distribution to compute the probability of 2 successes in 7 draws and then multiplying it by $\frac{3}{10}$ which is the probability of getting a face card on the 8th draw.
Jun 13, 2017 at 15:45 comment added lulu I wrote out some more details in my posted solution. Your idea is good: first look at the probability of getting exactly three faces in the first eight, then subtract those cases in which the eighth card is a non-face.
Jun 13, 2017 at 15:44 comment added Amelius @lulu why is it necessary to factor in the 8th draw if for that part I'm only looking at the chance of having all 3 in the first 7? Isn't the probability of drawing 3 face cards in the first 7 correctly listed above, with $7\choose3(\frac{3}{13})^3(\frac{10}{13})^4$ ?
Jun 13, 2017 at 15:44 answer added lulu timeline score: 1
Jun 13, 2017 at 15:40 comment added JMoravitz It is worth mentioning that $\binom{8}{3}(3/13)^3(10/13)^5-\binom{7}{3}(3/13)^3(10/13)^\color{red}{5}$ is the same answer as the book gives written in a different form. As such, we can try to use this to our advantage in trying to explain what went wrong. Having three successes in seven might still have been four successes in eight or three successes in eight, but we only wanted to consider those that were three successes in eight when we were subtracting.
Jun 13, 2017 at 15:40 comment added lulu For the term you are subtracting, you are missing a factor of $\frac {10}{13}$. That's because you need to require that the eighth choice be a non-face card.
Jun 13, 2017 at 15:36 review First posts
Jun 13, 2017 at 15:52
Jun 13, 2017 at 15:32 history asked Amelius CC BY-SA 3.0