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The problem is as follows:

Draw cards at random with replacement from a standard deck of cards. Given that face cards are Jack, Queen and King, what is the probability that the 3rd face card is drawn on the 8th draw?

The correct answer given is ${7 \choose 2}(\frac{3}{13} )^3(\frac{10}{13})^5 $ But I don't understand why taking the probability of getting 3 successes out of 8 which is ${8 \choose 3}(\frac{3}{13} )^3(\frac{10}{13})^5$ And subtracting the probability of having 3 successes in seven ${7 \choose 3}(\frac{3}{13} )^3(\frac{10}{13})^4$ Yielding ${8 \choose 3}(\frac{3}{13} )^3(\frac{10}{13})^5-{7 \choose 3}(\frac{3}{13} )^3(\frac{10}{13})^4$ is incorrect. It makes sense to me intuitively. Can someone help me understand this please?

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  • $\begingroup$ 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. $\endgroup$ – lulu Jun 13 '17 at 15:40
  • $\begingroup$ 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. $\endgroup$ – JMoravitz Jun 13 '17 at 15:40
  • $\begingroup$ @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 $7\choose3(\frac{3}{13})^3(\frac{10}{13})^4$ ? $\endgroup$ – Amelius Jun 13 '17 at 15:44
  • $\begingroup$ 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. $\endgroup$ – lulu Jun 13 '17 at 15:45
  • $\begingroup$ 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. $\endgroup$ – Wintermute Jun 13 '17 at 15:46
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The error in your computation is in the term you are subtracting.

The first term in your difference correctly computes the probability that you will get exactly three face cards in the first eight draws. Now, you want to subtract those cases in which the eighth draw is a non-face card. Thus you need exactly three face cards out of the first seven, and you need the eight draw to be a non-face card. Thus, your expression needs an extra factor of $\frac {10}{13}$.

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  • $\begingroup$ So it's basically like I'm multiplying the chance of having 3 in the first 7 by the probability of not having another face card on the 8th draw? So we're stepping outside normal Bernoulli trials and adding an extra term? $\endgroup$ – Amelius Jun 13 '17 at 15:48
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    $\begingroup$ I am not sure what you mean by "stepping outside" here. Like I said, your method is good: list the eight card draws with exactly three faces and then subtract those in which the eight card isn't a face. Excellent method. Works like a charm. $\endgroup$ – lulu Jun 13 '17 at 15:50

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