Probability of drawing a king on the 13th draw from a cardde

Probability theory and statistics

Probability of drawing a king on the 13th draw from a cardde

Postby flip101 » Tue Oct 09, 2018 8:54 am


Hello, i'm trying to solve question 2 of problem 2 by myself. The question + answer + explanation can be found in the link below.

https://github.com/theGreenJedi/The-Sci ... 0_%206.pdf

I will use the ncr() function of SpeedCrunch to denote combinations [tex]{a \choose b}[/tex].

My reasoning was as follows: when 4 kings can not be chosen there are ncr(48;12) ways to pick the first 12 cards. When any card can be chosen there are ncr(52;12) ways to pick the first 12 cards. So the probability that the first 12 cards are not kings is ncr(48;12) / ncr(52;12)

Then there are (52 - 12) cards left of which 4 are kings, so the probability that the next card is king is 4 / 40. So my answer is: ( ncr(48;12) / ncr(52;12) ) * 4/40 = 0,0337575 But this is numerically not the correct answer. Where did i go wrong in my reasoning?
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Re: Probability of drawing a king on the 13th draw from a ca

Postby flip101 » Wed Oct 10, 2018 8:05 am

This problem is solved. The given expression equates to the same numerical answer.

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