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Macro
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Well, on the first try, you have a $90/100$ probability of not drawing a red ball; if the first was not a red ball, then on the second try there are still 10 red balls left, but only 99 to choose from, so you have a $89/99$ chance of not drawing a red ball. Similarly, on the third draw, if the second draw was also not a red ball, then you have a $88/98$ chance of picking a red ball, and so on. In general, if you attempt $k$ times independently without replacement, the probability you seek is

$$ \prod_{i=1}^{k} \frac{ 90-i+1 }{100-i+1} $$

One important thing to note is that this probability actually isn'tdoesn't arise from a binomial distribution. You are not conducting independent trials with equal probability and counting the number of "successes". The trials are not independent because the success probability of a future trial depends on whether a past trial was a success, making it fundamentally different from the binomial distribution. If there was replacement, then you'd be correct in saying the number of success follows a binomial distribution.

Well, on the first try, you have a $90/100$ probability of not drawing a red ball; on the second try there are still 10 red balls left, but only 99 to choose from, so you have a $89/99$ chance of not drawing a red ball. Similarly, on the third draw, you have a $88/98$ chance, and so on. In general, if you attempt $k$ times independently without replacement, the probability is

$$ \prod_{i=1}^{k} \frac{ 90-i+1 }{100-i+1} $$

One important thing to note is that this actually isn't a binomial distribution. You are not conducting independent trials with equal probability and counting the number of "successes". The trials are not independent because the success probability of a future trial depends on whether a past trial was a success, making it fundamentally different from the binomial distribution.

Well, on the first try, you have a $90/100$ probability of not drawing a red ball; if the first was not a red ball, then on the second try there are still 10 red balls left, but only 99 to choose from, so you have a $89/99$ chance of not drawing a red ball. Similarly, on the third draw, if the second draw was also not a red ball, then you have a $88/98$ chance of picking a red ball, and so on. In general, if you attempt $k$ times independently without replacement, the probability you seek is

$$ \prod_{i=1}^{k} \frac{ 90-i+1 }{100-i+1} $$

One important thing to note is that this probability actually doesn't arise from a binomial distribution. You are not conducting independent trials with equal probability and counting the number of "successes". The trials are not independent because the success probability of a future trial depends on whether a past trial was a success, making it fundamentally different from the binomial distribution. If there was replacement, then you'd be correct in saying the number of success follows a binomial distribution.

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Macro
Macro
  • 45.8k
  • 12
  • 158
  • 158

Well, on the first try, you have a $90/100$ probability of not drawing a red ball; on the second try there are still 10 red balls left, but only 99 to choose from, so you have a $89/99$ chance of not drawing a red ball. Similarly, on the third draw, you have a $88/98$ chance, and so on. In general, if you attempt $k$ times independently without replacement, the probability is

$$ \prod_{i=1}^{k} \frac{ 90-i+1 }{100-i+1} $$

One important thing to note is that this actually isn't a binomial distribution. You are not conducting independent trials with equal probability and counting the number of "successes". The trials are not independent because the success probability of a future trial depends on whether a past trial was a success, making it fundamentally different from the binomial distribution.