A class has 70% females and 30% males. Students are selected at random, what is the probability that 7 students have to be selected before 2 males can be obtained?

Is this binomial distribution? Or not?

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    $\begingroup$ Either we know the size of the class, or we have to assume students can be picked twice. Which is it? Also, we need to know the precise meaning of "before 2 males can be obtained". $\endgroup$ – Jack M Apr 24 '13 at 22:21
  • $\begingroup$ The question says to assume a really big class. Sorry if my wording was a bit too opaque. The actual word for word question is 'We select students at random, one at a time. What is the probability that we need to select 7 students in order to obtain 2 males?' $\endgroup$ – tuba09 Apr 24 '13 at 22:24

Assuming the students are selected with replacement, i.e., every draw is independent, then the total sample size required to obtain a fixed number of failures (k = 2 males - no pun intended) is the sum of k and X, where X follows a negative binomial distribution.

This distribution has two parameter, the number of failures to observe (2) and the succes rate in each trial (70%).

See http://en.wikipedia.org/wiki/Negative_binomial_distribution

  • $\begingroup$ Thanks! Makes sense I suppose. Would you mind helping me with another similar one? It states that 'We select 5 students at random. What is the probability that we will have exactly 2 females in the sample? Assume the class size is 40 students.' I know that the distribution that should be used for this question is hypergeometric, but I don't know why. How come I can't use a regular binomial distribution here? $\endgroup$ – tuba09 Apr 24 '13 at 22:35
  • $\begingroup$ @tuba09, you could use the binomial if the sample was to be taken with replacement, to make each sample independent from the others. Sampling without replacement leads to a hypergeometric distribution for the number of successes. $\endgroup$ – Ferdinand.kraft Apr 24 '13 at 22:39
  • $\begingroup$ Given that you know the size of the class, you should consider extending the answer I gave for the Negative Hypergeometric Distribution. It is the analogue of the negative binomial for a finite population (i.e. sampling without replacement). $\endgroup$ – Ferdinand.kraft Apr 24 '13 at 22:45
  • $\begingroup$ Alright, I think I get it now. Thank you for the prompt response. $\endgroup$ – tuba09 Apr 24 '13 at 22:50

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