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In a group of 5 students, 2 are males and 3 are females. Two students are randomly selected (without replacement). Let X be the number of males in the two selected students. (a) Find the (probability) distribution of X (i.e., list all possible values of X and their corresponding probabilities). (b) Find the expected value of X, and the standard deviation of X.

let n = 2 = number of trials P = 2/5 = probability of success Q = 3/5 = probability of failure k = number of success

For a) I used the equation nCkP^kQ^(n-k), and got P(X=0)= 0.36 P(X=1)= 0.48 P(X=2)= 0.16

but the solution key tells me that P(X=0)= 0.3 P(X=1)= 0.6 P(X=2)= 0.1

And for b) E(X)= np = 0.8 SD = sqrt(npq) = 0.48

but the solution key tells me that E(X) = 0.8 SD = 0.6

And now i'm confused... I'm not sure if I should approach this question using the binomial probability or is it completely unrelated to binomial distribution. Please give me a hint on how to approach this question, thank you.

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    $\begingroup$ This is sampling WITHOUT replacement, not with replacement, so the Binomial distribution does not apply. Try thinking it through systematically, rather than blindly applying formulas. $\endgroup$ – Mark L. Stone Aug 14 '16 at 0:32
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    $\begingroup$ Hypergeometric distribution $\endgroup$ – Antoni Parellada Aug 14 '16 at 0:41
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    $\begingroup$ Please add the [self-study] tag (you will have to remove one of the existing tags) & read its wiki. $\endgroup$ – gung - Reinstate Monica Aug 14 '16 at 3:03
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Hi thank you for the question. It is useful to think of the possible set of values and determine it's probability distribution, as oppose to jumping into binomial formulas. As Mark has has mentioned inside the comments, binomial is only not suitable for when there is no replacement. For example, flipping a fair coin n times is equivalent to sampling a red or a blue ball from a bag with replacement (not without).

There are four possible situations describing which set of two students you have selected.Note:Sequence matters in the "without replacement" case.

  • $P(MM)= \frac{2}{5} \times \frac{1}{4} =0.1$
  • $P(FF)=\frac{3}{5}\times\frac{2}{4}=0.3$
  • $P(FM)=\frac{3}{5} \times \frac{2}{4}=0.3$
  • $P(MF)=\frac{2}{5} \times \frac{3}{4}=0.3$

In other words, the probability of sampling $X$ men with 2 draws from the original sample are as below

  • $P(X=2)=P(MM)=0.1$
  • $P(X=0)=P(FF)=0.3$
  • $P(X=1)=P(FM)+P(MF)=0.6$

All the probabilities ofthe outcomes of $X$ sum to 1, so we know it gives a valid probability distribution. This answer question a)

As for question b). Now, that you know the probability distribution of $X$, you can quite easily derive the expected value and the standard deviation with these formulas.

  • $E(X)= \sum{xP(X=x)}$
  • $SD= \sqrt{\sum{x^2P(X=x)}-E(X)^2}$

Attempt the rest by yourself. Good Luck :)

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