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10 students are in a class,

4 of these students are boys and 6 are girls

the teacher wanted to randomly select 3 students to represent the class in an event

what is the probability that the 3rd student is a boy?

Here is what I have tried

the probability of that 1st student is a boy = 4/10 the probability of that 1st student is a girl = 6/10

the probability of that 3rd student is a boy means one of these scenarios

1- previously 2 boys were selected

2- previously 2 girls were selected

3- previously a boy and a girl were selected

in case (1) the probability of having the 3rd is a boy would be 2/8

in case (2) the probability of having the 3rd is a boy would be 4/8

in case (3) the probability of having the 3rd is a boy would be 3/8

So which one of these is the right answer?!

asmgx
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    You can do it that way but there's no need to, each choice is like any other so the answer is $\frac 4{10}$. To do it the way you started, you need to weight each case by the probability of being in that scenario. – lulu Apr 28 '19 at 22:50
  • @lulu but when we select the 1st and 2nd student, number of students change and the distribution changed as well.. how come it is the same as the 1st pick? – asmgx Apr 28 '19 at 22:53
  • If you specify the first two choices then of course the answer changes, but if you don't specify them then there is no information from them, – lulu Apr 28 '19 at 22:56
  • Related: If you draw two cards, what is the probability the second card is a Queen – JMoravitz Apr 28 '19 at 22:59
  • @JMoravitz by applying the same logic in the answer of the draw of 2 cards question you mentioned we get the probability of the 3rd is a boy = 2/8 + 4/8 + 3/8 = 9/8 !!!! – asmgx Apr 28 '19 at 23:20
  • @asmgx absolutely not. Pay closer attention. – JMoravitz Apr 28 '19 at 23:44
  • If you are referring to the first answer in the linked question... yes, there is an addition symbol but that is as far as the analogy goes to your most recent comment. If you were to do the same thing as the first answer for this problem... you could see the answer as $\frac{4}{10}\times\frac{3}{9}\times\frac{2}{8}+\frac{4}{10}\times\frac{6}{9}\times\frac{3}{8}+\frac{6}{10}\times\frac{4}{9}\times\frac{3}{8}+\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}$... which is exactly what lulu alluded to when saying "You can do it that way...(but) you need to weight each case..." – JMoravitz Apr 28 '19 at 23:48
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    If you were to simplify that god-awful expression however... it very simply results in $\frac{4}{10}$... which as many of the answers and comments in the linked question as well as other answers and comments here already will tell you makes sense as being "the third" person to be selected is not different enough from being "the first" person selected, so naturally the probability the third is a boy is going to be the same as the probability the first is a boy which is obviously $\frac{4}{10}$ with no tedious calculations required. – JMoravitz Apr 28 '19 at 23:50
  • Yes, I got the right results now.

    (4/10 * 3/9 * 2/8) + (4/10 * 6/9 * 3/8) + (6/10 * 4/9 * 3/8) + (6/10 * 5/9 * 4/8)

    = 24/720 + 72/720 + 72/720 + 120/720

    = 288 / 720 = 0.4

     – asmgx Apr 29 '19 at 04:29
  • @JMoravitz plz put this as an answer so i mark it – asmgx Apr 29 '19 at 11:06

1 Answers1

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If the selection is done randomly, the probability of picking students $x, y, z$ (in that order) is always the same as the probability of picking $z, y, x$. So the probability that the third student is a boy is the same as the probability that the first student is a boy.

Robert Israel
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