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To use this question as an example: In a class there are four freshman boys, six freshman girls, and six sophomore boys. How many sophomore girls must be present if sex and class are to be independent when a student is selected at random?

Let $B = \rm boy$, $G = \rm girl$, $F = \rm freshman$, $S = \rm sophomore$, $x = \rm sophomore girls$. There are four cases:

  1. $P(FB) = P(F)P(B)$
  2. $P(FG) = P(F)P(G)$
  3. $P(SB) = P(S)P(B)$
  4. $P(SG) = P(S)P(G)$

These cases assume independence. Now if we solve for case 3 lets say, then x = 9.

The calculations are clear to me, but I don't understand why solving case 3 means that the other cases also hold when x = 9, it's not intuitive to me.

I tried showing myself that all other cases are independent as well, because we can use substitution to get
$P(SB) = P(S)P(B)$
$P(SB) = (P(SG)/P(G))*P(B)$
Which shows that case 4 is independent. The thing I don't understand here is that, this is assuming that case 4 is indeed independent. Why is this possible?

Question: So what I'm trying to ask is, Why is it that if one case holds, the others hold too? Is there an intuitive way to think about such problems? Because I can't picture it in my mind, I can just do the calculation here.

TShiong
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MrPuffer
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1 Answers1

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For your specific example you have complementary events so that $G =$ girl and $G^{c}= $ boy as well as $F=$ freshman and $F^{c} =$ sophomore. Then your question just comes down to showing that if two events $A$ and $B$ are independent then so are $A$ and $B^c$, $A^c$ and $B$, and $A^c$ and $B^c$.

Such a fact is well known. I would advise you to show it yourself but if you need guidance then consult this question or/and this question

Manifoldski
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