Solutions - 5190-07

(a)

Count the total number of groups of 3 from 7:

(\binom{7}{3}) = \frac{7!}{4! 3!} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2} ≫ ≫ 35

(b)

Choose 2 from 4 grad students and 1 from 2 undergrads:

(\binom{4}{2}) (\binom{2}{1}) = 6 \cdot 2 ≫ ≫ 12

(c)

Construct the joint PMF of $X$ and $Y$ :

$Y ↓ X \to$	0	1	2	3
0	0	0	$\frac{(\binom{4}{2})}{35} = \frac{6}{35}$	$\frac{(\binom{4}{3})}{35} = \frac{4}{35}$
1	0	$\frac{(\binom{4}{1}) (\binom{2}{1})}{35} = \frac{8}{35}$	$\frac{(\binom{4}{2}) (\binom{2}{1})}{35} = \frac{12}{35}$	0
2	$\frac{(\binom{2}{2})}{35} = \frac{1}{35}$	$\frac{(\binom{4}{1}) (\binom{2}{2})}{35} = \frac{4}{35}$	0	0

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