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Question Number 165673 by mkam last updated on 06/Feb/22

	Answered by mr W last updated on 06/Feb/22

	$t o s e l e c t 3 f r o m 10 s t u d e n t s t h e r e$ $a r e t o t a l l y C_{3}^{10} = 120 p o s s i b i l t i e s .$ $0 g i r l : C_{3}^{6} = 20 p o s s i b i l i t i e s, p (0) = \frac{20}{120}$ $1 g i r l : C_{1}^{4} C_{2}^{6} = 60, p (1) = \frac{60}{120}$ $2 g i r l s : C_{2}^{4} C_{1}^{6} = 36, p (2) = \frac{36}{120}$ $3 g i r l s : C_{3}^{4} = 4, p (3) = \frac{4}{120}$ $E [X] = 0 \times \frac{20}{120} + 1 \times \frac{60}{120} + 2 \times \frac{36}{120} + 3 \times \frac{4}{120}$ $= 1.2$ $V a r (X) = {(0 - 1.2)}^{2} \times \frac{20}{120} + {(1 - 1.2)}^{2} \times \frac{60}{120} + {(2 - 1.2)}^{2} \times \frac{36}{120} + {(3 - 1.2)}^{2} \times \frac{4}{120}$ $= 0.56$ $o r$ $E [X^{2}] = 0^{2} \times \frac{20}{120} + 1^{2} \times \frac{60}{120} + 2^{2} \times \frac{36}{120} + 3^{2} \times \frac{4}{120}$ $= 2$ $V a r (X) = E (X^{2}) - {(E (X))}^{2} = 2 - 1 . 2^{2} = 0.56$
	Commented by Tawa11 last updated on 06/Feb/22

	$Great sir$
	Commented by peter frank last updated on 09/Feb/22

	$nice$
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