<PRE><B>Question 1178477</B>
p(0 good) is (3/10)(2/9)=1/15
this is also 3C2*7C0/10C2=3/45=1/15
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p(1) is 3C1/7C1/10C2=21/45 or 7/15
or (3/10)*(7/9)+(7/10)*(3/9)=42/90=7/15
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p(2)=(7/10)(6/9)=42/90 or21/45=7/15
3C0*7C2/10C2=21/45=7/15
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the probabilities add up to 1.
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Expected value is 0*(1/15)+1(7/15)+2(7/15)=21/15, or 1.4 batteries
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variance is sum of squared deviations or (-1.4)^2*0+(-0.4)^2*(7/15)+(0.6)^2*(7/15)
=0+0.074666+0.168
sd is sqrt (0.24266)=0.493</PRE>