Let the mean and variance of 7 observations 2,4,10,x,12,14,y (with x>y) be 8 and 16 respectively. Two numbers are chosen one after another without replacement from {1,2,3,x−4,y,5}, then the probability that the smaller number among the two chosen numbers is less than 4, is:
Solutions · 1
Official solution
\[\text{Mean}=8 \Rightarrow \frac{2+4+10+x+y+12+14}{7}=8 \Rightarrow x+y=14\]
\[\text{Variance}=16 \Rightarrow \frac{4+16+100+x^2+y^2+144+196}{7}-64=16 \Rightarrow x^2+y^2=100\]
Solving, x=8,y=6. So the new set is {1,2,3,4,5,6} (since x−4=4).
\[P(\text{smaller}<4) = 1-P(\text{both}\ge 4) = 1-\frac{\binom{3}{2}}{\binom{6}{2}} = 1-\frac{3}{15}=\frac{4}{5}\]