Chapter 7 - Section 7.4 - Probability and Counting Techniques - Exercises - Page 493: 13

$\displaystyle \frac{4}{15}$

$P(E)=\displaystyle \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$ $n(S)=C(10,4)=210$ $ E$= (1 of 1 with highest yield) AND (0 of the lowest one) AND (3 of the remaining 8) $n(E)=C(1,1)\cdot C(1,0)\cdot C(8,3)=56$ $P(E)=\displaystyle \frac{56}{210}=\frac{4}{15}$