An urn contains ten balls numbered $1$ through $10$. Five balls are drawn at random and without replacement. Find probability that the second largest number drawn will be $8$.
The root of my confusion lies in whether the balls are drawn one at a time, or scooped up at once.
Best Answer
Addressing your comment as to whether the balls are drawn one at a time, or together: it doesn't matter.
If you draw the balls one at a time, you would probably think of choosing five numbers from ten, including the number $8$, with order being important. To count the number of favourable draws,
To count the total number of draws:
The probability is $$\frac{5\times2\times4\times7\times6\times5}{10\times9\times8\times7\times6}=\frac{5}{18}\ .$$
If you draw them all together you are probably thinking of order as being irrelevant. The number of favourable draws is $2C(7,3)$, the total number of draws is $C(10,5)$, and the probability is $$\frac{2C(7,3)}{C(10,5)}=\frac{2\times7\times6\times5}{3\times2\times1}\frac{5\times4\times3\times2\times1}{10\times9\times8\times7\times6}=\frac{5}{18}\ .$$