A box has 10 balls labeled 1 2 10 Suppose a random sample o

A box has 10 balls labeled 1, 2, ..., 10. Suppose a random sample of size 3 is selected. Find the probability that balls 1 and 6 are among the three selected balls.

Solution

Note that

P(at least one ball from 1-6) = 1 - P(no ball from 1-6)

There are 10C3 ways to pick any 3 balls, and 4C3 ways to pick 3 balls that are not 1-6.

Thus,

P(no ball from 1-6) = 4C3 / 10C3 = 0.033333333

Thus,

P(at least one ball from 1-6) = 1 - P(no ball from 1-6)

= 1 - 0.033333333

= 0.966666667 [answer]