Probability Math Question

Assume a team had 10 aluminum bats
and 4 magnesium bats. Three bats are selected at random.

(1) What is the probability that all 3 are aluminum?

(2) What is the probability that at least one is aluminum?

(1) (10/14)x(9/13)x(8/12) = (5/7)x(9/13)x(2/3) = (5/7)x(3/13)x2 = 30/91

(2) P(at least 1 aluminium) = 1 - P(no aluminium)

=> 1 - (4/14)(3/13)(2/12)

=> 1 - (2/7)(3/13)(1/6)

=> 1 - (1/7)(1/13)

i.e. 1 - 1/91 = 90/91

:)>

1. 32.967%
2. 98.001% :)