A box contains 10 twoinch screws of which 4 have a Phillips

A box contains 10 two-inch screws, of which 4 have a Phillips head and 6 have a regular head. Suppose that you select 3 screws randomly from the box. What is the probability that there will be more than one Phillips head screw (a) under a without replacement scenario? (b) under a with replacement scenario?

please explain and show all your work, thanks!

Solution

P(A)=P(more than one Phillips head screw without replacement)

=P(one philips head without replacement) + P(two philips head without replacement) +P(three philips head without replacement)

Therefore P(one philips head without replacement)

=3 x 4/10 x 6/9 x 5/8 [ since this can happen in 3 ways so multiply 3 ]

=1/2

And P(two philips head without replacement)

=3 x 4/10 x 3/9 x 6/8 [ since this can also happen in 3 ways so multiply 3 ]

=3/10

And P(three philips head without replacement)

= 4/10 x 3/9 x 2/8 [ since this can also happen in 1 way only ]

=1/30

Hence P(A)=