Question

Armando buys a bag of cookies that contains 6 chocolate chip cookies, 7 peanut butter cookies, 4 sugar cookies and 7 oatmeal cookies. What is the probability that Armando reaches in the bag and randomly selects 2 sugar cookies from the bag? Round your answer to four decimal places.

Answer 1

0.0476

Explanation:

Given:

Armando buys a bag of cookies that contains 6 chocolate chip cookies, 7 peanut butter cookies, 4 sugar cookies and 7 oatmeal cookies.

Question asked:

What is the probability that he chosen randomly 2 sugar cookies from the bag?

Solution:

As we know:

`Probability =(Favourable \ outcome)/(Total\ outcome)`

First of all we will find favorable outcome for choosing 2 sugar cookies from the bag randomly:

Favorable outcome for choosing 2 sugar cookies out of 6 chocolate chip cookies =
`^(6) C_(0)`

Favorable outcome for choosing 2 sugar cookies out of 7 peanut butter cookies =
`^(7) C_(0)`

Favorable outcome for choosing 2 sugar cookies out of 4 sugar cookies =
`^(4) C_(2)`

Favorable outcome for choosing 2 sugar cookies out of 7 oatmeal cookies =
`^(7) C_(0)`

Thus, total Favorable outcome for choosing 2 sugar cookies =

By using:
`^(n) C_(r)=(n!)/((n-r)!\ r!)`

`^(6) C_(0)*^(7) C_(0)*^(4) C_(2)*^(7) C_(0)`

`(6!)/((6-0)!\ 0!) *(7!)/((7-0)!\ 0!)*(4!)/((4-2)!\ 2!)*(7!)/((7-0)! \ 0!)`

`1*1*(4*3*2)/(2*2) *1=(24)/(4) =6`

Total outcome for choosing 2 sugar cookies out of 24 cookies =

`^(24) C_(2)` =
`(24!)/((24-2!\ 2!))`=
`(24!)/((24-2)!\ 2!) =(24*23*22!)/(22!\ 2!) =(552)/(2)`

Now,

`Probability =(Favourable \ outcome)/(Total\ outcome)`

`=(6)/((552)/(2) )\\ \\ = 6/(552)/(2)`

`=6*(2)/(552) \\\\ =(12)/(252) \\\\ =0.0476`

Thus, the probability that Armando reaches in the bag and randomly selects 2 sugar cookies from the bag is 0.0476