Probability: With and Without Replacement

1. A bag contains 20 blue balls and 16 red balls. A ball is picked and not replaced. What is the probability of picking at least one blue ball?

2. A box has 18 red toy cars, 30 green toy cars and 60 yellow toy cars. You select 15 toy cars from the box with replacement. Determine the probability the first toy car is green and the second toy car is also green.

 

1. A bag contains 20 blue balls and 16 red balls. A ball is picked and not replaced. What is the probability of picking at least one blue ball?

20 blue balls and 16 red balls->total 36 balls
to pick only one you have C(36,1) =36 ways

If you pick only one, the probability of picking a blue ball at random is 20/36 = 5/9.

2. A box has 18 red toy cars, 30 green toy cars and 60 yellow toy cars. You select 15 toy cars from the box with replacement. Determine the probability the first toy car is green and the second toy car is also green.
18 red toy cars, 30 green toy cars and 60 yellow toy cars->total 108 cars
select 15 toy cars

(C(18,15)+C(30,15)+C(60,15))/C(108.15)

(15/108)=5/36 is the probability the first toy car is any color
the probability the first toy car is green and the second toy car is also green
(5/36)(30/15)(30/15)=5/9=0.0107167352 or 1.07167352%

 

How about one more in this forum?

https://www.math-forums.com/threads/probability-without-replacement.441315/