This is a short blurb that explains the solution to practice problem 2.29.
Chips in a bag. Imagine you have a bag containing 5 red, 3 blue, and 2 orange chips. (a) Suppose you draw a chip and it is blue. If drawing without replacement, what is the probability the next is also blue? (b) Suppose you draw a chip and it is orange, and then you draw a second chip without replacement. What is the probability this second chip is blue? (c) If drawing without replacement, what is the probability of drawing two blue chips in a row? (d) When drawing without replacement, are the draws independent? Explain.
- for part a, we are told that we have 5 red, 3 blue, and 2 orange. We want the probability of the next being blue given that we drew a blue chip without replacement There are 2 blue chips left since we have one, and only 9 possible chips to pick from since it was not replaced
p<-choose(2, 1)/choose(9, 1)
p
## [1] 0.2222222
- We now want the probability that the second chip is blue, after we draw an orange without replacement There are three blue chips to choose from, and since we already have one orange, we have 9 chips left for our sample space
p<-choose(3, 1)/choose(9, 1)
p
## [1] 0.3333333
- Now we want the probability of drawing two blue in a row without replacement We pick a blue out of 10 chips, then we pick another blue out of the two left from a pool of 9 chips left
p<- (choose(3, 1)/choose(10, 1))*(choose(2, 1)/choose(9, 1))
p
## [1] 0.06666667
- If we draw without replacement, the draws are not independent. This changes the probability since we are changing the overall sample space with each draw.