Question 1165656: I am completely spacing on how you solve this type of problem
An internet research company surveyed 90 online shoppers, each of whom made one purchase today. The company recorded the type of purchase each shopper made. Here is a summary.
Type of purchase Number of shoppers
Clothing 25
Food 19
Electronics 25
Toys 18
Three shoppers from the survey are selected at random, one at a time without replacement. What is the probability that none of the three shoppers purchased clothing?
Do not round your intermediate computations. Round your final answer to three decimal places.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there are 90 online shoppers, each of which made a purchase today.
the first problem i see is that the individual categories don't add up to 90.
if every one of the shoppers made a purchase, then there should be a total of 90 purchases.
despite that, i believe the problem can still be solved.
let's just say that the purchases that weren't categorized fall under the category of something else.
you get
clothing = 25
food = 19
electronics = 25
toys = 18
something else = 3
now the total is 90.
3 shopp[ers are selected.
the probability that the first shopper doesn't select clothing is 65/90.
the probability that the second shopper doesn't select clothing is 64/89.
the probability that the third shopper doesn't select clothing is 63/88.
the probability that the first and the second and the third don't select clothing is therefore 65/90 * 64/89 * 63/88 = .372 rounded to 3 decimal places.
you don't know what they selected.
you only know they didn't select clothing.
therefore you can lump all the other categories into one category of other.
you get:
selected clothing = 25
selected other = 65
before any selection you have 25 clothing + 65 other = 90
after the first selection you have 25 clothing + 64 other = 89
after the second selection you have 25 clothing + 63 other = 88
the probability of picking other on the first selection is 65/90.
the probability of picking other on the second selection is 64/89.
the probability of picking other on the third selection is 63/88.
the probability of selecting other on all 3 selections is therefore 65/90 * 64/89 * 63/88 = .372 rounded to 3 decimal places.
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