SOLUTION: A case of 24 cans contains 1 can that is contaminated. Three cans are to be chosen randomly for testin.
A. How many different combinations of 3 cans could be selected?
B. What
Algebra ->
Probability-and-statistics
-> SOLUTION: A case of 24 cans contains 1 can that is contaminated. Three cans are to be chosen randomly for testin.
A. How many different combinations of 3 cans could be selected?
B. What
Log On
Question 211492This question is from textbook
: A case of 24 cans contains 1 can that is contaminated. Three cans are to be chosen randomly for testin.
A. How many different combinations of 3 cans could be selected?
B. What is the probability that the contaminate can is selected for testing? This question is from textbook
You can put this solution on YOUR website! A case of 24 cans contains 1 can that is contaminated. Three cans are to be chosen randomly for testin.
A. How many different combinations of 3 cans could be selected?
Ans: 24C3 = 2024
------------------------------
B. What is the probability that the contaminate can is selected for testing?
P(contam in 3) = 1 - P(no contam in 3)
= 1 - 23C3/2024 = 1 - (1771/2024) = 0.1250
===============================================
Cheers,
Stan H.
(