+556 Ben twice chooses a random integer between 1 and 50, inclusive (and he may choose the same integer both times). What is the probability that at least one of the numbers Ben chooses is a multiple of 3?
+33616 When you get a question with "at least" in, it is often a good idea to find the probability of "none" first, then the probability of "at least" = 1 - probability of "none".
Suppose there are n integers between 1 and 50 that are not multiples of 3. Then the probability of choosing a random integer that is not a multiple of 3 is n/50.
The probability of doinjg this twice is (n/50)^2. Hence the probability that at least one number is a multiple of 3 is 1 - (n/50)^2
+221 What difficulties are you having with this problem? Don't expect people to solve your math problems for you. That is called cheating.
+556 Okay. I don't get what you are suposed to fill in for these cases.
Case 1:
First pick multiple of 3
Second pick not multiple of 3.
Case 2:
First pick not multiple of 3
Second pick multiple of 3
Case 3:
Both picks multiples of 3
Isn't this like a math forum so.....
Why are ALL your questions about "Probability"? Is that all you know, or don't know?
+556 I am currently taking a course in Counting and Probability andi am posting the questions I don't understand onto here. I have also posted some Geometry questions too so.....
+33616 When you get a question with "at least" in, it is often a good idea to find the probability of "none" first, then the probability of "at least" = 1 - probability of "none".
Suppose there are n integers between 1 and 50 that are not multiples of 3. Then the probability of choosing a random integer that is not a multiple of 3 is n/50.
The probability of doinjg this twice is (n/50)^2. Hence the probability that at least one number is a multiple of 3 is 1 - (n/50)^2
