STATISTICS - Probability Rules. Please see IMG: http://i1084.photobucket.com/albums/j409/QRAWarrior/B22-IMG-004.png
@Callisto
P(M and A) is asking me what is the probability that they are BOTH accepted and that they are BOTH male
The circled number is the number of applicants who are male + accepted. Use it to calculate the probability..
But how do I get the probability for 2 at once?
I know how to get for one. If it was just P(male and accepted), I could do that, but I do not know how to do P(both male and both accepted)
How to do calculate the probability of both male?
P(male and male)? That is (660/800)*(659/799) <-- Sampling without replacement.
You are going to tell me this I believe: Find the P(male and male), find P(accepted and accepted), and then multiply them together?
No! Total number of male who are accepted = 480 P(both male and accepted) = (480/800) x (479/799)
What?
I believe you did it wrong.
Yes? Can you explain?
Because: P(both male) = (480/800)*(479/799) P(male and male) = P(male)*P(male) If you find one male applicant, then you do NOT want to pick out that same one again randomly, so you remove it.
But now, the condition is P(both male and accepted). We know that to total number of male who are accepted is 480 (from the chart) So, P(both male and accepted) = P(male and accepted) x P(male and accepted) = (480/800) x (479/799)
Here is a simple example: - Suppose there are 3 gold rooms, 2 silver rooms and 3 bronze rooms. You get to pick first and you end up getting the gold room: The probability of getting a gold room for you was 3/8 Now, the probability that your friend can get a good room is 2/7 <-- BECAUSE we HAVE 7 rooms left over now.
I understand what you said
But, you still have to use sampling without replacement there so you do not end up getting that same accepted male again
And you did it
I just checked now the last few numbers
OHHH
So you did this: P( (male and accepted) and (male and accepted))
No
I meant
P(both male and accepted) = P( (male and accepted) and (male and accepted)) ......
So THAT is how you rearrange those two. I will remember that.
Thank you for your help.
Welcome... Hope you understand it....
@Callisto I have another probability problem.
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