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Mathematics 16 Online
OpenStudy (anonymous):

assume there are 4 grey females, 4 grey males, 2 white females, and 3 white males. As in the book, the biologist selects two mice randomly. (1) What is the probability of selecting two males given that both are grey? (2) What is the probability of selecting 1 mouse of each gender given that both are grey?

OpenStudy (anonymous):

please help!!!

OpenStudy (anonymous):

there are 4 grey males and how many total?

OpenStudy (anonymous):

oh damn i answered some other question i should learn how to read, ignore that post

OpenStudy (anonymous):

for (1) I believe it is 4/8*3/7

OpenStudy (anonymous):

thats not right

OpenStudy (anonymous):

or 4C2/8C2

OpenStudy (anonymous):

like 4 choose 2 over 8 choose 2?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

thats not right

OpenStudy (anonymous):

put \(A\) as the event both chosen are males, \(B\) as the event both chosen are grey, and then compute \[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

OpenStudy (anonymous):

\[P(B)=\frac{8}{13}\times \frac{7}{12}=\frac{14}{39}\]

OpenStudy (anonymous):

14/39 isn't right

OpenStudy (anonymous):

yes, it is not right, we are not done we still have to compute \(P(A\cap B)\) \(A\cap B\) is the event both chosen are grey males, it is \[\frac{4}{13}\times \frac{3}{12}\]

OpenStudy (anonymous):

better known as \[\frac{1}{13}\]

OpenStudy (anonymous):

so your answer should be \[\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{13}}{\frac{14}{39}}=\frac{1}{13}\times \frac{39}{14}=\frac{3}{14}\]

OpenStudy (anonymous):

that's exactly the same as 4C2/8C2

OpenStudy (anonymous):

that is right!! do u know how to do the second part?

OpenStudy (anonymous):

@pizzapi that is probably a snappier way of doing it, yes

OpenStudy (anonymous):

but he/she said it was wrong

OpenStudy (anonymous):

maybe less intuitive, but definitely quicker

OpenStudy (anonymous):

how do you do the second part

OpenStudy (anonymous):

(4C1*4C1)/(8C2)

OpenStudy (anonymous):

thanks!!

OpenStudy (anonymous):

hold phone, that one might not be right

OpenStudy (anonymous):

Assume that the box contains 11 balls: 4 red, 4 blue, and 3 yellow. As in the text, you draw one ball, note its color, and if it is yellow replace it. If it is not yellow you do not replace it. You then draw a second ball and note its color. (1) What is the probability that the second ball drawn is yellow? (2) What is the probability that the second ball drawn is red?

OpenStudy (anonymous):

do you know how to do that?

OpenStudy (anonymous):

yes but hold on a second

OpenStudy (anonymous):

part 2 was right.. my homework tells you right off when you enter if correct

OpenStudy (anonymous):

yes, it is right i had a brain freeze last problem you posted is harder and we have to work through it step by step

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

a nice tree diagram might help let me see if i can draw one, because it is hard for me to draw here |dw:1348627260493:dw|

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