anonymous
  • anonymous
Assume a jar has 7 red marbles and 5 black marbles. Draw out 3 marbles with and without replacement. Find the requested probabilities. (a) P(3 red marbles) With replacement , without replacement . (b) P(3 black marbles) With replacement , without replacement . (c) P(one red and two black marbles) With replacement , without replacement . (d) P(red on the first draw and black on the second draw and black on the third draw) With replacement , without replacement .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
a) with replacement: P(3 red) = P(1st red). P(2nd red). P(3rd red)=(7/10)(7/10)(7/10)=343/1000
anonymous
  • anonymous
these are the answers I got I just wanted to get them checked and see if they are right? a) with 343/128 without 21/132 b)125/1728 without 60/132 c) I did not know how to do this one D) with 175/1728 without 14/132
anonymous
  • anonymous
wouldnt it be 7/12? not 7/10?

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anonymous
  • anonymous
Oh sorry over 12 :(
anonymous
  • anonymous
ok could you possibly tell me if my answers are right? and possibly how to do C?
anonymous
  • anonymous
for a with replacement: P(3 red)=(7/12)^3
anonymous
  • anonymous
without replacement: P(3 red)=(7/12)(6/11)(5/10)=....
anonymous
  • anonymous
Use the same method to solve B.
anonymous
  • anonymous
ya I did that, but I just wanted to make sure i got the right answers and simplified right?
anonymous
  • anonymous
a) with replacement: P(3 red)=343/1728 = 0.1984
anonymous
  • anonymous
without replacement: P(3 red)=7/44 = 0.159
anonymous
  • anonymous
what about the others?
anonymous
  • anonymous
B) with replacement: P(3 black)=(5/12)^3=125/1728
anonymous
  • anonymous
without replacement: P(3 black)=(5/12)(4/11)(3/10)=1/22
anonymous
  • anonymous
how did you get 1/22?
anonymous
  • anonymous
Calculator.
anonymous
  • anonymous
You can do it by hand though.
anonymous
  • anonymous
do you know how to do c?
anonymous
  • anonymous
Sorry, I had to go. In part (c), we want to get one red and two black marbles. There are three possibilities for this, red comes first, second or last. if R denotes red marbles and B denotes black marbles, then: i) with replacement: P(one red and two black)=P(R B B)+P(B R B)+P(B B R) =(7/12)(5/12)(5/12)+(5/12)(7/12)(5/12)+(5/12)(5/12)(7/12) =3(7/12)(7/12)(5/12)= (use a calculator).
anonymous
  • anonymous
Try the same way and do the "without replacement" part.

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