anonymous
  • anonymous
A box has 60 balls, of which 22 are red, 19 are blue, and 19 are yellow, as before. Balls are extracted at random but this time not returned to the box. What is the probability that after performing this test three times exactly two are blue and one is red?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
when i do the calculation i got the answer 1818/17110 but the book answer have 5643/17110, i think i did something wrong my calculation
anonymous
  • anonymous
this what i did to get the calculation \[\frac{ \left(\begin{matrix}19 \\ 2\end{matrix}\right)* \left(\begin{matrix}22 \\ 1\end{matrix}\right)*\left(\begin{matrix}19 \\ 0\end{matrix}\right)}{ \left(\begin{matrix}60 \\ 3\end{matrix}\right)}\]
amistre64
  • amistre64
what is the test being performed?

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anonymous
  • anonymous
combination
amistre64
  • amistre64
What is the probability that after performing ***this test*** three times exactly two are blue and one is red? the questions feels like its missing information ... prolly from a different question?
anonymous
  • anonymous
k A box has 50 balls, of which 16 are red, 19 are blue, and 15 are yellow. Balls are extracted at random then returned to the box. What is the probability that after performing this test three times exactly two are yellow and one is blue?
anonymous
  • anonymous
similar but a bit different w/ a number
amistre64
  • amistre64
experiment ... might be a better word
anonymous
  • anonymous
i see
amistre64
  • amistre64
A box has 60 balls, of which 22 are red, 19 are blue, and 19 are yellow, as before. Balls are extracted at random but this time not returned to the box. What is the probability that after performing this (process) three times exactly two are blue and one is red? rbb brb bbr 3*(22.19.18)/(60.59.58) maybe?
anonymous
  • anonymous
doing that the number will be large
amistre64
  • amistre64
1881/17110 is what that turns out to be
amistre64
  • amistre64
P(bbr) or P(brb) or P(rbb) are the only ways we can pull 2bs and 1r in 3 pulls right?
anonymous
  • anonymous
correct
amistre64
  • amistre64
and after we pull them out the remaining amounts decrease by 1 22 for red, 19 and 18 for blues and the ball count reduces 60,59,58 for the product of the denominator ...
anonymous
  • anonymous
is like this |dw:1444575916412:dw|
amistre64
  • amistre64
not factorial, 3 * (22 P 1) * (19 P 2) ------------------- (60 P 3)
amistre64
  • amistre64
but my solution is even less than yours, so i might not have a grasp of the information
anonymous
  • anonymous
i wonder what is p1 and p2 p3 use for ?
amistre64
  • amistre64
nPk is the permuations of n objects, k times it is used when the order matters abc not equal bac nCk is the combinations of n objects, k times it is used when the order doesnt matter abc = bac k! nCk = nPk
anonymous
  • anonymous
we are using the permutation instead of combination
amistre64
  • amistre64
yes, or at least as a basis for a model
amistre64
  • amistre64
there are 3 ways to pull bbr and each pull has a probability of (22*19*18)/(60*50*58)
amistre64
  • amistre64
this is assuming it means that we are pulling 3 balls starting at the first pull.
amistre64
  • amistre64
59, not 50 ... my fingers hate me
anonymous
  • anonymous
k i see the answer |dw:1444576774278:dw|
amistre64
  • amistre64
1881 is what i get for a numerator 5643 is what your book gives, which is 3 times bigger than my outcome. 3*3*(...) where are they coming up with 9 different ways to order it?
amistre64
  • amistre64
what does "this test" refer to?
anonymous
  • anonymous
i have 3*(7524/205320)
amistre64
  • amistre64
which reduces yes
anonymous
  • anonymous
as you said the experiment
amistre64
  • amistre64
that was my assumption, but my outcome differs from the books outcome. someones got to be wrong, and I assume its me :)
amistre64
  • amistre64
are you looking at the correct solution?
anonymous
  • anonymous
yes im
anonymous
  • anonymous
just said about the multi-step process
amistre64
  • amistre64
even your approach gets the same results as me
1 Attachment
TuringTest
  • TuringTest
I get the same answer as you guys. I am suspicious of the book's answer.
amistre64
  • amistre64
could it be possible that we are drawing more than 3 of them out for our sample space?
anonymous
  • anonymous
here the answer but do not show the step
1 Attachment
amistre64
  • amistre64
3*(15*15*19)/(60*60*60) = 513/5000 so its the same approach
amistre64
  • amistre64
the book solution is wrong then
anonymous
  • anonymous
thank you so much for the correcting the confusion
amistre64
  • amistre64
im trying to see how they could have gotten an answer 3 times bigger, but other than a simpler human error in inputing it to begin with i cant duplicate it.
anonymous
  • anonymous
k

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