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a bag contains 9 blue marbles, 8 white marbles and 3 red marbles. Three marbles are selected from the bag. What is the probability they are all red? What is the probability that they are all the same colours?

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P(3 reds) = C(3,3)/C(20,3)
how do u solve that please?
at first, the probability of selecting a red marble is \(\frac{3}{20}\), and then it is \(\frac{2}{19}\), then it's \(\frac{1}{18}\). multiply them all to get the probability of selecting all 3 red ones!

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Other answers:

so multiply all three fractions?
^ yes multiply all three together \[\frac{ 3 }{ 20 } \times \frac{ 2 }{ 19 } \times \frac{ 1 }{18 } = ???\]
to get answer part a?
whats for part b?
For probability that they're all the same colours, add up three separate probabilities - prob of getting three white, and prob of getting three blue, and prob of getting three red (which we just found)
@agent0smith and then do you add them together? i forgot how to do that part.
Yeah i think you just add them together
that is the answer i just found for multiplying all three?? which answer does that belong to there are two questions, first one is: What is the probability they are all red? and the second question is : What is the probability that they are all the same colours?
the answer you just found was the reds... please read through the explanation agent0smith and I just provided
more help:
the first part of the question the answer that i got was 0.00087719298..
so for part a my answer would be 0.001 and for part b my answer would be 0.31
Show us how you did part b, so someone can check it.
part b. this is how you do it. C(n,r) is a combination. P(all the same color) = [(C(3,1)*C(9,3)+C(2,1)*C(8,3)+C(1,1)*C(3,3)]/C(20,3) the idea is C(3,1) is choosing the first color. then you have to choose three from that color. I chose blue to start with C(9,3). then you have C(2,1) which is because you have removed one color from the possible colors. then you have to choose 3 from that color. I chose white C(8,3). then you only have one color left C(1,1). this color must be red C(3,3). Then you divide by all possible combinations C(20,3). done.
@tomo are you sure that's correct? Because you'll end up with different probabilities depending on the order you enter them... eg if you choose white first instead of blue.
And a 32% chance of getting all three the same colour seems kinda high. The probability of JUST pulling out one blue from the bag is only 9/20 = 45%. Chance of getting three blues = 7.4%, chance of getting three reds = 0.09%, chance of getting three white = 4.91% Also note that my method and yours get the same probability IF you remove the C(3,1), C(2,1) and C(1,1) from your equation. my method on the left, yours on the right: \[\frac{ (3 \times 2 \times 1) + (9 \times 8 \times 7) + (8 \times 7 \times 6) }{ 20 \times 19 \times 18 } = \frac{ C(9,3) + C(8,3) +C(3,3) }{ C(20,3) }\]
They simplify to \[\frac{ 846 }{ 6840 } = \frac{ 141 }{ 1140 }\] So the prob. of picking three the same colour = 0.124 = 12.4%

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