Easy probability question... A container holds 13 red marbles, 9 blue marbles, and 16 green marbles. You will randomly select two marbles without replacement. How many ways can you select the marbles? What is the probability that you select 2 red marbles? What is the probability that you select a green marble and then a red marble?
for the first one im guessing its 2/38
oh okay so what do i put for P?
Sorry, my bad.
lol so just forget about that equation?
TM total marbles = R + B +G First draw P(R) = R/TM, P(B) = B/TM, P(G) = G/TM Then second draw you have to consider condition probabilities such as P(G|G) = (G - 1)/(TM-1) PG|B) = G/(TM-1) ect then use Baye's formual
what does P stand for Probability?
P(G) means the probability of green P(G|B) means the probability of green given previous draw bllue
oh okay but there are a total of 38 marbles and for the first question is ask, How many ways can you select the marbles? it should be 2/38 cause you take 2 marbles out
right?
MT! is they were all different color For different colors MT!/(R!B!G!) It's divided by R! because it repeats that many times
so what does MT! equal?
why always marbles?
MT to marbles = #R + #B +#G= R+B+G
38! = 38X37X36 ... X1 very large and probably wont show up on calulator
if you care about the order then there are 38*37 ways to pick 2 marbles
Theres three colors
there are 13*12 ways to pick two reds so the prob is (13*12)/(38*37)
oh okay thanks both for taking the time to help me :)
i got 151.89
for what?
(13*12)/(38*37)
you might want to try that again
Probabilities are always less than or equal to 1
You can't get an answer greater than 1; at least not in this scope.
parentheses are important
so i have to put it in fraction
\(\frac{13\times 12}{38\times37}\) is what you have to compute.
156/1406
that's is correct. (156/1406 = (13*12)/(38/37))
oh okay thanks for the help :)
umm who wants the medal?
oh i didn't work as much as the others.. :p thanks
lol welcome, and i became of fan of the others who helped
*a fan
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