Hey
Probability is such a nightmare for me, here's a two part Q I'm struggling with.
A jar contains 5 red, 2 blue and 3 green marbles. What is the probablity of choosing:
a) at least one green and at least one red marble
b) at least one red given that at least one green is chosen
For part B I'm using the formula of Pr(A|B) = Pr(A int B) / (Pr B), but I cannot for the life of me figure out how to get the intersection of "at least one red" and "at least one green".
Help would be much appreciated, thanks all!
i believe this is only half the story.
yes, the chances of selecting a green marble, then a red marble is 1/6.
but the chance of selecting a red marble, then a green marble is (5/10)(3/9) = 1/6,
so the chance of either happening (as these are mutually exclusive chains of events) is 1/6 + 1/6 = 1/3.
also, there are 10!/(2!8!) = 45 ways to pick 2 marbles out of 10.
there are (5)(3) = 15 ways this pair could be a red and green pair. therefore, we have 15/45 = 1/3 chance of getting a red and green pair.
Hello, Oaky!
Didn't anyone notice that the problem is incomplete?
How many marbles are chosen?A jar contains 5 red, 2 blue and 3 green marbles.
What is the probablity of choosing:
(a) at least one green and at least one red marble.
(b) at least one red, given that at least one green is chosen.
Suppose 3 marbles are chosen.
Then there are: . possible outcomes.
(a) At least one G and at least one R.
GRR: .
GRG: .
GRB: .
Hence, there are: . ways with at least 1 G and at least 1 R.
(b) At least 1 Red, given at least 1 G.
We have: .
We found the numerator in part (a): .
The denominator is:.
. . The opposite is:.
. . There are: . ways to get no G.
. . Then:.
. . Hence:
Therefore: .