Question 1135955: There is a bag filled with 3 blue, 4 red and 5 green marbles.
A marble is taken at random from the bag, the colour is noted and then it is not replaced.
Another marble is taken at random.
What is the probability of getting 2 the same colour?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Answer in fraction form: 19/66
Answer in decimal form: 0.2879 (this value is approximate)
----------------------------------------------------------------------------------
Work Shown:
Define the three events
R = event of getting red
B = event of getting blue
G = event of getting green
For any single marble, we have these three probabilities
P(R) = 4/12 since there are 4 red out of 12 total
P(B) = 3/12 since there are 3 blue out of 12 total
P(G) = 5/12 since there are 5 green out of 12 total
Let's compute the probabilities for getting two of the same color, for any given color R, G or B
P(RR) = probability of getting two red in a row
P(RR) = (4/12)*(3/11) = 12/132
P(BB) = probability of getting two blue in a row
P(BB) = (3/12)*(2/11) = 6/132
P(GG) = probability of getting two green in a row
P(GG) = (5/12)*(4/11) = 20/132
The values step down because we do not replace the marble we selected.
Note how I did not reduce the fractions. This was to ensure the denominators stays the same, so we can then add the fractions as such
P(RR or BB or GG) = P(RR) + P(BB) + P(GG)
P(RR or BB or GG) = 12/132 + 6/132 + 20/132
P(RR or BB or GG) = (12 + 6 + 20)/132
P(RR or BB or GG) = 38/132
Extra info: the addition formula used is only possible because the events RR, BB and GG are mutually exclusive. For example, it is impossible for the events RR and BB to happen simultaneously. This is assuming we only select two marbles.
Now reduce the fraction as much as possible. We do so by dividing both parts (38 and 132) by the GCF 2
38/2 = 19
132/2 = 66
So 38/132 reduces to 19/66 which is roughly equivalent to 0.2879 (rounded to four decimal places)
|
|
|
