Siddharthjain07 wrote:
A bucket contains three different colours flowers - red, blue and yellow. There are 4 red, 5 blue and 3 yellow flowers. If 4 flowers are to be chosen at random, what is the probability that at least 3 of the 4 flowers chosen will be blue?
A.19/24
B.5/33
C.7/24
D.11/24
E.1/3
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The "at least" language in the question should stand out to you. When this problem says "at least 3 of the 4 are blue," it's actually referring to two separate situations that don't overlap:
the probability that exactly 3 of the 4 are blue, OR that exactly 4 of the 4 are blue.
It's much easier to count the number of ways something can happen exactly, compared to counting the number of ways "at least" something can happen. So, break the question down into these two situations, and deal with them one at a time.
The second one is easier. Since there are 5 blue flowers in total, if we pick 4 blue flowers, then we're leaving out 1 flower out of the 5. There are 5 different ways to do this. So, there are 5 ways to have 4 blue flowers.
Now, the first one. If we pick 3 blue flowers, there are (5*4*3)/(3*2*1) = 10 possible different sets of blue flowers we might have picked. For each of these sets, the last flower could be any one of the other 7 non-blue flowers. So, there are 70 different sets of flowers where exactly 3 of them are blue.
The total number of possibilities is 5 + 70 = 75.
Since this problem asks for a probability, you also need to figure out the total number of different sets of flowers that could be picked. There are 12 flowers in total, and we're picking 4 of them. So, the number of sets is (12*11*10*9)/(4*3*2*1) = (11*10*9)/(2*1) = 11*5*9.
Therefore, the probability is 75/(11*5*9) = 15/(11*9) = 5/33, and the answer is
B.