A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked at random, what is the probability that at least one is blue?

Question

FirstInTest · Accepted Answer

Correct answer: D. Total number of marbles = $4 (red) + 5 (blue) + 3 (green) = 12$ marbles.
We need to find the probability that at least one marble picked is blue when 3 marbles are picked at random.
It's easier to calculate the complement: the probability that NONE of the marbles picked are blue.

Total ways to pick 3 marbles from 12 = $^12C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 2 	imes 11 	imes 10 = 220$.

Number of ways to pick 3 marbles that are NOT blue:
Marbles that are not blue = Red + Green = $4 + 3 = 7$ marbles.
Number of ways to pick 3 non-blue marbles from 7 = $^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 7 	imes 5 = 35$.

Probability of picking no blue marbles = $\frac{	ext{Ways to pick no blue marbles}}{	ext{Total ways to pick 3 marbles}} = \frac{35}{220}$.

Probability of picking at least one blue marble = $1 - 	ext{P(no blue marbles)}$
$= 1 - \frac{35}{220}$
$= \frac{220 - 35}{220} = \frac{185}{220}$.

Simplify the fraction by dividing by 5:
$\frac{185 \div 5}{220 \div 5} = \frac{37}{44}$.

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