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

A. $\frac{7}{12}$

B. $\frac{37}{44}$

C. $\frac{5}{12}$

D. $\frac{7}{44}$

Solution:

Total number of marbles = (4 + 5 + 3) = 12
Let E be the event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 7 =

^{7} C_{3}

= \frac{7 \times 6 \times 5}{3 \times 2 \times 1}

= 35
And,

n (S) =^{12} C_{3}

= \frac{12 \times 11 \times 10}{3 \times 2 \times 1}

= 220

∴ P (E) = \frac{n (E)}{n (S)} = \frac{35}{220} = \frac{7}{44}

∴ Required probability
= 1 - P(E)
=

(1 - \frac{7}{44})

\frac{37}{44}

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