Test Index

Probability Part 1

Section: Mathematics

Question : 81 of 100

Marks: +1, -0

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

[AIEEE 2010]

$\frac{2}{7}$
$\frac{1}{21}$
$\frac{1}{23}$
$\frac{1}{3}$

Validate

Solution:

Out of nine balls three balls can be chosen

={}^{9}C_{3}

ways

\therefore

Sample space

={}^{9}C_{3}=\;\frac{9!}{3!\,6!}=\;\frac{9 \times 8 \times 7}{6}=84

According to the question, all three ball should be different. So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen.

\therefore

Total cases

={}^{3}C_{1} \times {}^{4}C_{1} \times {}^{2}C_{1}=3 \times 4 \times 2=24

\therefore

Probability

=\;\frac{24}{84}=\;\frac{2}{7}

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