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KEAM 2017 Math Paper

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Question : 97 of 120

Marks: +1, -0

From 4 red balls, 2 white balls and 4 black balls, four balls are selected. The probability of getting 2 red balls is

7/21
8/21
9/21
10/21
11/21

Solution:

We have, 4 red, 2 white and 4 black balls.

\therefore

Total balls

=4+2+4=10

Number of ways of selecting 4 balls from 10 balls

= {}^{10}C_{4}

\therefore \;\; n(S)={}^{10}C_{4}

Let

E=

Event getting 2 red balls

\therefore \;\; n(E)={}^{4}C_{2} \times {}^{6}C_{2}

\therefore

Required probability

=p(E)

=\;\frac{n(E)}{n(S)}

=\;\frac{{}^{4}C_{2} \times {}^{6}C_{2}}{{}^{10}C_{4}}

=\;\frac{\frac{4 \times 3}{2 \times 1} \times \frac{6 \times 5}{2 \times 1}}{\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}}

=\;\frac{6 \times 15}{10 \times 3 \times 7}

=\;\frac{3}{7}

=\;\frac{9}{21}

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