Test Index

BITSAT 2013 Paper

Section: Mathematics

Question : 89 of 150

Marks: +1, -0

Solution:

Atleast one black ball can be drawn in the following ways

(i) one black and two other colour balls

\,{}^{3}C_{1} \times \,{}^{6}C_{1}

= 3 × 15 = 45

(ii) two black and one other colour balls

\,{}^{3}C_{2} \times \,{}^{6}C_{1}

= 3 × 6 = 18

(iii) All the three are black

\,{}^{3}C_{3} \times \,{}^{6}C_{0}

= 1

∴ Req. no. of ways = 45 + 18 + 1 = 64

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