A bag contains 3 red, 4 white and 5 blue balls. If two balls are drawn at random, then the probability that they are different colours is

Question

A bag contains 3 red, 4 white and 5 blue balls. If two balls are drawn at random, then the probability that they are different colours is (A) 47/66 (B) 23/33 (C) 47/132 (D) 47/33

Tardigrade Admin · Accepted Answer

Total number of balls =3+4+5=12 Favourable number of cases =12 C2-( 3 C2+ 4 C2+ 5 C2) =66-(3+6+10)=66-19=47 Total number of cases = 12 C 2=66 Hence, the required probability =( text Favourable cases / text Total number of cases )=(47/66)

Q. A bag contains $3$ red, $4$ white and $5$ blue balls. If two balls are drawn at random, then the probability that they are different colours is

47/66

23/33

47/132

47/33

70/33

Total number of balls $= 3 + 4 + 5 = 12$
Favourable number of cases
$=^{12} C_{2} - (^{3} C_{2} +^{4} C_{2} +^{5} C_{2})$
$= 66 - (3 + 6 + 10) = 66 - 19 = 47$
Total number of cases $=^{12} C_{2} = 66$ Hence, the required probability
$= \frac{Favourable cases}{Total number of cases} = \frac{47}{66}$

Q. A bag contains 3 red, 4 white and 5 blue balls. If two balls are drawn at random, then the probability that they are different colours is