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14.Probability
easy
A bag contains $3$ red, $4$ white and $5$ blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is
A
$\frac{{47}}{{66}}$
B
$\frac{{10}}{{33}}$
C
$\frac{5}{{22}}$
D
None of these
Solution
(a) We have the following three pattern :
$(i)$ Red, white $P(A) = \frac{{3 \times 4}}{{{}^{12}{C_2}}}$
$(ii)$ Red, blue $P(B) = \frac{{3 \times 5}}{{{}^{12}{C_2}}}$
$(iii)$ Blue, white $P(C) = \frac{{4 \times 5}}{{{}^{12}{C_2}}}$
Since all these cases are exclusive, so the required probability
$ = \frac{{(12 + 15 + 20)}}{{{}^{12}{C_2}}} = \frac{{(47 \times 2)}}{{(12 \times 11)}} = \frac{{47}}{{66}}.$
Standard 11
Mathematics
