A) \[\frac{47}{66}\]
B) \[\frac{10}{33}\]
C) \[\frac{5}{22}\]
D) None of these
Correct Answer: A
Solution :
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 \[=\frac{(12+15+20)}{{}^{12}{{C}_{2}}}=\frac{(47\times 2)}{(12\times 11)}=\frac{47}{66}.\]
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