Probability Question 116
Question: 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
[EAMCET 1987]
Options:
A) $ \frac{47}{66} $
B) $ \frac{10}{33} $
C) $ \frac{5}{22} $
D) None of these
Show Answer
Answer:
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}. $
BETA
