Probability Question 30
Question: Twenty tickets are marked the numbers 1, 2, ….. 20. If three tickets be drawn at random, then what is the probability that those marked 7 and 11 are among them
Options:
A) $ \frac{3}{190} $
B) $ \frac{1}{19} $
C) $ \frac{1}{190} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ 7,11 $ have always to be in that group of three, therefore 3rd ticket may be chosen in 18 ways.
Hence required probability is $ \frac{18}{{}^{20}C_3}=\frac{18.3.2}{20.19.18}=\frac{3}{190} $ .
BETA
