Probability Question 73
Question: Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that $ a^{2}-b^{2} $ is divisible by 3 is
Options:
A) $ \frac{9}{87} $
B) $ \frac{12}{87} $
C) $ \frac{15}{87} $
D) $ \frac{47}{87} $
Show Answer
Answer:
Correct Answer: D
Solution:
The total number of ways of choosing two numbers out of $ 1,2,3,………,30 $ is $ {}^{30}C_2=435. $
Since $ a^{2}-b^{2} $ is divisible by 3 if either $ a $ and b both are divisible by 3 or none of a and b is divisible by 3.
Thus the favourable number of cases = $ ^{10}C_2+{{}^{20}}C_2=235 $ .
Hence the required probability = $ \frac{235}{435}=\frac{47}{87} $ .
BETA
