Statistics And Probability Question 242
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:
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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
