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14.Probability
hard
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
A
$\frac{9}{{87}}$
B
$\frac{{12}}{{87}}$
C
$\frac{{15}}{{87}}$
D
$\frac{{47}}{{87}}$
Solution
(d) 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 $= \frac{{235}}{{435}} = \frac{{47}}{{87}}$.
Standard 11
Mathematics
