First slide

Theorems of probability

Question

Two numbers x and y are chosen at random (without replacement) from amongst the numbers 1, 2, 3,……..2004. the probability that $x^{3} + y^{3}$ is divisible by 3 is

Difficult

A

$\frac{1}{3}$
B

$\frac{2}{3}$
C

$\frac{1}{6}$
D

$\frac{1}{4}$

Solution

We divide the number in three groups
$3 k + 1 t y p e {1, 4, 7, ..........., 2005}$

$3 k + 2 t y p e {2, 5, 8, ..........., 2006}$

$3 k + 3 t y p e {3, 6, 9, ..........., 2007}$

$x^{3} + y^{3}$ is divisible by 3. If x and y both belong to 3^rd group or one of them belongs to the first group and the other to the second group.

So favorable number of cases $= 669_{C_{2}} + 669 \times 669$

Total number of cases $= 2007_{C_{2}}$

$∴$ desired probability $\frac{\frac{669 \times 668}{2} + 669 \times 669}{\frac{2007 \times 2006}{2}} = \frac{669 \times 2006}{2007 \times 2006} = \frac{1}{3}$

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