A box has 6 black, 4 red, 2 white and 3 blue shirts. When 2 shirts are picked randomly, the probability that…
A box has 6 black, 4 red, 2 white and 3 blue shirts. When 2 shirts are picked randomly, the probability that either both are white or both are blue is
$4 / 105$
$1 / 35$
$1 / 105$
$1 / 15$
Solution
There are 6 black, 4 red, 2 white and 3 blue shirts.
2 shirts can be picked in ${ }^{15} C_2$. ways 2 white shirts can be picked in ${ }^2 C_2$ ways.
2 blue shirts can be picked in ${ }^3 C_2$ ways.
Probability of getting 2 white shirts $=\frac{{ }^2 C_2}{{ }^{15} C_2}$
Probability of getting 2 blue shirts $=\frac{{ }^3 C_2}{{ }^{15} C_2}$
$\therefore$ Required probability
$=\frac{{ }^2 C_2+{ }^3 C_2}{{ }^{15} C_2}=\frac{1+3}{\left(\frac{15 !}{2 ! 13 !}\right)}=\frac{4 \times 2}{15 \times 14}=\frac{4}{105}$