A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at random then the probability that all the five are blue color is:

Solution(By Examveda Team)

Total number of erasers in the box = 3 + 4 + 7 = 14
Let S be the sample space
Then, n(S) = number of ways of taking 5 out of 14
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{14}{C_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{14 \times 13 \times 12 \times 11 \times 10}}{{2 \times 3 \times 4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \times 13 \times 11 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2002 \cr} $$
Let E be the event of getting all the 5 blue erasers
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^7{C_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7 \times 6 \times 5 \times 4 \times 3}}{{2 \times 3 \times 4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21 \cr} $$
Now, the required probability
$$\eqalign{ & \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{21}}{{2002}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{286}} \cr} $$