An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 Is green, 2 are blue and 1 is red ?

Solution(By Examveda Team)

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 1 green, 2 blue and 1 red marble.
Then,
n (E) = $${}^2\mathop C\nolimits_1 \times {}^4\mathop C\nolimits_2 \times {}^6\mathop C\nolimits_1 $$   $$ = 2 \times \frac{{4 \times 3}}{{2 \times 1}} \times 6$$     = 72
And, n (S) = $${}^{15}\mathop C\nolimits_4 = $$   $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$     = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{72}}{{1365}}$$      $$ = \frac{{24}}{{455}}$$