5C3*9C3/14C3*14C3
a bag contains 5 white and 9 black balls 3 balls are drawn twice . balls are replaced before the second draw. what is the prob that in the first draw 3 white and in second draw 3 lack balls will be drawn
-
UP 0 DOWN 0 0 3
3 Answers
Lokesh Verma
·2009-03-13 05:39:54
No fo ways to draw 3 white balls is
4C3
No of ways to draw 4 black balls is 9C3
NUmerator inPriyam's solution gives the product.. (Possible cases)
14C3 is the no of ways to draw 3balls.
Denominator gives the no of ways to draw 3 balls twice.
Hence the answer