number of selections with no restrictions
= C(11,6) = 462
We can't have: zero red, 1 red, zero black, 1 black
which would be
C(5,0)xC(6,6) + C(5,1)xC(6,5) + C(6,0)xC(5,5) + C(6,1)xC(5,4)
= 1x1 + 5x6 + 1x1 + 6x5
= 62
prob(your event) = (462 - 62)62/462 = 200/231
a box contains 5 red balls and 6 black balls in how many ways can 6 balls be selected so that there are atleast two balls of the same colour
2 answers
i) 2 red,4 black => C(5,2)*C(6,4) =150
ii)3 red,3 black => C(5,3)*C(6,3) =200
iii)4red,2 black => C(5,4)*C(6,2) =75
Therefore total number of ways =
150+200+75 = 425 ways
Hope this helps! Cheers! :)
ii)3 red,3 black => C(5,3)*C(6,3) =200
iii)4red,2 black => C(5,4)*C(6,2) =75
Therefore total number of ways =
150+200+75 = 425 ways
Hope this helps! Cheers! :)