Question

A box contains 12 bulbs of which 4 are defective.if 3 are selected at random what is the probaiblty that atleast two bulbs are defective

Answer 1

Total bulbs= 12
Defective bulbs=4
Non defective = 12-4 = 8
Let S be sample space
and E be Event of selecting atleast 2 defective bulbs
Then, n(S)= selecting 3 from 12= 12C3
n(S)= (12*11*10)/(3*2*1)= 2*11*10= 220

Selecting At least 2 defective
Cases:
2 def+ 1 non def (4C2 + 8C1)
=6+8=14
2 def + 1 def = 3 def (4C3)
=4

n(E)= 14 + 4= 18
Required probability =n(E)/n(S)
=18/220
=9/110
Hope it helps
:)
Mark brainliest one if you find this helpful