Question

here are 3 lamps in a room. A bag contains 10 bulbs of which 4 are fused. A person draws 3 bulbs at random and arranges in the three lamps. Then the probability that the room gets lighted is

• A. 2/30
• B. 1/30
• C. 29/30
• D. 27/30

Answer 1

Answer: (C)

29/30

Answer 2

n(S) = ${ } ^ { 10 } \mathrm { C } _ { 3 } = \frac { 10 \times 9 \times 8 } { 3 \times 2 }$ = 10 x 3 x 4 = 120.

E = selecting atleast, 1 good bulb

$\overline { \mathrm { E } }$ = selecting all fused bulbs.

= 4

$P ( \bar { E } ) = \frac { 4 } { 120 } = \frac { 1 } { 30 }$

P(5) = 1 - $\frac { 1 } { 30 } = \frac { 29 } { 30 }$