Question

From a collection of 12 bulbs of which 6 are defective 3 bulbs are chosen at random for three sockets in a room.Find the probability that the room is lighted if one bulb is sufficient to light the room.; From a collection of 12 bulbs of which 6 are defective 3 bulbs are chosen at random for three sockets in a room.Find the probability that the room is lighted if one bulb is sufficient to light the room.

Answer 1

Answer:

7/8

Step-by-step explanation:

Here we are going to use Bernoulli's probability distribution

Total no. of bulbs = 12

No. of defective bulbs = 6

∴ No. of good bulbs = 12 - 6 = 6

Let 'selecting a good bulb' be considered as a success

∴ P(success) = p = No. of good bulbs/ Total no. of bulbs

      =   6/12   = 1/2

No. of selections(trials), n = 3

P(failure) = q = 1 - 1/2 = 1/2

By Bernoulli's distribution formula,

P ( x successes in n trials) = ⁿCₓ (p)ˣ (q)ⁿ⁻ˣ

Since one bulb is enough to light the room, we need one success or more

P( ≥1 successes) = 1 - P( 0 successes)

= 1 - ³C₀(1/2)⁰(1/2)³

= 1 - (1)(1)(1/8)

= 1 - 1/8

= 7/8

∴ Probability that room will be lighted is 7/8

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