Question

2. Four of the light bulbs in a box of ten bulbs are burnt out or otherwise defective. If two bulbs are selected at random without replacement and tested, (i) what is the probability that exactly one defective bulb is found? (ii) What is the probability that exactly two defective bulbs are found?​

Answer 1

Answer:

Here probability of bulb is defective is 4/10 = 0.40 This is based on Binomial distribution P(X= x) = nCx p^x (1-p)^(n-x) here p = 0.40, n = 2 (i) Exactly one is defective means x = 1...

Answer 2

The probability of selecting exactly one defective bulb is $4/9.$

(i) The probability that exactly one defective bulb is found:

Let's assume that the defective bulbs are labeled D and the functional bulbs are labeled F.

There are two ways that exactly one defective bulb can be selected: either DF or FD.

The number of ways to select DF is 4 (number of defective bulbs) × 6 (number of functional bulbs) = 24.

The number of ways to select two bulbs without replacement is 10 (total number of bulbs) choose 2, which is

$10!/(2!(10-2)!) = 45.$

So the probability of selecting exactly one defective bulb is $24/45 = 4/9.$

(ii) The probability that exactly two defective bulbs are found:

The number of ways to select two defective bulbs is 4 choose 2,

which is $4!/(2!(4-2)!) = 6.$

So the probability of selecting exactly two defective bulbs is$6/45 = 2/15.$

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