Question

A box contains 9bulbs out of which 4 bulbs are defective .If four bulbs are chosen random.find the probability that exactly three bulbs are good ​

Answer 1

Solution

Since, box contains 9bulbs out of which 4 bulbs are defective.

It means, box contain 4 defective bulb and 5 non - defective bulb.

Now,

• Number of ways of selecting 4 bulbs from 9 bulbs  is given by

$\rm : \implies \:\:^9 C_4 = \dfrac{9!}{4! \times 5!} = \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

Now,

We have to pick 4 bulbs in such a way that it will exactly have 3 non - defective bulbs.

• So number of ways of selecting 3 non defective bulbs out of 5 and 1 defective bulb out of 4 is given by

$\rm : \implies \:\:^5 C_3 \times \:^4 C_1$

$\rm : \implies \:\dfrac{5!}{3! \times 2!} \times \dfrac{4!}{3! \times 1!}$

$\rm : \implies \:\dfrac{5 \times 4}{2 \times 1} \times \dfrac{4}{1}$

$\rm : \implies \:40$

Hence,

required probability that exactly three bulbs are good out of 4 are taken out randomly is

$\rm : \implies \:\dfrac{40}{126}$

$\rm : \implies \:\dfrac{20}{63}$