Question

A box contains 20 lightbulbs, of which 5 are defective. If 4 lightbulbs are picked from the box randomly, the probability that at most 2 of them are defective is . NextReset

Answer 1

Number of light bulbs in the box = 20

Number of defective light bulbs= 5

So, non defective light bulb= 20-5=15

Probability of an event
`=\frac{\text{total favorable outcome}}{\text{total possible outcome}}`

Now, 4 light bulbs are picked randomly,the probability that at most 2 of them are defective is

`=\frac{_(3)^(15)\textrm{C}* _(1)^(5)\textrm{C}+_(2)^(15)\textrm{C}* _(2)^(5)\textrm{C}+_(4)^(15)\textrm{C}* _(0)^(5)\textrm{C}}{_(4)^(20)\textrm{C}}\\\\= ((15!)/(12!*3!)* 5 +(15!)/(13!*2!)* 10+1365)/((20!)/(16! * 4!))\\\\ =(3325+1365)/(4845)\\\\\=(4690)/(4845)`

=0.9680

Required probability = 0.97 or 97 %

Answer 2

The probability that at most 2 of them are defective is 0.9492 or 94.92%

Explanation:

A box contains 20 light bulbs, out of which 5 are defective.

So the probability of selecting a defective bulb is

`=(5)/(20)=0.25`

Now we have to pick 4 lightbulbs the box randomly, so that at most 2 of them are defective.

As we can have only two possibilities i.e either defective or not defective, so we can treat this as Binomial distribution.

In Binomial distribution,

`P(X=r)=\dbinom{n}{r}p^r(1-p)^(n-r)`

where,

p = probability of success

n = number of trials

At most 2 defective means, either 0 defective or 1 defective or 2 defective bulbs. So the probability that at most 2 defectives from a random draw of 4 bulbs is,

`=P(X=0)+P(X=1)+P(X=2)`

`=\dbinom{4}{0}(0.25)^0(1-0.25)^4+\dbinom{4}{1}(0.25)^1(1-0.25)^3+\dbinom{4}{2}(0.25)^2(1-0.25)^2`

`=\dbinom{4}{0}(0.25)^0(0.75)^4+\dbinom{4}{1}(0.25)^1(0.75)^3+\dbinom{4}{2}(0.25)^2(0.75)^2`

`=0.9492=94.92\%`