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

Answer:

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

Step-by-step explanation:

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

So the probability of selecting a defective bulb is

$=\dfrac{5}{20}=0.25$

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,

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

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

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

$=0.9492=94.92\%$

Answer 2

Solution:

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

$=\frac{_{3}^{15}\textrm{C}\times _{1}^{5}\textrm{C}+_{2}^{15}\textrm{C}\times _{2}^{5}\textrm{C}+_{4}^{15}\textrm{C}\times _{0}^{5}\textrm{C}}{_{4}^{20}\textrm{C}}\\\\= \frac{\frac{15!}{12!\times3!}\times 5 +\frac{15!}{13!\times2!}\times 10+1365}{\frac{20!}{16! \times 4!}}\\\\ =\frac{3325+1365}{4845}\\\\\=\frac{4690}{4845}$

=0.9680

Required probability = 0.97 or 97 %