Question

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 .

Answer 1

938/969

plato answer

Explanation:

Answer 2

Probability that at most 2 of them are defective = 94.92%

Explanation:

There are 20 lightbulbs and 5 are defective.

Probability of picking a defective ball = 5/20 = 0.25

We need to randomly pick 4 lightbulbs so that at most 2 are defective

We can treat this as a binomial distribution since there are two possibilities: either a defective or a non-defective lightbulb

`P(X = r) = \left(\begin{array}{ccc}n\\r\end{array}\right)p^(r)(1-p)^(n-r)`

where p = probability of success,

n = number of trials

Since at most 2 lightbulbs are defective, we can either have 0 defective or 1 defective or 2 defective bulbs.

The probability that at most 2 defective lightbulbs from a random draw of 4 lightbulbs is,

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

`= \left(\begin{array}{ccc}4\\0\end{array}\right)0.25^(0)(1-0.25)^(4-0) + \left(\begin{array}{ccc}4\\1\end{array}\right)0.25^(1)(1-0.25)^(4-1) + \left(\begin{array}{ccc}4\\2\end{array}\right)0.25^(2)(1-0.25)^(4-2)\\\\\\= \left(\begin{array}{ccc}4\\0\end{array}\right)0.25^(0)(0.75)^(4) + \left(\begin{array}{ccc}4\\1\end{array}\right)0.25^(1)(0.75)^(3) + \left(\begin{array}{ccc}4\\2\end{array}\right)0.25^(2)(0.75)^(2)\\`

= 0.9492 = 94.92%