Question

The probability that a given computer chip will fail is 0.02. find the probability that of 5 delivered chips, exactly 2 will fail.

Answer 1

Since the chip either passes or fails, this problem is one in binomial probability, with p(fail)=0.02 and n=5.
                 
If your calculator features probability functions such as binompdf(n, p), use it.
Mine is a TI-83 Plus.

The probability that exactly two chips will fail is

binompdf(5,0.02,2) = 0.0038.  This is quite a small probability.

Answer 2

Probability of an event is the measure of its chance of occurrence. The probability that of 5 delivered chips, exactly 2 will fail is 0.003764768 or approx 0.0004

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

For the given case, as failing or working of chips usually doesn't depend on other chips' working condition, so they're independent(assuming). There are two cases for each chip's working(random experiment) that it is either working or failed.

If we take Success of the considered binomial experiment(working status of chip) = Chip working fine

Then, as per given data, we have

P(chip working fine ) = 1  - P(chip not working) = 1 - 0.02 = 0.98 = Probability of success in Bernoulli experiment (chip's working status).

There are n = 5 chips, p = probability of success = 0.98, q = 1-p = 0.02

If we take X = number of chips of of 5 chips working, then:

$X \sim B(n = 5, p = 0.98)$

Then, the needed probability is P(Exactly two chips fail) = P(Exactly 5-2 = 3 chips work) = P(X  = 3)

And thus,  the needed probability is calculated as:

$P(X = 3) = \: ^5C_3(0.98)^3(0.02)^2 = \dfrac{5.4.3}{3.2.1} \times 0.941192 \times 0.0004 = 0.003764768$

(using the probability mass function of binomial distribution).

Thus, the probability that of 5 delivered chips, exactly 2 will fail is 0.003764768 or approx 0.0004

Learn more about binomial distribution here:

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