Question

if 0.8% of the fuses delivered to an arsenal are defective ,use the poisson approximation to determine the probability that four fuses will be defective in a random sample of 400​

Answer 1

Probability that four fuses will be defective in a random sample of 400​ is 0.1781.

Step-by-step explanation:

We are given that 0.8% of the fuses delivered to an arsenal are defective.

A random sample of 400​ fuses is selected.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 400 fuses

            r = number of success = four are defective

           p = probability of success which in our question is % of fuses 

                 delivered to an arsenal that are defective, i.e; 0.8%

LET X = Number of fuses that are defective

So, it means X ~ $Binom(n=400, p=0.008)$

Here, we can't use binomial distribution to find the required probability because the sample size is very large (n > 30). So, we will use Poisson approximation here.

The Probability distribution function for Poisson distribution is given by;

$P(Y=y) = \frac{e^{-\lambda} \times \lambda^{y} }{y!} ;y=0,1,2,3,......$

where, $\lambda$ = rate of defectiveness of fuses

AS, we know that mean of Poisson distribution = $\lambda$

For Poisson approximation;  Mean of Binomial distribution = Mean of Poisson distribution

                       $\lambda$ = $n \times p$

                          = $400 \times 0.008$ = 3.2

So, Y ~ Poisson( $\lambda = 3.2$ )

Now, probability that four fuses will be defective is given by = P(Y = 4)

           P(Y = 4) =  $\frac{e^{-3.2} \times 3.2^{4} }{4!}$

                         = $\frac{e^{-3.2} \times 104.8576}{24}$

                         = 0.1781

Hence, the probability that four fuses will be defective in a random sample of 400​ is 0.1781.