Question

. 20% of items produced from a factory are defective. Find the probability that in a sample of 5 chosen at random (i) none is defective (ii) one is defective (iii)p(1<x<4). 

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Answer 1

(i) P(X = 0) = 0.328

(ii) P(X = 1) = 0.409

(iii) P(1 < X < 4) = 0.256

Step-by-step explanation:

We are given that 20% of items produced from a factory are defective.

A sample of 5 items is chosen at random.

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 = 5 items

            r = number of success

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

                 produced from a factory that are defective, i.e; 20%

LET X = Number of items that are defective

So, it means X ~ $Binom(n=5, p=0.20)$

(i) Probability that none is defective is given by = P(X = 0)

                P(X = 0)  =  $\binom{5}{0}\times 0.20^{0} \times (1-0.20)^{5-0}$

                               =  $1 \times 1 \times 0.80^{5}$

                               = 0.328

(ii) Probability that one is defective is given by = P(X = 1)

                P(X = 1)  =  $\binom{5}{1}\times 0.20^{1} \times (1-0.20)^{5-1}$

                               =  $5 \times 0.20 \times 0.80^{4}$

                               = 0.409

(iii) Probability of P(1 < X < 4) is given by = Probability that number of items that are defective is between 1 and 4.

    So, P(1 < X < 4) = P(X = 2) + P(X = 3)

          =  $\binom{5}{2}\times 0.20^{2} \times (1-0.20)^{5-2}+\binom{5}{3}\times 0.20^{3} \times (1-0.20)^{5-3}$

          =  $10 \times 0.20^{2} \times 0.80^{3}+10 \times 0.20^{3} \times 0.80^{2}$

          = 0.2048 + 0.0512 = 0.256