Question

:: If 10% of the screws produced by an automatic machine are
defective, find the probability that out of 20 screws selected
at random, there are (i) exactly 2 defective (ii) atmost 3
defective (iii) at least 2 defectives and (iv) between 1 and 3
defectives (inclusive)​

Answer 1

Answer:

,

Step-by-step explanation:

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Answer 2

Answer:

Probability of 20 screws selected randomly :

(i) exactly 2 defective = 0.2851

(ii) at most 3 defective = 0.8670

(iii) at least 2 defective = 0.6082

(iv) between 1 and 3 defectives (inclusive)​ = 0.7454

Step-by-step explanation:

Given, the probability of defective screws produced  = 10%

$p=\frac{10}{100}=0.1$

The probability of undefective screws, $q = 1-p=1-0.1=0.9$

The number of screws selected randomly, n = 20

From the binomial distribution:

$P(r)=^{n}C_{r}p^{r}q^{n-r}$

(i)  Probability of exactly 2 defective screws:

    $P(X=2)=\; ^{20}C_{2}(0.1)^{2}(0.9)^{20-2}$

    $P(X=2)=\; ^{20}C_{2}(0.1)^{2}(0.9)^{18}$

    $P(X=2)=\; \frac{20!}{(18!)*(2!)} *(0.01)(0.15)$

    $P(X=2)=\; 0.2851$

(ii)  Probability of at most 3 defective screws:

$=P(X=0)+ P(X=1)+P(X=2)+P(X=3)$

$=\; ^{20}C_{0}(0.1)^{0}(0.9)^{20}\; +\; ^{20}C_{1}(0.1)^{1}(0.9)^{19}\; +\; ^{20}C_{2}(0.1)^{2}(0.9)^{20-2}\; + \; ^{20}C_{3}(0.1)^{3}(0.9)^{17}$

$=0.1216+0.2702+0.2851+0.1901$

$=0.8670$

(iii) Probability of at least 2 defective screws:

$P(X\geq 2)=1-[P(X=0)+P(X=1)]$

$P(X\geq 2)=1-(0.1216+0.2702)$

$P(X\geq 2)=1-0.3918\\P(X\geq 2)=0.6082$

(iv) Probability of between 1 and 3 defective screws  (inclusive)​:

$P(1\leq x\leq 3)= P(X=1)\; +\; P(X=2)\; + \;P(X=3)\;$

$P(1\leq x\leq 3)=\; 0.2702\; +\; 0.2851\; +\; 0.1901$

$P(1\leq x\leq 3)=0.7454$