Question

A packet of 10 electronic components is known to include 2 defectives. If a sample of 4 components is selected at random from the packet, what is the probability that the sample does not contain more than 1 defective?

Answer 1

Hint: We use the concept of probability to find the probability of cases where there is only 1 defective piece and cases where there is no defective piece. Calculate the total number of pieces that are not defective. Case having no defective piece will have all four components from a non-defective set and a case with one defective component will have one from each defective and non-defective set. Use the method of combinations to find a number of ways to choose components.
* Probability of an event is given by dividing the number of favorable outcomes by total number of outcomes.
* Formula of combination is given by $${}^n{C}_r={\displaystyle \frac{n!}{(n-r)!r!}}$$, where factorial is expanded by the formula $$n!=n\times (n-1)!=n\times (n-1)\times (n-2)!....=n\times (n-1)\times (n-2)....3\times 2\times 1$$

Complete step-by-step answer:
We are given a total number of components as 10.
Number of defective components $$=2$$
So, number of non-defective components$$=10-2$$
$$\Rightarrow$$Number of non-defective components$$=8$$
We select a sample of 4 components at random.
We have to find the probability that a sample of 4 components has not more than 1 defective component.
We take two cases, one having no defective components and other having 1 defective component. Use formula of combination to find suitable values i.e.$${}^n{C}_r={\displaystyle \frac{n!}{(n-r)!r!}}$$...........… (1)
CASE 1: Zero defective
Number of ways to choose 4 components from 8 components $${=}^8{C}_4$$
Number of ways to choose 4 components from total 10 components $${=}^{10}{C}_4$$
Probability of choosing zero defective components is given by dividing the number of ways to choose 4 components from 8 non- defective components by the number of ways to choose 4 components from total given components. Let the probability of zero defective component be P(0).
$$\Rightarrow$$Probability $$={\displaystyle \frac{{}^8{C}_4}{{}^{10}{C}_4}}$$
From equation (1), expand the terms in numerator and denominator
$$\Rightarrow$$Probability $$={\displaystyle \frac{{\displaystyle \frac{8!}{4!4!}}}{{\displaystyle \frac{10!}{6!4!}}}}$$
$$\Rightarrow$$Probability $$={\displaystyle \frac{8!\times 6!\times 4!}{10!\times 4!\times 4!}}$$
Write factorial in simpler terms using$$n!=n\times (n-1)!=n\times (n-1)\times (n-2)!....=n\times (n-1)\times (n-2)....3\times 2\times 1$$
$$\Rightarrow$$Probability $$={\displaystyle \frac{8!\times 6\times 5\times 4!\times 4!}{10\times 9\times 8!\times 4!\times 4!}}$$
Cancel same terms from numerator and denominator
$$\Rightarrow$$Probability $$={\displaystyle \frac{6\times 5}{10\times 9}}$$
$$\Rightarrow$$Probability $$={\displaystyle \frac{1}{3}}$$
$$\therefore P(0)={\displaystyle \frac{1}{3}}$$............… (2)
CASE 2: One defective
We choose 3 components from 8 non-defective set and 1 component from 2 defective set
Number of ways to choose 3 components from 8 components $${=}^8{C}_3$$
Number of ways to choose 1 component from 2 components $${=}^2{C}_1$$
Probability of choosing one defective component is given by dividing the number of ways to choose 4 components from 10 components having 1 defective component by the number of ways to choose 4 components from the total given components. Let the probability of zero defective component be P(1).
$$\Rightarrow$$Probability $$={\displaystyle \frac{{}^8{C}_3{\times }^2{C}_1}{{}^{10}{C}_4}}$$
From equation (1), expand the terms in numerator and denominator
$$\Rightarrow$$Probability $$={\displaystyle \frac{{\displaystyle \frac{8!}{5!3!}}\times {\displaystyle \frac{2!}{1!1!}}}{{\displaystyle \frac{10!}{6!4!}}}}$$
$$\Rightarrow$$Probability $$={\displaystyle \frac{8!\times 6!\times 4!\times 2!}{10!\times 5!\times 3!}}$$
Write factorial in simpler terms using$$n!=n\times (n-1)!=n\times (n-1)\times (n-2)!....=n\times (n-1)\times (n-2)....3\times 2\times 1$$
$$\Rightarrow$$Probability $$={\displaystyle \frac{8!\times 6\times 5!\times 4\times 3!\times 2}{10\times 9\times 8!\times 5!\times 3!}}$$
Cancel same terms from numerator and denominator
$$\Rightarrow$$Probability $$={\displaystyle \frac{2\times 4}{5\times 3}}$$
$$\Rightarrow$$Probability $$={\displaystyle \frac{8}{15}}$$
$$\therefore P(1)={\displaystyle \frac{8}{15}}$$............… (3)
Now we find sum of probabilities by adding equations (2) and (3)
Probability that sample does not contain more than 1 defective component is:
$$\Rightarrow P=P(0)+P(1)$$
Substitute values from equations (2) and (3)
$$\Rightarrow P={\displaystyle \frac{1}{3}}+{\displaystyle \frac{8}{15}}$$
Take LCM
$$\Rightarrow P={\displaystyle \frac{5+8}{15}}$$
$$\Rightarrow P={\displaystyle \frac{13}{15}}$$

$$\therefore$$Probability that a sample does not contain more than 1 defective component is $${\displaystyle \frac{13}{15}}$$.

Note: Probability of any event always lies between 0 to 1. The sum of probability of an event to occur and the probability of the event not occurring will always be equal to 1. Many students make mistakes in forming the second case as they tend to choose 2 from two defective sets, this is wrong as we have to take 1 defective piece which implies other pieces should be non-defective, and non-defective as to be chosen from the sample obtained after deducting defective components from total.