Question

A box contains 10 pens, of which three are defective. If a random sample of five pens is drawn, find the probability that the sample contains exactly two defective pens.

Answer 1

Hint: Use permutations and combinations to get the number of favourable outcomes and the total number of outcomes. For finding the total number of favourable outcomes find the number ways of selecting 2 pens out of the three defective pens and 3 pens out of the 7 working pens.

Complete step by step solution:
Before moving to the question, let us talk about probability.
Probability in simple words is the possibility of an event to occur.
Probability can be mathematically defined as $=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .
Now, let’s move to the solution to the above question.
It is given that there are 10 pens out of which 3 are defective. This means that there are 7 working pens. Now according to the question, the number of favourable outcomes are those outcomes in which 2 pens are selected from the collection of 3 defective pens and 3 pens are selected from the remaining 7 working pens.
So, the number of ways of selecting two pens out of the three defective pens is $^{3}{{C}_{2}}$ and the number of ways of selecting 3 pens out of the seven working pens is $^{7}{{C}_{3}}$ . So, the number of favourable outcomes is $^{7}{{C}_{3}}^{3}{{C}_{2}}$ . Also, we know that the total number of possible outcomes is equal to the number of ways of selecting 5 pens out of the given ten pens, i.e., $^{10}{{C}_{5}}$ .
Now the probability that 2 pens are defective out of the 5 drawn pens is:
$\text{Probability}=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
$\text{Probability}=\dfrac{^{7}{{C}_{3}}^{3}{{C}_{2}}}{^{10}{{C}_{5}}}=\dfrac{\dfrac{7\times 6\times 5\times 3\times 2}{3\times 2\times 2}}{\dfrac{10\times 9\times 8\times 7\times 6\times 5}{5\times 4\times 3\times 2\times 1}}=\dfrac{1}{12}$
Therefore, the answer to the above question is $\dfrac{1}{12}$ .

Note: It is preferred that while solving a question related to probability, always cross-check the possibilities, as there is a high chance you might miss some or have included some extra or repeated outcomes. Formula for selecting r things out of n is $^{n}{{C}_{r}}$ = $\dfrac{n!}{r!.(n-r)!}$ . Also, when a large number of outcomes are to be analysed then permutations and combinations play a very important role as we see in the above solution.