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Question: \[12\] defective pens are accidentally mixed with \[132\] good ones. It is not possible to just look...

1212 defective pens are accidentally mixed with 132132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen is taken out is a good one.

Explanation

Solution

We have to find the probability to take out a good pen. The given in the question is the number of defective pens and the number of good pens. Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
Probability = (The number of wanted outcomes)(The number of possible outcomes){\text{Probability = }}\dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}

Complete step-by-step answer:
It is given that 1212 defective pens are accidentally mixed with 132132 good ones.
It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot .We need to determine the probability that the pen is taken out is a good one.
Total number of pens in the list =12+132=144 = 12 + 132 = 144
So, the possible outcome of choosing one pen from the total pens, is 144C1^{144}{C_1} .
Thus, the possible outcome of choosing one pen and it is a good one, is 132C1^{132}{C_1}.
The probability that the pen is taken out is a good one,
⇒(The number of wanted outcomes)(The number of possible outcomes)\Rightarrow \dfrac{{\left( {{\text{The number of wanted outcomes}}} \right)}}{{\left( {{\text{The number of possible outcomes}}} \right)}}
⇒132C1144C1\Rightarrow \dfrac{{^{132}{C_1}}}{{^{144}{C_1}}}
Using the combination formula,
⇒132!1!131!144!1!143!\Rightarrow \dfrac{{\dfrac{{132!}}{{1!131!}}}}{{\dfrac{{144!}}{{1!143!}}}}
Simplifying the terms in above equation,
⇒132×131!131!144×143!143!\Rightarrow \dfrac{{\dfrac{{132 \times 131!}}{{131!}}}}{{\dfrac{{144 \times 143!}}{{143!}}}}
⇒132144\Rightarrow \dfrac{{132}}{{144}}
So we finally get the probability that the pen that is taken out is a good one is 132144\dfrac{{132}}{{144}}.

Note: Students make mistakes by considering the total number of pens as 132. You have to add the defective pens with good ones to get a total number of pens.
A combination is a grouping or subset of items.
For a combination,
⇒C(n,r)=nCr=n!(n−r)!r!\Rightarrow C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
Where, factorial n is denoted by n!n! and defined by
n!=n(n−1)(n−2)(n−4).......2.1n! = n(n - 1)(n - 2)(n - 4).......2.1