Question

A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs?

Answer 1

Hint: We start solving the problem by assuming the random variable to represent the total number of defective bulbs present in the drawn 3 bulbs. We then write the sample space for the assumed random variable. We then find the probability for all the elements present in the sample space using the fact that the probability is the ratio of the favorable cases to the total number of cases. We list the elements of sample space along with their probabilities to get the required answer.

Complete step by step answer:
According to the problem, we are given that a box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. We need to find the probability distribution of the number of defective bulbs.
Let us assume X be the random variable representing the total number of defective bulbs present in the drawn 3 bulbs.
Now, let us write the possibilities for X. Since we have 5 defective bulbs in the box, we get the possibilities of getting zero, one, two, and three defective bulbs.
Now, let us find the probability to draw three non-defective bulbs.
 $ \Rightarrow P\left( X=0 \right)=\dfrac{\text{no}\text{. of ways of drawing three bulbs from 8 non-defective bulbs}}{\text{no}\text{. of ways of drawing three bulbs from 13 bulbs in a box}} $ .
We know that the no. of ways of drawing ‘r’ objects out of ‘n’ objects $ \left( n\ge r \right) $ is $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ ways, where $ n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 3\times 2\times 1 $ .
 $ \Rightarrow P\left( X=0 \right)=\dfrac{{}^{8}{{C}_{3}}}{{}^{13}{{C}_{3}}} $ .
 $ \Rightarrow P\left( X=0 \right)=\dfrac{\dfrac{8!}{3!5!}}{\dfrac{13!}{3!10!}} $ .
 $ \Rightarrow P\left( X=0 \right)=\dfrac{8\times 7\times 6}{13\times 12\times 11} $ .
 $ \Rightarrow P\left( X=0 \right)=\dfrac{28}{143} $ ---(1).
Now, let us find the probability to draw two non-defective bulbs and one defective bulb.
 $ \Rightarrow P\left( X=1 \right)=\dfrac{\text{no}\text{. of ways of drawing two bulbs from 8 non-defective bulbs and one defective bulb from 5 defective bulbs}}{\text{no}\text{. of ways of drawing three bulbs from 13 bulbs in a box}} $ .
 $ \Rightarrow P\left( X=1 \right)=\dfrac{{}^{8}{{C}_{2}}\times {}^{5}{{C}_{1}}}{{}^{13}{{C}_{3}}} $ .
 $ \Rightarrow P\left( X=1 \right)=\dfrac{\left( \dfrac{8!}{2!6!} \right)\times \left( \dfrac{5!}{1!4!} \right)}{\dfrac{13!}{3!10!}} $ .
 $ \Rightarrow P\left( X=1 \right)=\dfrac{\left( \dfrac{8\times 7}{2\times 1} \right)\times \left( 5 \right)}{\dfrac{13\times 12\times 11}{3\times 2\times 1}} $ .
 $ \Rightarrow P\left( X=1 \right)=\dfrac{\left( 28 \right)\times \left( 5 \right)}{13\times 2\times 11} $ .
 $ \Rightarrow P\left( X=1 \right)=\dfrac{70}{143} $ ---(2).
Now, let us find the probability to draw one non-defective bulbs and two defective bulbs.
 $ \Rightarrow P\left( X=2 \right)=\dfrac{\text{no}\text{. of ways of drawing one bulbs from 8 non-defective bulbs and two defective bulb from 5 defective bulbs}}{\text{no}\text{. of ways of drawing three bulbs from 13 bulbs in a box}} $ .
 $ \Rightarrow P\left( X=2 \right)=\dfrac{{}^{8}{{C}_{1}}\times {}^{5}{{C}_{2}}}{{}^{13}{{C}_{3}}} $ .
 $ \Rightarrow P\left( X=2 \right)=\dfrac{\left( \dfrac{8!}{1!7!} \right)\times \left( \dfrac{5!}{2!3!} \right)}{\dfrac{13!}{3!10!}} $ .
\[\Rightarrow P\left( X=2 \right)=\dfrac{\left( 8 \right)\times \left( \dfrac{5\times 4}{2\times 1} \right)}{\dfrac{13\times 12\times 11}{3\times 2\times 1}}\].
 $ \Rightarrow P\left( X=2 \right)=\dfrac{\left( 8 \right)\times \left( 10 \right)}{13\times 2\times 11} $ .
 $ \Rightarrow P\left( X=2 \right)=\dfrac{40}{143} $ ---(3).
Now, let us find the probability to draw three defective bulbs.
 $ \Rightarrow P\left( X=3 \right)=\dfrac{\text{no}\text{. of ways of drawing three defective bulb from 5 defective bulbs}}{\text{no}\text{. of ways of drawing three bulbs from 13 bulbs in a box}} $ .
 $ \Rightarrow P\left( X=3 \right)=\dfrac{{}^{5}{{C}_{3}}}{{}^{13}{{C}_{3}}} $ .
 $ \Rightarrow P\left( X=3 \right)=\dfrac{\dfrac{5!}{3!2!}}{\dfrac{13!}{3!10!}} $ .
\[\Rightarrow P\left( X=3 \right)=\dfrac{5\times 4\times 3}{13\times 12\times 11}\].
 $ \Rightarrow P\left( X=3 \right)=\dfrac{5}{143} $ ---(4).
Now, let us write the probability distribution using the values obtained from equations (1), (2), (3) and (4) which is as shown below:

X	$ P\left( X \right) $
0	$ \dfrac{28}{143} $
1	$ \dfrac{70}{143} $
2	$ \dfrac{40}{143} $
3	$ \dfrac{5}{143} $

Note:
Whenever we get this type of problem, we first try to write the possibilities present in the sample space. We should keep in mind that the sum of all the probabilities present in the probability distribution is equal to 1. We have considered the discrete random variable in this variable as we can see that assuming the bulbs with fractions is not meaningful. Similarly, we can expect problems t