logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Probability
0 votes

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

$\begin{array}{1 1}\begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{256}{625} &\large \frac{256}{625} &\large\frac{96}{625} &\large\frac{16}{625}&\large\frac{1}{625}\end{matrix} \\ \begin{matrix} \textbf{X} & 0 &1 &2&3\\ \textbf{P(X)} &\large \frac{256}{625} &\large \frac{256}{625} &\large\frac{96}{625} &\large\frac{16}{625}\end{matrix} \\ \begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{1}{625} &\large \frac{256}{625} &\large\frac{96}{625} &\large\frac{16}{625}&\large\frac{1}{625}\end{matrix} \\\begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{16}{625} &\large \frac{96}{625} &\large\frac{256}{625} &\large\frac{16}{625}&\large\frac{1}{625}\end{matrix}\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • The number of ways of selecting n items out of a sample of m can be determined as $^{m}C_n$
We have a lot of 30 bulbs, 6 are defective $\rightarrow$ 30 - 6 = 24 are non-defective.
P (drawing a defective bulb) = $\large\frac{6}{30} = \frac{1}{5}$.
P (drawing a non-defective bulb) = 1 - P (drawing a defective bulb) = 1 - $\large\frac{1}{5} = \frac{4}{5}$
Note: 4 bulbs can be drawn from a lot of 30 bulbs in $\large {^{30}c_4}$ ways.
Let X be the random variable of drawing 4 defective bulbs from the lot.
P (X = 0) = P (no defective bulb in the sample) = $\large\frac{^{24}C_4}{^{30}C_4} = \frac{256}{625}$
Note: $^{24}C_4$ is the number of ways of drawing 4 bulbs from the non-defective lot.
P (X = 1) = P (one defective bulb from the lot) = P(one defective, 3 non-defective bulbs) =\(\large\frac{^{6}C_1\;^{24}C_3}{^{30}C_4}\) = $\large\frac{256}{625}$
P (X = 2) = P (two defective bulbs from the lot) = P(2 defective, 2 non-defective bulbs) =\(\large\frac{^{6}C_2\;^{24}C_2}{^{30}C_4}\) = $\large\frac{256}{625}$
P (X = 3) = P (three defective bulbs from the lot) = P(3 defective, 1 non-defective bulb) =\(\large\frac{^{6}C_3\;^{24}C_1}{^{30}C_4}\) = $\large\frac{16}{625}$
P (X = 4) = P (four defective bulbs from the lot) =\(\large\frac{^{6}C_4\;^{24}C_0}{^{30}C_4}\) = $\large\frac{1}{625}$
Therefore the required probability distribution is as follows: $\begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{256}{625} &\large \frac{256}{625} &\large\frac{96}{625} &\large\frac{16}{625}&\large\frac{1}{625}\end{matrix}$
answered Jun 20, 2013 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...