# Two bad oranges are accidentally mixed with ten good ones. Three oranges are drawn at random without replacement from this lot. Obtain the probability distribution for the number of bad oranges. and find the expected value of the distribution.

## 1 Answer

Toolbox:
• If $S$ is a sample space with a probability measure and $X$ is a real valued function defined over the elements of $S$, then $X$ is called a random variable.
• Types of Random variables :
• (1) Discrete Random variable (2) Continuous Random variable
• Discrete Random Variable :If a random variable takes only a finite or a countable number of values, it is called a discrete random variable.
• Continuous Random Variable :A Random Variable X is said to be continuous if it can take all possible values between certain given limits. i.e., X is said to be continuous if its values cannot be put in 1 − 1 correspondence with N, the set of Natural numbers.
• The probability mass function (a discrete probability function) p(x) is a function that satisfies the following properties :
• (1) $P(X=x)=p(x)=p_x$
• (2) $P(x)\geq 0$ for all real $x$
• (3) $\sum p_i=1$
Step 1:
Total number of oranges=12
Number of bad oranges=2
Let $X$ be the discrete RV denoting the number of bad oranges when 3 oranges are drawn from the lot of 12
$\Rightarrow\:X$ takes the values $0,1,2$
Step 2:
$P(X=0)$=probability of no bad oranges
$\qquad\quad\;\;=\large\frac{10C_3}{12C_3}$
$\qquad\quad\;\;=\large\frac{10\times 9\times 8/1\times 2\times 3}{12\times 11\times 10/1\times 2\times 3}$
$\qquad\quad\;\;=\large\frac{36}{66}$
$\qquad\quad\;\;=\large\frac{6}{11}$
$P(X=1)$=probability of drawing 1 bad orange
$\qquad\quad\;\;=\large\frac{2\times 10}{12\times 11/1\times 2}$
$\qquad\quad\;\;=\large\frac{20}{66}$
$\qquad\quad\;\;=\large\frac{10}{33}$
$P(X=2)$=probability of drawing 2 bad oranges
$\qquad\quad\;\;=\large\frac{2C_2\times 10C_1}{12C_2}$
$\qquad\quad\;\;=\large\frac{1\times 10}{12\times 4/1\times 2}$
$\qquad\quad\;\;=\large\frac{10}{66}=\frac{5}{33}$
Step 3:
The probability distribution is given by
Step 4
Expected value of $X$ is Mean $=\sum_{x=0}^{2}\:x.P(x)$
Mean$=(0\times\large\frac{6}{11})$$+(1\times\large\frac{10}{33})$$+(2\times\large\frac{5}{33})$
$=\large\frac{20}{33}$
answered Sep 16, 2013
edited Feb 11, 2014

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