Probability

$2$ bad oranges are accidentally mixed with $10$ good ones. $3$ oranges are drawn at random without replacement from this lot. Obtain the probability distribution for the number of bad oranges.

$3$ oranges can be drawn, without replacement, from $12$ oranges in ${^{12}}{P}_{3} = \dfrac{12!}{9!}$ ways

Let $X$ denote a random variable which is drawing bad oranges when $3$ oranges are selected without replacement.

$X$ can take values $0, \; 1, \; 2$

Now,

$P \left(X = 0\right) = P \left(\text{No bad orange is selected}\right)$

$ = \dfrac{{^{10}}{P}_{3}}{{^{12}}{P}_{3}} = \dfrac{\dfrac{10!}{7!}}{\dfrac{12!}{9!}} = \dfrac{10! \times 9!}{12! \times 7!} = \dfrac{9 \times 8}{12 \times 11} = \dfrac{6}{11}$

$P \left(X = 1\right) = P \left(1\text{ bad orange is selected}\right)$

$ = \dfrac{3 \times {^{2}}{P}_{1} \times {^{10}}{P}_{2}}{{^{12}}{P}_{3}} = \dfrac{3 \times 2 \times\dfrac{10!}{8!}}{\dfrac{12!}{9!}} = \dfrac{3 \times 2 \times 10! \times 9!}{12! \times 8!} = \dfrac{3 \times 2 \times 9}{12 \times 11} = \dfrac{9}{22}$

$P \left(X = 2\right) = P \left(2\text{ bad oranges are selected}\right)$

$ = \dfrac{3 \times 2 \times {^{10}}{P}_{1}}{{^{12}}{P}_{3}} = \dfrac{3 \times 2 \times\dfrac{10!}{9!}}{\dfrac{12!}{9!}} = \dfrac{3 \times 2 \times 10! }{12!} = \dfrac{3 \times 2}{12 \times 11} = \dfrac{1}{22}$

Hence the probability distribution of the random variable $X$ is given by

$X$	$0$	$1$	$2$
$P \left(X\right)$	$\dfrac{6}{11} = \dfrac{12}{22}$	$\dfrac{9}{22}$	$\dfrac{1}{22}$

Probability

Find the probability distribution of the number of sixes in throwing three dice once.

Let $X$ denote a random variable which is the number of sixes obtained when three dice are thrown at once.

$X$ can take values $0, \; 1, \; 2, \; 3$

Probability of getting a $6$ when a dice is thrown $= \dfrac{1}{6}$

Probability of NOT getting a $6$ when a dice is thrown $= \dfrac{5}{6}$

Now,

$P \left(X = 0\right) = P \left(\text{No } 6 \text{ is obtained in any throw}\right) = \left(\dfrac{5}{6}\right)^3 = \dfrac{125}{216}$

$P \left(X = 1\right) = P \left(\text{One } 6 \text{ is obtained in } 3 \text{ throws}\right) = 3 \times \dfrac{1}{6} \times \left(\dfrac{5}{6}\right)^2 = \dfrac{75}{216}$

$P \left(X = 2\right) = P \left(\text{Two } 6's \text{ are obtained in } 3 \text{ throws} \right) = 3 \times \left(\dfrac{1}{6}\right)^2 \times \dfrac{5}{6} = \dfrac{15}{216}$

$P \left(X = 3\right) = P \left(6's \text{ are obtained in all throws}\right) = \left(\dfrac{1}{6}\right)^3 = \dfrac{1}{216}$

Hence the probability distribution of the random variable $X$ is given by

$X$	$0$	$1$	$2$	$3$
$P \left(X\right)$	$\dfrac{125}{216}$	$\dfrac{75}{216}$	$\dfrac{15}{216}$	$\dfrac{1}{216}$

Probability

Find each probability if three coins are tossed:

1. P($3$ heads or $3$ tails)
2. P(exactly $2$ tails)

When a coin is tossed,

probability of getting head $= p = \dfrac{1}{2}$

probability of getting tail $= q = 1- p = \dfrac{1}{2}$

Number of coins $= 3$

1. P($3$ heads or $3$ tails)
   
   $= P \left(3 \text{ heads}\right) + P \left(3 \text{ tails}\right)$
   
   $= {^{3}}{C}_{3} \times \left(\dfrac{1}{2}\right)^3 \times \left(\dfrac{1}{2}\right)^0 + {^{3}}{C}_{0} \times \left(\dfrac{1}{2}\right)^0 \times \left(\dfrac{1}{2}\right)^3$
   
   $= 2 \times \left(\dfrac{1}{2}\right)^3$
   
   $= \dfrac{1}{4}$
   
   
2. P(exactly $2$ tails)
   
   $= P \left(\text{exactly }1 \text{ head}\right)$
   
   $= {^{3}}{C}_{1} \times \left(\dfrac{1}{2}\right)^1 \times \left(\dfrac{1}{2}\right)^2$
   
   $= 3 \times \dfrac{1}{8} = \dfrac{3}{8}$

Probability

A lady carries lipstick tubes in a bag in her purse. The probability of pulling out the color she wants is $\dfrac{2}{3}$. Suppose she uses her lipstick $4$ times a day. Find the probability

1. P(never the correct color)
2. P(no more than $3$ times correct)

Let $p$ be the probability of selecting the right color lipstick.

Let $p$ be the probability of not selecting the right color lipstick.

Given: $\;$ $p = \dfrac{2}{3}$ $\;$ $\therefore \;$ $q = 1 - p = 1 - \dfrac{2}{3} = \dfrac{1}{3}$

Total number of times lipstick is selected $= 4$

1. P(never selecting the correct color) $= q^4 = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}$



2. P(no more than $3$ times correct)
   
   $\begin{aligned} P\left(\text{no more than 3 times correct}\right) & = 1 - P \left(\text{all times correct}\right)\\\\ & = 1 - \left(p\right)^4 \\\\ & = 1 - \left(\dfrac{2}{3}\right)^4 \\\\ & = 1 - \dfrac{16}{81} = \dfrac{65}{81} \end{aligned}$

Probability

A weather reporter for a channel is forecasting a $30\%$ chance of rain for today and the next four days. Find the probability of having rain on exactly one day.

Let $p$ be the probability of raining on a given day.

Let $q$ be the probability of not raining on a given day.

Given: $\;$ $p = 30\% = 0.3$; $\;$ $q = 1- p = 1 - 0.3 = 0.7$

Total number of days $= 4$

$\begin{aligned} P \left(\text{It rains exactly one day}\right) & = {^{4}}{C}_{1} \times \left(p\right)^{1} \times \left(q\right)^{4 - 1} \\\\ & = \dfrac{4!}{1! \times 3!} \times 0.3 \times \left(0.7\right)^{3} \\\\ & = 4 \times 0.3 \times \left(0.7\right)^{3} \\\\ & = 0.4116 \approx 41.2\% \end{aligned}$

Probability

Three urns are given each containing red and white chips as given below:

Urn $I$ : $6$ red $4$ white ; Urn $II$ : $3$ red $5$ white ; Urn $III$ : $4$ red $6$ white

An urn is chosen at random and a chip is drawn from the urn.

1. Find the probability that it is white.
2. If the chip is white find the probability that it is from urn $II$.

Let $A_1$ be the event of selecting urn $I$.

Let $A_2$ be the event of selecting urn $II$.

Let $A_3$ be the event of selecting urn $III$.

Let $B$ be the event of selecting a white chip.

1. To find: $\;$ P(of selecting a white chip) $= P \left(B\right)$
   
   Events $A_1$, $A_2$ and $A_3$ are mutually exclusive and exhaustive events.
   
   $\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right) + P \left(A_3\right) \times P \left(B | A_3\right)$ $\;\;\; \cdots \; (1)$
   
   Probability of selecting urn $I$ $= P \left(A_1\right) = \dfrac{1}{3}$
   
   Probability of selecting urn $II$ $= P \left(A_2\right) = \dfrac{1}{3}$
   
   Probability of selecting urn $III$ $= P \left(A_3\right) = \dfrac{1}{3}$
   
   P(selecting a white chip when urn $I$ is selected) $= P \left(B | A_1\right)$
   
   $P \left(B | A_1\right) = \dfrac{{^{4}}{C}_{1}}{{^{10}}{C}_{1}} = \dfrac{4}{10} = \dfrac{2}{5}$
   
   P(selecting a white chip when urn $II$ is selected) $= P \left(B | A_2\right)$
   
   $P \left(B | A_2\right) = \dfrac{{^{5}}{C}_{1}}{{^{8}}{C}_{1}} = \dfrac{5}{8}$
   
   P(selecting a white chip when urn $III$ is selected) $= P \left(B | A_3\right)$
   
   $P \left(B | A_3\right) = \dfrac{{^{6}}{C}_{1}}{{^{10}}{C}_{1}} = \dfrac{6}{10} = \dfrac{3}{5}$
   
   Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(A_3\right)$, $P \left(B | A_1\right)$, $P \left(B | A_2\right)$ and $P \left(B | A_3\right)$ in equation $(1)$ we have,
   
   $P \left(B\right) = \left(\dfrac{1}{3} \times \dfrac{2}{5}\right) + \left(\dfrac{1}{3} \times \dfrac{5}{8}\right) + \left(\dfrac{1}{3} \times \dfrac{3}{5}\right) = \dfrac{13}{24}$
   
   
2. To find: $\;$ P(of selecting urn $II$ when a white chip is selected) $= P \left(A_2 | B\right)$
   
   Since events $A_1$, $A_2$ and $A_3$ are mutually exclusive and exhaustive events, therefore we have by Bayes' theorem,
   
   $\begin{aligned} P \left(A_2 | B\right) & = \dfrac{P \left(A_2\right) \times P \left(B | A_2\right)}{P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right) + P \left(A_3\right) \times P \left(B | A_3\right)} \\\\ & = \dfrac{\dfrac{1}{3} \times \dfrac{5}{8}}{\dfrac{1}{3} \times \dfrac{2}{5} + \dfrac{1}{3} \times \dfrac{5}{8} + \dfrac{1}{3} \times \dfrac{3}{5}} = \dfrac{5/8}{13/8} = \dfrac{5}{13} \end{aligned}$

Probability

A factory has two Machines - $I$ and $II$. Machine-$I$ produces $25\%$ of items and Machine-$II$ produces $75\%$ of the items of the total output. Further $3\%$ of the items produced by Machine-$I$ are defective whereas $4\%$ produced by Machine-$II$ are defective. If an item is drawn at random, what is the probability that it is defective?

Let $A_1$ be the event that the items are produced by machine $I$

Let $A_2$ be the event that the items are produced by machine $II$

Let $B$ be the event of selecting a defective item

To find: $\;$ P(of selecting a defective item) $= P \left(B\right)$

Events $A_1$ and $A_2$ are mutually exclusive and exhaustive events.

$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right)$ $\;\;\; \cdots \; (1)$

Probability that the items are produced by machine-$I$ $= P \left(A_1\right) = \dfrac{25}{100}$

Probability that the items are produced by machine-$II$ $= P \left(A_2\right) = \dfrac{75}{100}$

P(getting a defective item when the item is selected from machine-$I$)

$= P \left(B | A_1\right) = \dfrac{3}{100}$

P(getting a defective item when the item is selected from machine-$II$)

$= P \left(B | A_2\right) = \dfrac{4}{100}$

Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(B | A_1\right)$ and $P \left(B | A_2\right)$ in equation $(1)$ we have,

$P \left(B\right) = \left(\dfrac{25}{100} \times \dfrac{3}{100}\right) + \left(\dfrac{75}{100} \times \dfrac{4}{100}\right) = \dfrac{3}{80}$