Probability Distributions
Probability  Basic Concepts
Learning Objectives:

Understand basic terminology related to probability such as sample space, event, complement, boolean combination and dependency of multiple events.
An event is a set of outcomes of an experiment. Sample space is defined as set of all possible outcomes.
Probability of an event is defined as ratio of number of favourable outcomes to the number of outcomes in the sample space.
Probabilities can be expressed as proportions that range from 0 to 1, and they can also be expressed as percentages ranging from 0% to 100%.

Tossing a coin. The sample space is S = H,T.

Tossing a die. The sample space is S = 1,2,3,4,5,6.

Tossing a coin twice. The sample space is S = HH,HT,TH,TT.

Tossing a die twice. The sample space is S = (i,j) : i,j = 1,2,...,6, which contains 36 elements.
Conditional probability is a measure of the probability of an event given that another event has already occurred.
Intersection of two events is denoted by A \(\cap\) B and is shown in the following Venn Diagram.
Union of two events is denoted by A \(\cup\) B and is shown in the following Venn Diagram.
In Permutations, order matters.
\[P(n,r) = \dfrac{n!}{nr!}\] In Combinations, order does not matter.
\[C(n,r) = \dfrac{n!}{r!(nr)!}\]
Solved Example: 7101
From a pack of regular playing cards, two cards are drawn at random. What is the probability that both cards will be Kings, if first card in NOT replaced?
A. $\dfrac{1}{26}$
B. $\dfrac{1}{52}$
C. $\dfrac{1}{169}$
D. $\dfrac{1}{221}$
Given : Total number of cards = 52 and two cards are drawn at random. Number of kings in playing cards = 4 So the probability that both cards will be king is given by, \[P = \dfrac{4}{52} \times \dfrac{3}{51} = \dfrac{1}{221}\]
Correct Answer: D
Solved Example: 7102
Two dice are thrown. What is the probability that the sum of the numbers on the two dice is eight?
A. $\dfrac{5}{36}$
B. $\dfrac{5}{18}$
C. $\dfrac{1}{4}$
D. $\dfrac{1}{3}$
We know that a dice has 6 faces and 6 numbers so the total number of cases
(outcomes) = 6$\times$6 = 36
And total ways in which sum of the numbers on the dices is eight,
$\left\{(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)\right\}$
So, the probability that the sum of the numbers eight is, p =$\dfrac{5}{36}$
Correct Answer: A
Solved Example: 7103
A box contains 5 black and 5 red balls. Two balls are randomly picked one after another form the box, without replacement. The probability for balls being red is:
A. $\dfrac{1}{90}$
B. $\dfrac{1}{2}$
C. $\dfrac{19}{90}$
D. $\dfrac{2}{9}$
\[\mathrm{Black\ balls} = 5,
\mathrm{Red\ balls} = 5,
\mathrm{Total\ balls} = 10\]
Here, two balls are picked from the box randomly one after the other without replacement. So the probability of both the balls are red is:
$P =\dfrac {5C_{0}\times 5C_{2}}{10C_{2}} =\dfrac {\dfrac {5!}{0!5!}\times \dfrac {5!}{3!2!}}{\dfrac {10!}{3!2!}} =\dfrac {1\times 10}{45} =\dfrac {2}{9}$
Alternate Method: The probability of drawing a red ball,
$P_1 = \dfrac{5}{10} = \dfrac{1}{2}$
If ball is not replaced, then box contains 9 balls.
So, probability of drawing the next red ball from the box. $P_2= \dfrac{4}{9}$
Hence, probability for both the balls being red is,
$P = P_1 \times P_2 = \dfrac{1}{2} \times \dfrac{4}{9} = \dfrac{2}{9}$
Correct Answer: D
Solved Example: 7104
A single die is thrown twice. What is the probability that the sum is neither 8 nor 9?
A. $\dfrac{1}{9}$
B. $\dfrac{5}{36}$
C. $\dfrac{1}{4}$
D. $\dfrac{3}{4}$
We know a die has 6 faces and 6 numbers so the total number of ways = 6$\times$6 = 36. And total ways in which sum is either 8 or 9 is 9, i.e.
$(2,6), (3,6) (3,5) (4,4) (4,5) (5,4) (5,3) (6,2) (6,3)$
Total number of tosses when both the 8 or 9 numbers are not come = 36  9 = 27. Then probability of not coming sum 8 or 9 is, 36
= $\dfrac{27}{36}$ = $\dfrac{3}{4}$
Correct Answer: D
Solved Example: 7105
A box contains 20 defective items and 80 nondefective items. If two items are selected at random without replacement, what will be the probability that both items are defective ?
A. $\dfrac{1}{5}$
B. $\dfrac{1}{25}$
C. $\dfrac{20}{99}$
D. $\dfrac{19}{495}$
Sample Space= Any two items can be selected from 100 items = $100_{C_2}$
Favourable case= Defective two items can be selected from 100 items = $20_{C_2}$
Probability of the required event is given by:
$= \dfrac{20_{C_2}}{100_{C_2}}
= \dfrac{\dfrac{20!}{18!2!}}{\dfrac{100!}{98!2!}}
= \dfrac{20 \times 19}{100 \times 99}
= \dfrac{19}{495}$
Correct Answer: D
Solved Example: 7106
A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is:
A. $\dfrac{2}{315}$
B. $\dfrac{1}{630}$
C. $\dfrac{1}{1260}$
D. $\dfrac{1}{2520}$
The box contains:
Number of washers = 2, Number of nuts = 3, Number of bolts = 4
Total objects = 2 + 3 + 4 = 9
First two washers are drawn from the box which contain 9 items. So the probability of drawing 2 washers is,
\[P_{1} =\dfrac {2_{C_{2}}}{9_{C_{2}}}
=\dfrac {1}{\left( \dfrac {9!}{7!2!}\right)}
=\dfrac {7!2!}{9\times 8\times 7!}
=\dfrac {2}{9\times 8}
=\dfrac{1}{36}\]
After this box contains only 7 objects and then 3 nuts drawn from it. So the probability of drawing 3 nuts from the remaining objects is,
\[P_{2} =\dfrac {3_{C_{3}}}{7_{C_{3}}}
=\dfrac {1}{\left( \dfrac {7!}{4!3!}\right) }
=\dfrac {4!3!}{7\times 6\times 5\times 4!}
=\dfrac {1}{35}\]
After this box contain only 4 objects, probability of drawing 4 bolts from the box,
$P_{3}=\dfrac {4_{C_{4}}}{4_{C_{4}}} =\dfrac {1}{1} =1$
Therefore the required probability is,
$P = P_1P_2P_3 = \dfrac{1}{36} \times \dfrac{1}{35} \times 1 = \dfrac{1}{1260}$
Correct Answer: C
Solved Example: 7107
A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is:
A. $\dfrac{1}{20}$
B. $\dfrac{1}{12}$
C. $\dfrac{3}{10}$
D. $\dfrac{1}{2}$
No. of Red balls = 4, No. of Black ball = 6
3 balls are selected randomly one after another, without replacement.
1 red and 2 black balls are will be selected as following
\[R B B: \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} = \dfrac{1}{6}, \]
\[B R B: \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} = \dfrac{1}{6},\]
\[B B R: \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} = \dfrac{1}{6}\]
Hence Total probability of selecting 1 red and 2 black ball is:
\[P = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{2}\]
Correct Answer: D
Solved Example: 7108
In a company, 35% of the employees drink coffee, 40% of the employees drink tea and 10% of the employees drink both tea and coffee, what % of the employees drink neither tea nor coffee? (GATE Civil 2021)
A. 15
B. 35
C. 25
D. 40
Percent of employees drink neither tea nor coffee = 100  25  10  30 = 35Correct Answer: B
Binomial Distribution
Learning Objectives:

Calculate probabilities for Binomial distributions.
The problems relating to tossing of coins or throwing of dice or drawing cards from a pack of cards with replacement lead to binomial probability distribution.
Only two possibilities are in Binomial distribution.
p = Probability of success
q = Probability of failure = 1 p .. ... .. .. .. Since only two outcomes, success and failure, are possible
n = no. of total attempts
x = no. of successful attempts = 0,1,2,3,.. .. .. ,n
Then probability of success P(x) is given by:
\[P(x)= C(n,x)p^{x}q^{nx} = \dfrac{n!}{x!(nx)!}p^{x}q^{nx}\]
Conditions for Binomial Distribution:
We get the Binomial distribution under the following experimental conditions.

The number of trials (n) is finite.

The trials are independent of each other.

The probability of success (p) is constant for each trial.

Each trial must result in a success or a failure.
Solved Example: 7201
A probability distribution function is a rule of correspondence or equation that:
A. Finds the mean value of the random variable.
B. Assigns values of x to the events of a probability experiment.
C. Assigns probabilities to the various values of x.
D. Defines the variability in the experiment.
A probability distribution function is some function or rule that may be used to define a particular probability distribution.
Correct Answer: C
Solved Example: 7202
An unbiased coin is tossed three times. The probability that the head turns up in exactly two cases is:
A. $\dfrac{1}{9}$
B. $\dfrac{1}{8}$
C. $\dfrac{2}{3}$
D. $\dfrac{3}{8}$
In a coin probability of getting head p = $\dfrac{1}{2}$ and probability of getting tail, q = 1 $\dfrac{1}{2}$= $\dfrac{1}{2}$
When unbiased coin is tossed three times, then total possibilities are:
{HHH, TTH, THT, THH, HTT, HTH, HHT, TTT}
From these cases, there are three cases, when head comes exactly two times. So, the probability of getting head exactly two times, when coin is tossed 3 times is, $\dfrac{3}{8}$.
Correct Answer: D
Solved Example: 7203
An unbiased coin is tossed five times. The outcome of each toss is either a head or a tail. The probability of getting at least one head is:
A. $\dfrac{1}{32}$
B. $\dfrac{13}{32}$
C. $\dfrac{15}{32}$
D. $\dfrac{31}{32}$
Here n = 5 = no. of attempts
r = 0 = no. of successes....we are calculating for NONE.
p = 0.5 = probability of success
q = 0.5 = probability of failure
\begin{align*} P(\mathrm{At\ least\ one}) &= 1  P(\mathrm{none})\\ &= 1  \left [ \binom{n}{r} p^{r} q^{nr}\right ]\\ &= 1  \left [ \binom{5}{0} \left ( \dfrac{1}{2}\right )^{0} \left (\dfrac {1}{2} \right )^{5}\right]\\ &= 1  \dfrac{1}{32}\\ &= \dfrac{31}{32} \end{align*}Correct Answer: D
Solved Example: 7204
A fair coin is tossed infinite times then find the probability of 4$^{th}$ head occurring exactly at 10$^{th}$ time?
A. 0.208
B. 0.029
C. 0.082
D. 0.016
P(4$^{th}$ head occurring exactly at 10$^{th}$ time)
=P(3 heads occurring in first 9 attempts) AND P(4$^{th}$ head occurring on the 10$^{th}$ time)
= $\left [ \binom{9}{3} \left ( \dfrac{1}{2}\right )^{3} \left (\dfrac {1}{2} \right )^{6}\right ] \times \dfrac{1}{2}$ = 0.082
Correct Answer: C
Solved Example: 7205
A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is:
A. 0.0036
B. 0.1937
C. 0.2234
D. 0.3874
Let, p = defective items and q = nondefective items
10% items are defective, then probability of defective items, p = 0.1
Probability of nondefective item, q = 1  0.1 = 0.9
The Probability that exactly 2 of the chosen items are defective is:
$10_{C_2} \times (0.1)^2 \times (0.9)^8 = 0.1937$
Correct Answer: B
Solved Example: 7206
Gender inequality is real ! There are still so many couples who prefer having a male child rather than a female child for whatsoever reasons ! Mr. J.M. has two businesses and as per his retirement planning, he wishes to have exactly two sons. He does not mind having any number of children till he has two sons. He will not have any more children after having two sons. Assume Mr. J.M.’s wife is equally likely to have a male or female child. What is the probability that he has k daughters? [Adapted from J. L. Devore, 2010.]
A. $k (0.5)^{k+1}$
B. $(k+1) (0.5)^{k+2}$
C. $k (0.5)^{k+2}$
D. $(k+2) (0.5)^{k}$
If Mr. J.M. has k daughters. He has total (k+2) children including 2 sons.Mr. J.M. stopped having children at this stage means, the most recent child is his second son.
So he has one son from from first (k+1) children AND the last child is son.
Combined AND probability = $(k+1) (0.5)^{k+1} \times 0.5$ = $(k+1) (0.5)^{k+2}$
Correct Answer: B
Normal Distribution
Learning Objectives:

Describe what a normal distribution is.

Understand properties of Normal distributions and able to calculate probability by finding area under the curve for a given scenario.
Normal distribution has a perfectly symmetrical bell shaped curve in which Mean = Median = Mode
Normal distribution is also referred as Gaussian distribution. Population parameters such as height and weight usually follow normal distribution. A standard normal distribution is a special case of normal distribution where mean = 0 and standard deviation = 1.
Here, first calculate Z value using, \[Z = \dfrac{x  \mu }{\sigma }\] Then refer to the table of values.
Total area under the curve is = 1. Since the curve is symmetrical about mean line, left area = 0.5 and right area = 0.5.
Solved Example: 7301
Total area under the normal probability curve is:
A. Less than one
B. Unity
C. Greater than one
D. Zero
The total area under the normal curve is equal to 1.
Correct Answer: B
Solved Example: 7302
A standard normal distribution has:
A. Mean = variance
B. Mean = 1, variance = 1
C. Mean = 0, variance = 1
D. Mean = 0, standard deviation = 0
Normal distribution is a function that represents the distribution of a random variable as a symmetrical bellshaped graph. It is an arrangement of a data set in which most values cluster in the middle of the range and the rest taper off symmetrically toward either extreme.
The standard normal distribution is a special case of the normal distribution . It is the distribution that occurs when a normal random variable has a mean of zero and a standard deviation of one.
Correct Answer: C
Solved Example: 7303
If a normal distribution has mean 200 and standard deviation 20, find K so that the probability that a sample value is less than K is .975.
A. 239
B. 204
C. 210
D. 215
For 97.5% probability, z = 1.96
Using the formula,
\[\dfrac{K\mu}{\sigma} = z, \]
\[\dfrac{K200}{20} = 1.96, \]
\[K = 239\]
Correct Answer: A
Solved Example: 7304
The lengths of a large stock of titanium rods follow a normal distribution with a mean ($\mu$)of 440 mm and a standard deviation ($\sigma$) of 1 mm. What is the percentage of rods whose lengths lie between 438 mm and 441 mm?
A. 81.85%
B. 68.4%
C. 99.75%
D. 86.64%
\[Z(L= 438) = \dfrac{438440}{1} = 2,\] \[Z(L= 441) = \dfrac{441440}{1} = +1\] Refer standard tables of normal distribution to calculate area percentages.Left (Orange) area = 47.72%,
Right (Blue) area = 34.15%
Total area = 81.87%
Correct Answer: A
tdistribution
Learning Objectives:

Interpret the tdistribution and use a tdistribution table.

Construct and interpret confidence intervals for a population mean when \(\sigma\) is not known.

Form confidence intervals for, and test hypotheses about, \(\mu\) using the tdistribution.
tdistribution is used instead of zdistribution when population standard deviation is unknown. The height of the t distribution differs depending on the sample size. As n approaches infinity (\(\infty\)) the t distribution approaches the standard normal distribution. The height of the t distribution is determined by the number of degrees of freedom. For a distribution based on one group, the degrees of freedom (df) is n1.
\[t = \dfrac{\bar{x}  \mu}{\dfrac{s}{\sqrt{n}}}\] Where,
n = sample size
n1 = Degrees of Freedom (DOF or D.F.)
\(\bar{x}\) = Sample mean, \(\mu\) = Population mean, s = sample standard deviation

Subtract one from your sample size. This will be your df, or degrees of freedom.

Look up the df in the left hand side of the tdistribution table. Locate the column under your alpha level (the alpha level is usually given to you in the question.
Solved Example: 7401
Find the 95% confidence interval for the population mean for grades for people going on to graduate school. You have a sample of 11 Mechanical Engineering majors that have been accepted to graduate school and their mean grade on a 4 point system is 3.52 with a sum of squared deviations equal to 0.26.
A. 3.40 to 3.64
B. 3.41 to 3.63
C. 3.42 to 3.62
D. 3.43 to 3.61
$\bar{X}$ = 3.52, Degrees of freedom = N1 = 111 = 10
\[\hat{s} = \sqrt{\dfrac{SS}{N1}} = \sqrt{\dfrac{0.26}{10}} = 0.16125\] \[\hat{s}_{\bar{X}} = \dfrac{\hat{s}}{\sqrt{N}} = \dfrac{0.161245}{\sqrt{11}} = 0.04862\](From table) tscore from 2.228 to 2.228 contains 95% of the scores for a students t distribution with df = 10.
The Confidence Interval (CI) can be calculated by:
Lower Confidence Limit (LCL) = $\bar{X}  t \times \hat{s}_{\bar{X}} = 3.52  2.228 \times 0.04862 = 3.4117$
Upper Confidence Limit (UCL) = $\bar{X} + t \times \hat{s}_{\bar{X}} = 3.52 + 2.228 \times 0.04862 = 3.6283$
In this case the CI = 3.4117 to 3.6283.So, we are 95% sure that the population mean that this sample came from is between 3.41 and 3.63.
(Reference: This question is derived from Colgate University's Psychology 309 Class Fall 2000)
Correct Answer: B
Chisquared Distribution
Learning Objectives:

Conduct a \(\chi^2\) test to determine if two qualitative variables are dependent.
When to Use \(\chi^2\) Test for Independence?
The test procedure described in this lesson is appropriate when the following conditions are met:

The sampling method is simple random sampling.

The variables under study are each categorical.

If sample data are displayed in a contingency table,

the expected frequency count for each cell of the table is at least 5.

Degrees of freedom. The degrees of freedom (DF) is equal to: \[DF = (r  1) \times (c  1)\] where r is the number of levels for one categorical variable, and c is the number of levels for the other categorical variable.

Expected frequencies. The expected frequency counts are computed separately for each level of one categorical variable at each level of the other categorical variable. Compute r * c expected frequencies, according to the following formula. \[E_{r,c} = (n_r \times n_c) / n\] where E\(_{r,c}\) is the expected frequency count for level r of Variable A and level c of Variable B, n\(_r\) is the total number of sample observations at level r of Variable A, n\(_c\) is the total number of sample observations at level c of Variable B, and n is the total sample size.

Test statistic. The test statistic is a \(\chi\)squared random variable (\(\chi^2\)) defined by the following equation. \[\chi ^2 = \Sigma [ (O_{r,c}  E_{r,c})^2 / E_{r,c}]\] where O\(_{r,c}\) is the observed frequency count at level r of Variable A and level c of Variable B, and E\(_{r,c}\) is the expected frequency count at level r of Variable A and level c of Variable B.

Pvalue. The Pvalue is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chisquare, use the \(\chi ^2\) Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above.
Solved Example: 7501
In order to use a $\chi ^2$ test, what type of data values are generally expected?
A. Probability values
B. Categorical values
C. Interval values
D. Fractional values
The Chi Square statistic is commonly used for testing relationships between categorical variables.
Correct Answer: B