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A probability distribution is a statistical concept describing the likelihood of each random experiment or occurrence outcome. It gives the odds of several probable outcomes.
- Probability distribution is occurrence of a random phenomenon in the sample space.
- Its two important functions are the probability density function and the cumulative distribution function.
- The value of probability will form a shape when all random variables are aligned to a graph.
- It is assumed that the probability of an event is greater than zero.
- A probability distribution is represented by graphs and tables.
- For example, consider the case of tossing a coin.
- There is a probability of getting either a head or a tail.
- This distribution of an experiment could be defined with an uncertain or unpredictable outcome.
- The function for probability distribution is given as:
P(a < X ≤ b) = F(b) - F(a)
Key Terms: Probability, Probability Distribution, Random Variable, Domain, Function, Cumulative Distribution Function, Probability Density Function, Poisson Probability Distribution, Discrete Probability Distribution
Random Variables
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Random Variables are defined as a real-valued function where the domain represents the sample space of an experiment. It is also known as random quantity, aleatory variable, or stochastic variable.
- Random variables also represent the uncertainty of an event, like the measurement of error.
- There is a chance that two random variables are identical, but they are independent.
- George Mackey and Pafnuty Chebyshev were the first ones to study about random variables.
- The concept is used in the fields of graph theory, natural language processing and machine learning.
- It assigns a real number to every event of a probability distribution.
Example of Random VariablesExample: Assume that when we draw a card from a deck of 52 cards, we have no way of knowing whether it will be a heart or an ace, or a spade. If an experiment has more than one possible conclusion, it is said to be random, and if it only has one, it is said to be deterministic. |
Probability Distribution of Random Variables
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Probability distribution of random variables defines how the probabilities are distributed over the random variable's values. A random variable is a numerical representation of a statistical experiment's outcome.
- A discrete random variable can only take one of two values: a finite number or an infinite succession of values.
- Continuous random variable can take any value in any interval on the real number line.
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The required probability distribution function of random variable X if and only if its values, f(x), satisfy the conditions:
f(x) ≥ 0 for each value within its domain
∑x f(x) = 1
- where the summation extends over all the values within its domain
Example of Probability Distribution of Random VariablesExample 1: A random variable reflecting the number of cars sold at a specific dealership on a given day would be discrete. Example 2: A random variable expressing a person's weight in kilograms (or pounds) would be continuous. |
Types of Probability Distribution
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There are two types of statistical probability distributions, namely discrete and continuous distributions. All integers greater than 0 (which would include indefinitely values such as pi = 3.14159265...) form a continuous distribution.
- The analysis give the ideas of discrete and continuous probability distribution.
- Distinct probability distributions describe different data-generating processes and serve different objectives.
Cumulative Probability Distribution
Cumulative probability distribution for a random variable x is defined as a function whose value is equal to the required variable x. It is used to compare the values of probability distribution under specific conditions.
- Cumulative Probability Distribution is also known as Continuous Probability Distribution.
- The normal distribution represents the behaviour of most circumstances in the universe.
- Normal distributions are important in statistics.
- The big sum of (small) random variables is frequently found to be regularly distributed.
- The value is contributed to common use.
- It is often used in the natural and social sciences.
- The distribution represent real-valued random variables whose values are unknown.
- The cumulative probability distribution is given as:
- where Fx(x) = function of X (real value variable)
- P is the probability that the value of X is less than or equal to x
Example of Cumulative Probability DistributionExample: Suppose you are rolling a six-sided die, where X is the random variable. The probability of getting an outcome by rolling a die is given as:
|
Discrete Probability Distribution
Discrete distributions are used to model the probabilities of random variables with discrete outcomes. It takes into consideration those events that are finite and countable.
- In discrete probability distribution, the random variable has specific values.
It is based on two conditions, which are as follows:
- The value of the probability of a discrete random variable should lie between 0 and 1.
- Sum value of all probability functions is equal to 1.
Example of Discrete Probability DistributionExample 1: The set of potential values for the random variable X, which indicates the number of heads that can occur when a coin is tossed twice, is 0 1, 2 and not any value between 0 and 2, such as 0.1 or 1.6. Example 2: The binomial distribution analyses the likelihood of a "yes" or "no" outcome occurring over a specified number of trials, given the event's probability in each trial—for example, flipping a coin 100 times and having the outcome be "heads." |
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Probability Distribution Formulas
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The important probability distibution formulas are as follows:
| Category | Formula | Description |
|---|---|---|
| P(X) = nCx.ax.bn-x | Where a = probability of success, b=probability of failure, n= number of trials and x=random variable denoting success | |
| Cumulative Distribution Function | FX(x)=∫x−∞fX(t)dt | x=random variable |
| Discrete Probability Distribution | P(x)=n!r!(n−r)!⋅pr(1−p)n−r and P(x)=C(n,r)⋅pr(1−p)n−r | x=random variable |
What is Negative Binomial Distribution?
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The negative binomial distribution is a discrete probability distribution that models the number of wins in a series of independent and identically distributed Bernoulli trials before a predetermined (non-random) number of failures (denoted r).
- Sometimes we want to see how many Bernoulli trials we'll need to achieve a result.
- We specify the intended outcome in advance and continue the experiment until we obtain it.
- The negative binomial distribution could also be used to model the number of days a machine works before breaking.
Example of What is Negative Binomial Distribution?Example: we can designate rolling a 6 on a die as a failure and any other number as a success and then calculate the number of successful rolls before the third failure (r = 3). The probability distribution of the number of non-6s appearing will be a negative binomial distribution in this scenario. |
What is Poisson Binomial Distribution?
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The Poisson distribution is a probability distribution that shows how many times an event is expected to happen over a certain amount of time. To put it another way, it's a count distribution.
- The Poisson distribution is a discrete function.
- It means that the variable may only take specific values from a (potentially infinite) set of options.
- To put it another way, the variable can't take all of the possible values in any continuous range.
- The variable can only take the values 0, 1, 2, 3, etc., with no fractions or decimals.
Example of What is Poisson Binomial Distribution?Example 1: A call center, for example, receives 180 calls every hour on average, 24 hours a day. The calls are unrelated; getting one does not affect the likelihood of receiving the next. The number of calls received in any given minute follows a Poisson probability distribution: the most likely values are 2 and 3, but 1 and 4 are also possible, and there's a tiny chance it might be as low as zero and a very small chance it could be 10. Example 2: Another example is the number of decay events emitted by a radioactive source during a set length of time. |
Read More:
| Class 11 Mathematics Related Concept | ||
|---|---|---|
| Difference Between Mean and Median | Relation Between Mean Median and Mode | Irrational Numbers |
| Number Systems | Frequency Distribution Table | Euclid's Proof |
Probability Distribution Table
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The probability distribution table for definite set of random variables are as follows:
| Random Variable (X) | Probability(P(X)) |
|---|---|
| X1 | P(X1) |
| X2 | P(X2) |
| X3 | P(X3) |
| ... | ... |
| Xn | P(Xn) |
Posterior Probability
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In Bayesian statistics, a posterior probability is the revised or updated likelihood of an event happening after new information is taken into account. Using Bayes' theorem, the posterior probability is derived by updating the prior probability.
- In statistical terminology, it is the likelihood of event A happening after event B has occurred.
- The posterior probability distribution is the probability distribution of an unknown quantity.
- It is regarded as a random variable and is based on data from an experiment or survey.
- Posterior probability is used in the field of biology and psychology.
- The required formula for posterior probability is given as:
Posterior Probability = Prior Probability + New Evidence
Prior Probability
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Prior Probability is a type of probability that is used in bayesian statistics. It is used for assigning the value to the probability of an event before happening of an event.
- It will predict the outcomes before collecting all the observation for an outcomes.
- The probability will use the updated information to calculate posterior probability distribution.
- It is also based on the experience of an expert.
- Prior probability is divided into number of levels known as hierarchical priors.
Example of Prior ProbabilityExample: Prior Probability indicates the number of voter voting for a politician before actual result are released in public. |
Things to Remember-
- A probability distribution is a statistical concept that describes the likelihood of each outcome of a random experiment or occurrence.
- It gives the odds of several probable outcomes which is indicated by P(X) = nCx.ax.bn-x where a = probability of success and b=probability of failure.
- For practice, individuals can try out Probability Important Questions and MCQ on Probability.
- Cumulative Distribution Function and discrete probability distribution are two types of probability distribution.
- It is based on the probability of subsets of sample space.
Sample Questions
Ques. Suppose that a coin is tossed twice so that the sample space is S = {HH, HT, TH, TT}. Let X represent the number of heads that can come up. So, with each sample point, we can associate a number for X so for HH the value of X is 2 as in HH there are 2 heads, for HT the value of X is 1 as there is only one head in HT. similarly for TH, X=1 and for TT value of X = 0 as there are no heads in TT. (3 marks)
Ans. So, in the sample space S = {HH, HT, TH, TT} assuming that the coin is fair,
P(HH)= ¼, P(HT)= ¼, P(TH)= ¼, P(TT)= ¼
Therefore, P(X=0) = P(TT) = ¼
P(X=1) = P (HT ∪ TH) = P(HT) + P(TH) = ¼ + ¼ = ½
P(X=2) = P(HH) = ¼
Thus, the table formed is shown below:
| X | P(X) |
| 0 | ¼ |
| 1 | ½ |
| 2 | ¼ |
Ques.Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of several aces. (3 marks)
Ans. Let X= Number of aces
Since two cards are drawn, so X can take values 0,1 and 2
Now, probability of getting an ace= 4/52 = 1/13
And probability of not getting an ace = 1- 1/13 = 12/13
P(X=0) = P (Not getting an ace card)
= 12/13 * 12/13 = 144/169
P(X=1) = P (getting one ace card)
= 1/13 * 12/13 + 12/13 * 1/13 = 24/169
P(X=2) = P (getting tow ace cards)
= 1/13 * 1/13 = 1/169
The probability distribution is as follows:
| X | P(X) |
| 0 | 144/169 |
| 1 | 24/169 |
| 2 | 1/169 |
Ques. In a set of 10 coins, 2 coins with heads on both sides. A coin is selected at random from this set and tossed five times. Of all the five times, the result was head, find the probability that the selected coin had heads on both sides. (3 marks)
Ans. Let the events be :
E1 : Selecting a coin with two heads
E2 : Selecting a normal coin and A : The coin falls head all the times.
Since E1 and E2 are mutually exclusive and by the data given in the problem, we have :
P(E1) = 2/10=1/5
P(A/E2) = 8/10=4/5 = P(A/E1) = 1
P(A/E1) = 1
P(A/E2) = ½ × ½ × ½ × ½ × ½ = 1/32
Now P(A) = P(A ∩ E1) + P(A ∩ E2)
= P(E1) P(A/ E1) + P(E2) P(A/ E2)
= 1/5 + 1/40 = (8+1)/40 = 9/40
Ques. Two numbers are selected at random (without replacement) from first 7 natural numbers. If X denotes the smaller of the two numbers obtained, find the probability distribution of X. (4 marks)
Ans. Let ‘X’ be the smaller of the two numbers obtained.
Thus, X takes values 1, 2, 3, 4, 5 and 6.
P(X=1) = 6/ 7C2 = 6/21 = 2/7
P(X=2) = 5/ 7C2 = 5/21 = 5/21
P(X=3) = 4/ 7C2= 4/21
P(X=4) = 3/ 7C2 = 3/21 = 1/7
P(X=5) = 2/ 7C2 = 2/21
P(X=6) = 1/ 7C2 = 1/21
Hence, Probability Distribution is:
| xi | 1 | 2 | 3 | 4 | 5 | 6 |
| Pi | 2/7 | 5/21 | 4/21 | 1/7 | 2/21 | 1/21 |
Ques. Two groups are competing for the positions of the Board of Directors of a corporation. The probabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group. (3 marks)
Ans. Let E1 = First group wins,
E2 = Second group wins
H = Introduction of new product.
P(E1) = 0.6,
P(E2) = 0.4
P(H/E2) = 0.3
P(H/E1) = 0.7
Now,
P(E2/H) = P
Ques. From a lot of 20 bulbs which include 5 defectives, a sample of 3 bulbs is drawn at random, one by one with replacement. Find the probability distribution of the number of defective bulbs. Also, find the mean of the distribution. (2 marks)
Ans. Let X denote the number of defective bulbs.
X = 0, 1, 2, 3.
P(X = 0) = =
P(X = 3 )=(5/20)3 =1/64
Mean = ΣXP(X) = 27/64+18/64+3/64=¾
Ques. From a lot of 10 bulbs containing 4 defective ones, 4 bulbs are drawn at random. If X is a random variable that denotes the number of defective bulbs. Find the probability distribution of X. (4 marks)
Ans. Since, X denotes the number of defective bulbs and there is a maximum of 4 defective bulbs, hence X can take values 0, 1, 2, 3 and 4. Since 4 bulbs are drawn at random, the possible combination of drawing 4 bulbs is given by 10C4.
- P(Getting No defective bulb) = P (X = 0) = 6C4/ 10C4. = 15/210 = 5/70
- P(Getting 1 Defective Bulb) = P (X = 1) = 4C1 × 6C3/ 10C4. = 80/70
- P(Getting 2 defective Bulb) = P(X = 2) = 4C2 × 6C2/ 10C4. = 12/70
- P(Getting 3 Defective Bulb) = P(X =3) = 4C3 × 6C1/ 10C4. = 8/70
- P(Getting 4 Defective Bulb) = P(X =4) = 4C4 × 6C0/ 10C4.= 1/70
- Hence Probability Distribution Table is given as follows:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | 5/70 | 80/70 | 12/70 | 8/70 | 1/70 |
Ques. A box contains 4 blue balls and 2 green balls. Find the probability distribution of the number of green balls in a random draw of 2 balls. (3 marks)
Ans. Given that the total number of balls is 6 out of which 2 have to be drawn at random. On drawing 2 balls the possibilities are all 2 are green, only 1 is green, and no green. Hence X = 0, 1, 2, 3.
- P(No ball is green) = P(X = 0) = 4C3/6C3 = 4/20 = 1/5
- P(1 ball is green) = P(X = 1) = 2C1 × 4C2/6C3 = 12/20
- P(2 balls are green) = P(X = 2) = 2C2 × 4C1/6C3 = 1/5
- Hence, the probability distribution for this problem is given as follows
| X | 0 | 1 | 2 |
| P(X) | 1/5 | 12/20 | 1/5 |
Ques. In a game of darts suppose you have a 35% chance that you will hit the bullseye. If you take a total of 10 shots then what is the probability that you will hit the bullseye 5 times. (3 marks)
Ans. n = 10, p = 35 / 100 = 0.35, x = 5
We have to use the Binomial probability distribution given by P(X = x) = (nx)px(1−p)n−x
P(X = 5) = (155)0.355(1−0.35)15−5
= 0.265
Ques. Suppose a die is tossed multiple times. What is the probability that the die will land on a number smaller than 5. (2 marks)
Ans. P(X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
P(X < 5) = 1 / 6 + 1 / 6 + 1 / 6 + 1 / 6 = 4 / 6
Ques. The weight of a pot of water chosen is a continuous random variable. The following table gives the weight in kg of 100 containers recently filled by the water purifier. It records the observed values of the continuous random variable and their corresponding frequencies. Find the probability or chances for each weight category. (3 marks)
| Weight W | Number of Containers |
|---|---|
| 0.900−0.925 | 2 |
| 0.925−0.950 | 6 |
| 0.950−0.975 | 20 |
| 0.975−1.000 | 30 |
| 1.000−1.025 | 20 |
| 1.025−1.050 | 2 |
| Total | 80 |
Ans. We first divide the number of containers in each weight category by 80 to give the probabilities.
| Weight W | Number of Containers | Probability |
|---|---|---|
| 0.900−0.925 | 2 | 0.025 |
| 0.925−0.950 | 6 | 0.075 |
| 0.950−0.975 | 20 | 0.25 |
| 0.975−1.000 | 30 | 0.375 |
| 1.000−1.025 | 20 | 0.25 |
| 1.025−1.050 | 2 | 0.025 |
| Total | 80 | 1.00 |
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