Probability: Definition, Formula, Types and Problems

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Probability is the branch of mathematics that states how likely an event is to occur in mathematical context. There are events whose outcomes can not be predicted with full certainty. In this case, the best we can say is how likely they are to happen by using the idea of probability.  In this chapter, we will study about the concepts of conditional ptobability of any event given that the other event has occured. It will also explain the probability distribution of random variables and the mean and the variance of probability distribution. 

Read More: Determinant Formula

Key Terms:  Probability, probability of any event, conditional probability, occurrence of event, theorem of Total Probability


Definition

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The measure of the possibility of any event is known as probability. Event is any phenomenon that happens or is bound to happen in the future. 

The probability line

The probability line

To understand probability, we need to understand various concepts. Let us know about them in detail.

  • Experiment: Any phenomenon that occurs is known as an experiment. Eg - tossing a coin, rolling dice, etc.
  • Sample Space: The set of all the possible outcomes of an experiment is known as a sample space.
  • Event: The subset of a given sample space is defined as an event. It is associated with a random experiment. An example of an event can be getting head or tail in an experiment of tossing a coin.
  • If none of the events is expected to occur in preference to the other, the events are known as equally likely events. For example - when a dice is rolled, all 6 faces of the dice are equally likely to come.
  • If the happening of any one event excludes the possibility of occurrence of any other event, the events are known as mutually exclusive events. (A ∩ B) = Φ
  • If the performance of an experiment results in the occurrence of at least one of the events, the events are said to be exhaustive events
  • Any two events are called independent events if the probability of occurrence or non-occurrence of any one event is not affected by the other event
  • Outcome: Outcome can be defined as the result of any experiment. Example - getting a six at the roll of a dice.

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Probability of an event

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The probability of happening of any event A can be given by -

P(A) =Number of favourable outcomesTotal number of outcomes= mn

  • The probability of any event A is greater than or equal to zero and less than or equal to 1 i.e.

 0 ≤ P(A) ≤ 1.

  • The probability of a possible event is 1.
  • The probability of an impossible event is 0.
Figure: Probability of events 

Figure: Probability of events 

  • Conditional Probability: If E and F are two events that are associated with a random experiment’s same sample space. Then, the probability of the occurrence of any event E, when the event F has already occurred, is known as the conditional probability of the event E over F. This is denoted by P(E/F).

P(E/F) = P(E

F)P(F) (Where P(F)0)

Like this, the conditional probability of event F over E can be given as

P(F/E) = P(E

F)P(E)  (where P(E)≠0)

Conditional probability

Figure: Conditional probability 

  • Properties of Conditional Probability: If E and E are two events of sample space S and G is an event of S which has already occurred such that P(G) ≠ 0, then

(i) P[(E ∪ F)/G] = P(F/G) + P(F/G) – P[(F ∩ F)/G], P(G) ≠ 0

(ii) P[(E ∪ F)/G] = P(F/G) + P(F/G), if E and F are disjoint events.

(iii) P(F’/G) = 1 – P(F/G)

(iv) P(S/E) = P(E/E) = 1

  • Multiplication theorem: P (A ∩ B) = P (A). P(B)

For example,in the event of rolling a dice,

Where A is the event of a number less than 4 occurring on a roll of dice, A = {1,2,3}

And B is the event of a multiple of 2 occurring on a roll of dice B ={ 2,4,6}

Hence (A ∩ B) = {2}, and P(A ? B)= 16

  • Event(A OR B)

P(A U B) = P(A) + P(B) – P(A ∩ B)

If both A and B are two mutually exclusive events, then P (A ∩ B) = 0 and hence 

P(A U B)= P (A) + P (B)

  • Event(NOT A)

If the probability of occurrence of any event A is denoted as P(A), then the probability of the non-occurrence of that same event A is P(A’).

  • Event (A BUT NOT B)

If there are two events A and B, then Event (A BUT NOT B) gives us the remainder after the common elements in A and B gets eliminated from A.

  • Event (B BUT NOT A)

If there are two events A and B, then Event (B BUT NOT A) gives us the remainder after the common elements in A and B gets eliminated from B.

  • Bayes Theorem
  • Partition of Sample Space: A set of events E1, E2,…,En is said to represent a partition of the sample space S, if it satisfies the following conditions:

(i) Ei ∩ Ej = Φ; i ≠ j; i, j = 1, 2, …….. n

(ii) E1 ∪ E2 ∪ …… ∪ En = S

(iii) P(Ei) > 0, ∀ i = 1, 2,…, n

  • Theorem of Total Probability: Let events E1, E2, …, En form a partition of the sample space S of an experiment.If A is any event associated with sample space S, then
Figure: Division of sample space S

Figure: Division of sample space S

  • Bayes Theorem: If E1, E2,…,En are n non-empty events which constitute a partition of sample space S, i.e. E1, E2,…, En are pairwise disjoint E1 ∪ E2 ∪ ……. ∪ En = S and P(Ei) > 0, for all i = 1, 2, ….. n Also, let A be any non-zero event, the probability

Probability Class 12 Notes Maths Chapter 13 6

To know more about the chapter, 


Important Topics for JEE Main 

As per JEE Main 2024 Session 1, important topics included in the chapter probability are as follows:

  • Formula for Probability
  • Types of Probability
  • Properties of Probability
  • Probability of an Event

Some memory based important questions asked in JEE Main 2024 Session 1 include:

  1. Bag A contains 7 white balls & 3 red balls. Bag B contains 3 white balls & 2 red balis. A ball is chosen randomly & found to be red then find the probability that it is taken from bag A.
  2. In a paper there are 3 sections A, B and C which has 8, 6 and 6 questions each. A student have to attempt 15 questions such that they have to attempt atleast 4 questions out of each sections, then number of ways of attempting these questions are
  3. An urn contains 15 red. 10 white, 60 orange, and 15 green balls. If 2 balls are taken with replacement, find the probability 1 ball is red and the other ball is white.
  4. If the mean of 15 observations is 12 and the standard deviation is 3. If 12 is replaced by 10 in data, then the new mean is µ and variance is o² then what is the value of 15(μ + μ² + σ²) = ?

Sample Questions on Probability

Q1. What is probability?

Ans. The measure of the possibility of any event is known as probability. This is given by the ratio of the number of favourable outcomes for an event to the total possible outcomes.

Q2. State the basic rules of probability.

Ans. - P (A U B) = P (A) + P (B) – P (A ∩ B)

P (A ∩ B) = P (A). P (B)

Q3. Find the probability of getting an even number greater than or equal to 3 in a roll of a dice.

Ans. The sample space for rolling of dice is S = {1,2,3,4,5,6} and the probability of the event of greater than or equal to 3 in a roll of a dice is given by P(E)

So E = {3,4,5,6} and S = {1,2,3,4,5,6}

P(E) = n(E)/n(S)

= 4/6

P(E) = 2/3

Q4. In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl?

Ans. Let E denote that the student chosen randomly studies in class XII, and F denote that the randomly chosen student is girl.

P (E|F) = ?

P(F) = 430/1000= 0.43

P(E∩ F) = 43/1000 = 0.043

P(E/F) = P(E∩F)/P(F)

0.043/0.43 = 0.1

Q5. A die thrown thrice. Events A and B are defined as shown.
A : Dice shows 4 the 3rd time it is thrown
B : Dice shows six the 1st time it is thrown and 5 on the second time it is 
Find P(B|A)

Ans. Number of possible outcomes = 6x6x6 = 216

Outcomes of A

B = {(6,5,1) (6,5,2) (6,5,3) (6,5,4) (6,5,5) (6,5,6)}

A∩ B ={6,5,4}

P(B) = 6/216. P(A∩B) = 1/216

Probability of A and B

Q6. A Mother, father and son are lining up to click a family picture.
Let E denote that the son is on one end
And F denote that the Father in centre
Find the probability of E, given that F has already occured.

Ans. S = {ms, msf, fms, fsm, smf, smf}

E = {mfs, fms, smf, sfm}

F ={mfs, sfm}

E∩F = {mfs, sfm}

Q7. There are 3 boxes which are identical in nature : I, II and III. Each of the boxes contains two coins. In the first box : both of the coins are gold coins, in the second box : both of the coins are silver coins and in the third box : one coin is gold and the other coin is gold. If a person chooses a random box and randomly selected a coin from that box and the coin is a gold coin, calculate the probability that the other coin inside the box is also made out of gold.

Ans. let E1, E2 and E3 denote that boxes 1 , 2 and 3 are chosen. 

P (E1) = P (E2) = P (E3)

let A be the event a gold coin was drawn

p of E

p (A|E2) = 0 

P of E3

p of E3

value

Q8. Suppose that the reliability of a HIV test is specified as follows :of people having HIV, 90% of the test detect the disease but 10% go undetected of individuals freed from HIV, 99% of the test are Judged HIV – tive but 1% are diagnosed as showing HIV +tive. From an outsized population of which only 0.1% have HIV, one person is chosen randomly , given the HIV test, and therefore the pathologist reports him/her is HIV +tive what's the probability that the person actually has HIV.

Ans. let E denote the event that the person selected is actually having HIV and A the event that the person’s HIV test is diagnosed as + tive.

let E’ not having HIV.

P(E) = 0.1% = 0.1/150 = 0.001

P(E) = 1 - P(E) = 0.999

P(A|E)) = 90% = 90/100 = 0.9

P(A|E’) = 1% = 1/100 = 0.01

probability of  E and A

= 0.083

Q9. Calculate the distribution of probability of the number of doublets when a pair of dice is rolled thrice.

Ans. X is the number of doublets, so obviously, X equals 0,1,2,3

Probability of getting doublet = 6/36= 1/6

Probability of not getting doublet = 1- (1/6) = 5/6 

X 0 1 2 3
P(X) 125/216 75/216 15/216 1/216

Q10. Calculate the variance of the number obtained when an unbiased die is thrown.

Ans. S = {1, 2, 3, 4, 5, 6}

p (1) = p (2) = p (3) = p (4) = p (5) =p (6) =  1/6 

sum

sum

result

Q11. If A and B are two independent events, then prove that the probability of occurrence of at least one of A and B is given by 1 – P ( ) P ( ).

Ans. P (at least one of A and B) = P (A∪B)

= P (A) + P (B) – P (A∩B)

= P (A) + P (B) – P (A). P (B)

= P (A) + P (B) [1 – P (A)]

= P (A) + P (B) P (A’)

= 1 – P (A’) + P (B) P (A’)

= 1 – P (A’) [1 – P (B)]

= 1 – P (A’) P ((B’))

sum

Q 12. An unbiased coin is tossed and an unbiased dice is thrown. A denotes the event that  the coin showed heads and B represents the event that the dice showed three. Show that A and B are independent events.

Ans. Let A denote that the coin showed heads

And B denote that the dice showed three

A = {(H1), (H2), (H3), (H4), (H5), (H6)}

B = {(H, 3)}, (T,3)}

A∩B = {(H3)}

P(A) = 6/12 = ½, P(B) = 2/12=?

P(A∩B) = 1/12

multiply= 1/16

= P(A∩B)

Therefore, we can conclude that A and B are events which are independent

Read More: Integration


Previous Years’ Questions

Very Short Answer Questions

Q 1. Two cards are drawn random and one-by-one without any replacement from a well shuffled pack of 52 playing cards. Find the probability that one card is red and the other is black. (CBSE 2020)

Ans. P (one red and one black) = P (first red and second black) + P ( first black and second red)

= 26/ 52 x 26/ 51 + 26/ 52 x 26/ 51   [ without replacement]

= 13/ 51 + 13/ 51

= 26/ 51

Short Answer Questions

Q 1. A die is thrown 6 times. If ‘ getting an odd number ‘ is a ‘success’, what is the probability of (i) 5 successes? (ii) atmost 5 successes? (CBSE 2019)

Ans. The repeated tosses of a die are Bernoulli trials. Let X be the number of successes of getting odd numbers in an experiment of 6 trials.

The probability of getting an odd number in a single throw of a die is p = 3/ 6 = ½ 

therefore, q = 1 – p = ½

X has a binomial distribution

therefore, P ( X = x) 

6 cr

(i) P ( 5 successes) 

6 c5

(ii) P (atmost 5 successes) P (x ≤ 5)

successess

= 63/ 64

Q 2. The random variable X has a probability distribution P (X) of the following form where ‘k’ is some number.?
p of x
Find the value of ‘k’.  (CBSE 2019)

Ans. In the probability distribution Σ XP (X) =1

value of k

Q 3. A die whose faces are marked 1, 2, 3 in red and 4, 5, 6 in gree, is tossed. Let A be the event ‘ number obtained is even’ and  B be the event ‘number obtained is red’. Determine if A and B are independent events. (CBSE 2017)?

Ans. Let event A: Number obtained is even

And event B: Number obtained is red

independent events

Long Answer Questions

Q 1. A couple has two children. What is the probability that both are boys, if it is known that,
(i) one of them is a boy
(ii) the older child is a boy.(Delhi 2014C, ‘08 C; All India 2014, ‘10)

Ans. Let B represents the older child which is a boy and b represents younger child which is also a boy. At same time, let G represents older child which is a girl and g represents a younger chcild which is a girl as wee. The sample space of the given question is,

S = { Bb, Bg, Gg, Gb}

therefore, n(S) = 4

Now, let A be the event that both children are boys.

Then, A = {Bb}

therefore, n(A) = 1

(i) Let b – at least one of the children is a boy.

therefore, B = {Bb, Bg, Gb} and n (B) = 3

and P (B) = n (B)/ n(S) = ¾    …… (i)

probablity

(ii) Let B – let older child is a boy.

Then, B = {Bb, Bg}

therefore, n(B) = 2

and P (B) = n(B)/ n(S) = 2/4 = ½  ……. (iv)

probablity

Q 2. A speaks truth in the 75% of the cases whereas, B in 90% of the cases. In what % of the cases, are they likely to conradict each other in stating the same fact? Do you think that statement of B is true? (All India 2013)

Ans. Let event that A speaks truth br AT and,

Event that B speaks truth be BT

Given P (AT) = 75/ 100 

then, 

p of A

= 25/ 100

And, P(BT) = 90/ 100 

then,

p of B

Now, P (A and B are contradict to each other) 

Contradict

Therefore, we think that the statement of B may be false. 

Q 3. The probabilities of two students A and B coming to the school in time are 3/7 and 5/7 respectively. Considering that the events ‘A comin in time’ and ‘B coming in time’ are independent, find the probability of only one of them coming to the school in time. Write at least one advantage of coming to school in time. ( Value based question; Delhi 2013)

Ans. Given,the probability of student A coming to school in time,

P (A) = 3/7

Then probability of student A not coming in time,

probability of A

and the probability of student B coming in time,

P (B) = 5/7

Then probability of student B not coming in time,

Probability of B

Now, the required probability,

Probability

An advantage of coming to school in time: regular attendence and punctuality are important, if the students have to take the full advantage of learning opportunities offered by the school. Secondly, it is important to the educational process which encourages ahealthy pattern of work.

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CBSE CLASS XII Related Questions

  • 1.
    Let the polished side of the mirror be along the line \[ \frac{x}{1} = \frac{1 - y}{2} = \frac{2z - 4}{6}. \] A point \( P(1, 6, 3) \), some distance away from the mirror, has its image formed behind the mirror. Find the coordinates of the image point and the distance between the point \( P \) and its image.


      • 2.
        If \[ \begin{bmatrix} 4 + x & x - 1 \\ -2 & 3 \end{bmatrix} \] is a singular matrix, then the value of \( x \) is:

          • 0
          • 1
          • -2
          • -4

        • 3.
          If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ such that $|\vec{a}| = 3$, $|\vec{b}| = 5$, $|\vec{c}| = 7$, then find the angle between $\vec{a}$ and $\vec{b}$.


            • 4.
              Let \( A = \begin{bmatrix} 1 & -2 & -1 \\ 0 & 4 & -1 \\ -3 & 2 & 1 \end{bmatrix}, B = \begin{bmatrix} -5 \\ -2 \end{bmatrix}, C = [9 \ \ 7], \) which of the following is defined?

                • Only AB
                • Only AC
                • Only BA
                • All AB, AC and BA

              • 5.
                Consider the Linear Programming Problem, where the objective function \[ Z = x + 4y \] needs to be minimized subject to the following constraints: \[ 2x + y \geq 1000, \] \[ x + 2y \geq 800, \] \[ x \geq 0, \quad y \geq 0. \] Draw a neat graph of the feasible region and find the minimum value of $Z$.


                  • 6.
                    A spherical ball has a variable diameter $\frac{5}{2}(3x + 1)$. The rate of change of its volume w.r.t. $x$, when $x = 1$, is:

                      • $225\pi$
                      • $300\pi$
                      • $375\pi$
                      • $125\pi$
                    CBSE CLASS XII Previous Year Papers

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