Random Experiment
An experiment, whose all possible outcomes are known in advance but the outcome of any specific performance can not predicted before the completion of the experiment, is known as random experiment.
Sample space and Sample Point
A set of all possible outcomes associated with same random experiment is called sample space and is usually denoted by ‘S’. Each element of a sample space is called a sample point.
Event
An event is a subset of sample space.
Simple Event and Compound Event
If an event is a set containing only one element of the sample space, then it is called a
simple event.
A compound event is one that can be represented as a union of simple events.
Probability
If a random experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes favour A, then the probability of event A, P(A) = n / N =
no. of favourable cases / total number of cases
0 ≤ P(A) ≤ 1, P(ϕ) = 0 and P(S) = 1.
Ex.: In a single cast with two fair dice the chance of throwing two 4’s is
(A) 1/18
(B) 1/9
(C) 1/36
(D) 1/12
Solution: (C).
There are 6 × 6 equally likely cases (as any face of any die may turn up) ⇒ 36 possible outcomes. For this event, only one outcome (4-4) is favourable.
∴ Probability = 1/36.
Mutually Exclusive Events
A set of events is said to be mutually exclusive if the occurrence of one of them precludes the occurrence of any of the other events.
Independent Events
Events are said to be independent if the occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of other.
Exhaustive Event
A set of events is said to be exhaustive if the performance of random experiment always result in the occurrence of atleast one of them.
SET THEORETIC PRINCIPLES
If ‘A’ and ‘B’ be any two events of the sample space then
- 1. A ∪ B would stand for occurrence of at least one of them.
- 2. A ∩ B stands for simultaneous occurrence of A and B.
- 3. &overline;A (or A') stands for non-occurrence of A
- 4. &overline;A ∩ &overline;B (or A' ∩ B') stands for non-occurrence of both A and B.
- 5. If A and B are any two events, Then P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
- 6. P(A') = 1 - P(A)
- 7. P(A' ∪ B') = 1 - P(A ∩ B)
- 8. If A, B, C are any three events of the sample space, then P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
- 9. If out of m + n equally likely, mutually exclusive and exhaustive cases, m cases are favourable to an event A and n are not favourable to the event A, m : n is called odds in favour of A, n : m is called odds against the event A.
Ex.: From a pack of 52 cards two cards are drawn at random. Then the probability that one card is of spade and one card is of diamond is
(A) 3/17
(B) 2/17
(C) 13/102
(D) none of these
Solution:(C).
Let E2 be the event that one card is of spade and one is of diamond then n(E2) = number of elements in E2 = number of ways in which one card of spade can be selected out of 13 spade cards and one card of diamond can be selected out of 13 diamond cards. = 13C1 × 13C1 = 13 × 13 = 169
∴ P(E2) = n(E2)/n(S) = 169/1326 = 13/102
CONDITIONAL PROBABILITY
The probability of occurrence of an event B when it is known that some event A has occurred is called a conditional probability and is denoted by P(B/A). The symbol P(B/A) is usually read '‘the probability that B occurs given that A occurs” or ”simply probability of B, given A”.
Consider two events ‘A’ and ‘B’ of sample space S. When it is known that event ‘A’ has occurred, it means that sample space would reduce to the sample points representing event A. Now for P(B/A) we must look for the sample points representing the simultaneous occurrence of A and B i.e. sample points in A∩B.
⇒ P(B/A) = n(A ∩ B)n(A) = n(A ∩ B)n(S)n(A)n(S) = P(A ∩ B)P(A)
Thus P(B/A) = P(A ∩ B) P(A) , where 0 < P(A) ≤ 1; Similarly, P(A/B) = P(A ∩ B) P(B) , 0 < P(B) ≤ 1
Hence, P(A ∩ B) = { P(A) · P(B / A), P(A) > 0 P(B) · P(A / B), P(B) > 0 }
Independent Events
Two events A and B are said to be independent if occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of the other,
i.e. P(B/A) = P(B), P(A) ≠ 0 similarly P(A/B) = P(A), P(B) ≠ 0
P(B/A) = P(A ∩ B) / P(A) = P(B) ⇒ P(A ∩ B) = P(A) · P(B)
Ex.: An event A1 can happen with probability 0.4 and event A2 can happen with probability 0.3, then the probability that exactly one of them happens is
(A) 0.24
(B) 0.7
(C) 0.46
(D) none of these
Solution: (C).
The probability that A1 happens is p1 = 0.4
∴ The probability that A1 fails is 1 − p1 = 0.6
Also the probability that A2 happens is p2 = 0.3
Now, the chance that A1 happens and A2 fails is p1(1 − p2) and the chance that A1 fails and A2 happens is p2(1 − p1)
The probability that one and only one of them happens is
p1(1 - p2) + p2(1 - p1) = p1 + p2 - 2p1p2 = 0.46
Total Probability Theorem and Bayes' Theorem
Partition of Sample Space
Consider a sample space ‘S’. Let A1, A2, ⋯, An be the set of mutually exclusive and exhaustive subsets of sample space S.
These A1, A2, ⋯, An events are said to partition the sample space into n parts.
We have Ai ∩ Aj = ∅ for i ≠ j, 1 ≤ i, j ≤ n
And ∑ni=1 P(Ai) = 1
Total Probability Theorem
Let ‘A’ be any event of S. We can write A = (A1 ∩ A) ∪ (A2 ∩ A) ∪ … ∪ (An ∩ A). As A1, A2, …, An are mutually exclusive, (A1 ∩ A), (A2 ∩ A), …, (An ∩ A) would also be mutually exclusive.
This is known as the total probability of the event A.
Bayes’ Theorem
This theorem at times is also called inverse probability theorem.
Let us consider any event ‘A’ of sample space ‘S’ (as in the previous section). This event would have occurred due to the different causes (or due to the occurrence of any of the event A₁, A₂, ⋯ Aₙ).
Now, let us say that event A is found to have occurred and we have to find the probability that it has occurred to the occurrence of cause, say Ai.
That means we are interested in P(Ai/A). These types of problems are solved with the help of Bayes’ theorem.
From total probability theorem we get P(A) = ∑ni=1 P(Ai) · P(A | Ai)
Also, P(Ai/A) = [P(Ai ∩ A)] / P(A) = [P(Ai) · P(A | Ai)] / P(A)
⇒ P(Ai/A) = [P(Ai) · P(A | Ai)] / [∑ni=1 P(Ai) · P(A | Ai)]
This result is known as Bayes’ theorem.
BINOMIAL TRIALS AND BINOMIAL DISTRIBUTION
A probability distribution representing the binomial trials is said to be binomial distribution.
Let us consider a Binomial experiment which has been repeated ‘n’ times. Let the probability of success and failure in any trial be p and q respectively and we are interested in the probability of occurrence of exactly ‘r’ successes in these n trials. Now number of ways of choosing ‘r’ success in ‘n’ trials = nCr. Probability of ‘r’ successes and (n-r) failures is pr×qn-r. Thus probability of having exactly r successes = nCr×pr×qn-r.
Formulas and Concepts
- If a random experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes correspond to A, then the probability of event A,
P(A) = n / N. - If ‘A’ and ‘B’ be any two events of the sample space then
- A ∪ B would stand for occurrence of at least one of them.
- A (or A′ ∩ B′) stands for non-occurrence of both A and B.
- A ⊆ B stands for ‘the occurrence of A implies the occurrence of B’.
- A̅ (or A′) stands for non-occurrence of A.
- A̅ ∩ B̅ – P(A ∩ B)
- P(A′ ∪ B′) = 1 – P(A ∩ B)
- If out of m + n equally likely, mutually exclusive and exhaustive cases, m cases are favourable to an event and n are not favourable to the event A,
m : n is called odds in favour of A, n : m is called odds against the event A.
- The probability of occurrence of an event B when it is known that some event A has occurred is called a condition probability and is denoted by P(B/A).
P(B/A) = P(A ∩ B) / P(A), 0 < P(A) ≤ 1,P(A ∩ B) =P(A) · P(B/A), P(A) > 0 P(B) · P(A/B), P(B) > 0
- Let us consider a Binomial experiment which has been repeated ‘n’ times. Let the probability of success and failure in any trial be p and q respectively and we are interested in the probability of occurrence of exactly ‘r’ successes in these n trials. Now number of ways of choosing ‘r’ success in ‘n’ trials = nCr. Probability of ‘r’ successes and (n–r) failures is pr·qn–r. Thus probability of having exactly r successes = nCr·pr·qn–r
- Let us consider any event ‘A’ of sample space ‘S’ (as in the previous section). Let us say that event A is found to have occurred and we have to find the probability that it has occurred to the occurrence of cause, say Ai.
SOLVED EXAMPLES
A number of six digits is written down at random. Probability that sum of digits of the number is even is
(A)1/2
(B) 3/8
(C) 3/7
(D) none of these
Sol. (A).
As we are considering all possible six digit numbers, half of these would have sum of their digits to be even and half of these would have sum of their digits to be odd. Hence the required probability would be 1/2.
Two small squares on a chess board are chosen at random. Probability that they have a common side is,
(A) 1/3
(B) 1/9
(C) 1/18
(D) none of these
Sol. (C)
There are 64 small squares on a chessboard.
⇒ Total number of ways to choose two squares = 64C2 = 32.63
For favourable ways we must choose two consecutive small squares for any row or any column. ⇒ Number of favourable ways = (7×8)×2
favourable ways = (7×8)×2 ⇒ Required probability = 7×8×2 ⁄ 32.63 = 1 ⁄ 18
3. Fifteen coupons are numbered 1, 2, 3, … 15. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on the selected coupon is 9, is
(A) (9⁄16)6
(B) (8⁄15)7
(C) (3⁄5)7
(D) none of these
Sol. (D) Total ways = 157
For favorable ways, we must have 7 coupons numbered from 1 to 9 so that '9' is selected at least once. Thus total number of favorable ways are, 97 – 87
⇒ Required probability = 97 – 87⁄157
4. A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to getting 9 heads, then the probability of getting 2 heads is,
(A) 15/28
(B) 2/15
(C) 15/213
(D) none of these
Sol. (C). Let coin be tossed 'n' times and X be the random variable representing the number of heads appearing in 'n' trials.
P(X = 7) = P(X = 9); ⇒ nC7(1/2)n-7 = nC9(1/2)n-9 × (1/2)9
⇒ nC7 = nC9 ⇒ n = 16; Now P(X = 2) = 16C2(1/2)2(1/2)14
= 16C2/216 = (16×15)/217 = 15/213.
5. If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers is
(A) 2/7
(B) 32/343
(C) 12/49
(D) none of these
Sol. (D) Total number of ways in which 4 papers can be distributed among 7 teachers = 74
Now exactly 2 teachers out of 7 can be chosen in 7C2 ways. And total number of ways in which 4 papers can be given to these 2 teachers (each one getting at least one) = (24–2) =14
⇒ Total number of ways in which exactly 2 teachers check all four papers = 7C2 × 14 = 21 × 14
⇒ Required probability = 21 ⋅ 14 7 4 = 3 ⋅ 2 7 2 = 6 49 .
6. Events A, B, C are mutually exclusive events such that 
.
The set of possible values of x are in the interval
(A) [0, 1]
(B) [1/3,1/2)
(C) [1/3,2/3)
(D) [1/3,13/3)
Sol. (B). Since probability of an event is a non-negative number not more than 1, therefore, we must have
0≤ 3x+1/3 ≤ 1 …..(1) 0≤ 1-x/4 ≤1 …(2) and 0≤ 1-2x/2 ≤1 …..(3)
Also 0 ≤ P(A) + P(B) + P(C) ≤ 1 ( Events are M.E.)
⇒ 0≤ 3x+1/3 + 1-x/4 + 1-2x/2≤1 ⇒ 0 ≤ 13 - 3x ≤ 12 ⇒ -13 ≤ -3X ≤ -1
⇒1/3≤ x ≤13/3 …..(4)
From (1), (2) and (3), we have
and 
Hence 
7. There are three events A, B and C and three statements are made
(i) P(B) = 3/4
(ii) P(Ā◠B◠C̄)=1/3
(iii) P(Ā◠B◠ )=1/3,then value of P(B◠C) is equal to
(A) 1/12 (B) 7/12
(C) 5/12 (D) 11/21
Sol. (A). From the diagram it is clear that
P(B)=P(Ā◠B◠C̄)+ P(A◠B◠C̄)+ P(B◠C)
⇒3/4=1/3+1/3+P(B◠C)
⇒P(B◠C)=1/12
8. A bag contains 5 white and 3 black balls. 4 balls are successively drawn out and not replaced. The probability that they are alternately of different colours is
(A) 1/196 (B) 2/7
(C) 1/7 (D) 13/56
Sol. (C). Required probability
= P(WBWB) + P(BWBW)
= 5/8 x 3/7 x 4/6 x 2/5 + 3/8 x 5/7 x 2/6 x 4/5
= 1/7.
9. From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24 is
(A) 6/25 (B)1/4
(C) 1/6 (D) 1/5
Sol. (D). Let A: ‘’number is divisible by 6’’ and B: ‘’number is divisible by 8’’ then A ◠ B: ‘’number is divisible by 24’’.
⸫Required probability
= P(A OR B) – P(A ◠ B)
= P(A) + P(B) – P(A◠B) – P(A◠B)
=16/100+12/100-2(4/100)=20/100=1/5.
(Note that a number is divisible both by 6 and 8 iff it is divisible by L.C.M. of 6 and 8 i.e. 24. From 1 to 100, there are 16 multiples of 6, 12 multiples of 8 and 4 multiples of 24)
10. Statistics show that a specific disease is fatal in 10% cases. Find the probability that out of six patients in a hospital, only 3 will die is
(A) 1458 x 10-5 (B) 1458 x 10-6
(C) 41 x 10-6 (D) 8748 x 10-5
Sol. ( )It is a case of bernoullian trials where n = 6 and success may be defined as‘’ disease is fatal’’ so that
p=10/100=1/10 and q = 1 -1/10=9/10.
Required probability = P (3 successes)
=6C3p2q3 =20(1/10)3(9/10)3=14580/106 .
ASSIGNMENT
1. A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of the determinant chosen is negative, is
(A) 1/8 (B) 3/16
(C) 1/4 (D) 1/2
2. India play two matches each with West Indies and Australia .In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points is
(A) 0.8750 (B) 0.0875
(C) 0.0625 (D) 0.0250
3. A speaks the truth in 70% cases and B speaks the truth in 80% cases. The probability that they will contradict each other in describing a single event is
(A) 0.38 (B) 0.56
(C) 0.62 (D) 0.94
4. A man is known to speak the truth 3 out if 4 times. He throws a die and reports that it is a six. The probability that it is actually a six is
(A) 3/8 (B) 1/5
(C) 3/4 (D) none of these
5. A fair coin is tossed 99 times. Let X be the number of times heads occurs. Then P(X=r) is maximum when r is
(A) 49 (B) 52
(C) 51 (D) None of these
6. If a number of two digits is formed with the digits 2, 3, 5, 7, 9 without repetition of digits, then the probability that the number formed is 35 is
(A) 1/10 (B) 1/20
(C) 1/30 (D) none of these
7. If A and B are two events, the probability that exactly one of them occurs is given by
(A) P(A) + P(B) – 2 P(A◠B) (B) P(A◠B') – P(A'◠B)
(C) P(A◠B) + P(A◠B) (D) P(A') + P(B') + 2 P(A'◠B')
8. Suppose n(≥ 3) persons are sitting in a row. Two of them are selected at random. The probability that they are not together is
(A) 1 - 2/n (B) 2/n-1
(C) 1 - 1/n (D) none of these
9. A natural number x is chosen at random from the first 100 natural numbers. The probability that x + 100/x > 50 is
(A) 1/10 (B) 11/50
(C) 11/20 (D) none of these
10. If A and B are independent events such that P (A) = 0.3, P (A ◡ B̄ ) = 0.8, then P (B) equals to
(A) 1/7 (B) 2/7
(C) 3/4 (D) 1/4
11. A bag contains 4 tickets numbered 1, 2,3,4 and another bag contains 6 tickets numbered 2,4,6,7,8,9 .One bag is chosen and a ticket is drawn . The probability that the ticket bears the number 4 is
(A) 1/48 (B) 1/8
(C) 5/24 (D) none of these
12. A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let Ei (i = 1, 2, 3) denote the event that the digit on the ticket is
2. Then
(A) E1 and E2 are independent (B) E2 and E3 are independent
(C) E3 and E1 are independent (D) all of these
13. A four figure number is formed of the figures 1,2,3,5 with no repetitions. The probability that the number is divisible by 5 is
(A) 3/4 (B) 1/4
(C) 1/8 (D) none of these
14. Three natural numbers are taken at random from the set A= {x : 1 ≤ x ≤ 100, x∈N. then the probability that AM of the numbers taken is 25 is
15. A student appears for tests I, II and III. The student is successful, if he passes either in tests I and II or in tests I and III. The probabilities of the student passing in tests I, II and III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2 , then
(A) p = q = 1 (B) p = 0, q = 1
(C) p = 1, q = 0 (D) p = 1, q = 1/2
ANSWER TO ASSIGNMENT PROBLEMS
1. B 2. B 3. A
4. A 5. A 6. B
7. A 8. A 9. C
10. D 11. C 12. C
13. B 14. C 15. C