PROBABILITY
It is very easy to understand the concept of probability through the example of tossing a coin. A coin when tossed, gives either heads or tails. The result of a toss cannot be certain. So, we estimate the chances of a particular result. The chance of getting a head when a coin is tossed is one in two or 50% or 1/2. This chance is known as probability.
A random experiment is an action that gives one or more results. Each result is called an outcome.
The sample space (S) is the set of all possible outcomes of an event.
The number of elements in the sample space is denoted by n(S).
Thus, tossing of a coin is a random
experiment. Getting heads or tails is the outcome. The sample space S
{Heads, Tails} and n(S) = 2.
Outcomes of a random experiment are equally likely when the occurrence of any one result is not expected over the other. A random experiment having equally likely outcomes is called an unbiased experiment.
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Find the number of elements in the sample space for the following experiments: (i)Tossing two coins together. (ii)Tossing three coins together. (iii)Tossing a coin four times. (iv)Rolling an unbiased die. (v)Rolling two unbiased dice and three unbiased dice. (vi)Drawing two cards at random from a pack of 52 well-shuffled cards.
Solution: i. Tossing one coin gives 2 outcomes—either Heads (H) or Tails (T).
S
ii. Tossing three coins gives 2
S
iii. When a coin is tossed four times, the possible outcomes for each toss = 2 Total number of possible outcomes = 2 It is difficult to list the elements of the sample space. But knowing the number of elements in the sample space is enough.
iv. Rolling an unbiased die gives 6 outcomes.
S
v. As rolling one die gives 6 outcomes,
rolling two dice gives 6
S (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Consider the case where a die is rolled three times. The possible outcomes for first, second and third throw of the dice each = 6. ∴ n(S) = 6 × 6 × 6 = 216
vi. Two cards can be drawn from a pack of 52 well shuffled cards in 52C2 ways.
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Consider the sample space of the experiment of rolling an unbiased die.
S
{1, 2, 3, 4, 5, 6}
Any subset of S is called an event.
For example, the event E of getting an odd number from the throw of a die is
E
{1, 3, 5}
n(E) = 3
If E
{ }, then the event is an impossible event.
For example, getting 7 by throwing a die is an impossible event.
If E
S, then the event is a certain event.
For example, getting a natural number less than 7 by throwing a die is a certain event.
Occurrence of Two or More Events:
The union of the events A and
B of a sample space (A 
B) is the event that either A or B or both take
place.
The intersection of the events
A and B of a sample space (A 
B) is the event that both A and B take
place.
Two events A and B of a sample space are mutually exclusive if they cannot occur simultaneously.
If A and B are mutually
exclusive, A
B
{ }
n(A
B) = 0
If two events A and B of a
sample space S are such that A
B
S, then A and B are called exhaustive
events.
If A and B are exhaustive,
A
B
S
n(A
B) = n(S)
An event (A) and its complement (A’) are exhaustive events.
Probability is the chance of occurrence of an event.
If S is the sample space of an unbiased experiment and if E is an event, then the probability that E takes place is
Consider tossing an unbiased coin. Let E be the event that the result is heads. Let E’ be the event that the result is tails. Then,
n(S) = 2, n(E) = 1 and n(E’) = 1
P(E) = 1/2 and P(E’) = 1/2
So when a coin is tossed, the probability of occurrence of heads is 1/2.
Similarly, the probability of occurrence of tails is 1/2.
Finding the probability of any event involves the following steps:
Step I: Finding the number of elements in the sample space.
Consider the event of rolling of a die.
S
{1, 2, 3, 4, 5, 6}
n(S) = 6
Step II: Finding the number of outcomes satisfying the required event.
Let A be the event that the die shows a multiple of 3. Then, the outcomes which satisfy this event A can be written as
A
{3, 6}
n(A) = 2
Step III: Finding the probability of the occurrence of the event, P(A).
In this case, P(A) = 2/6 = 1/3
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Find the probability that a number selected at random from the first 10 natural numbers is prime.
Solution: Since we are considering only the first 10 natural numbers (1 to 10), n(S) = 10. There are 4 numbers less than 10, which are prime, namely 2, 3, 5 and 7.
random from the first 10 natural numbers is prime is
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REMEMBER
- 0
P(E)
1
- If P(E) = 0, the event is an impossible event.
- If P(E) = 1, the event is a certain event.
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Two unbiased coins are tossed. What is the probability that both the tosses give different results?
Solution: When a coin is tossed we get either heads or tails. Let heads be denoted by H and tails be denoted by T. Tossing two coins gives 4 outcomes.
Hence, n(S) = 4 Let A be the event that both the tosses give different results. i.e. The first coin shows heads and the second shows tails OR the first coin shows tails and the second shows heads.
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If a die is tossed two times, what is the probability that the sum of the two throws is less than or equal to 4?
Solution: When a die is thrown twice, the sample space can be given as
S
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} Hence, n(S) = 36. Let B be the event that sum of the two throws is less than or equal to 4.
Hence, n(B) = 6
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A coin is tossed 7 times. What is the probability of getting exactly four heads?
Solution: Possible outcomes for each toss = 2 Hence, total possible outcomes n(S) = 27 = 128
Let H be the event of getting exactly 4 heads. The number of ways in which one can get 4 heads in 7 tosses = n(H) = 7C4 = 35.
Hence, the probability of getting exactly 4 heads is given by
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There are 6 people who stand together for a photograph. Find the probability that two of them stand together.
Solution: n(S) = The total number of ways in which 6 people can be arranged together for a photograph = 6P6 = 6! = 720 Let E be the event that two people always stand together. Let us assume that the two people who always stand together are one entity. These 2 people can be arranged amongst themselves in 2 ways. Now we have 5 people who can be arranged amongst themselves in 5P5 = 5! = 120 ways. Thus, the total number of ways in which 2 people
will stand together = 120 Hence, n(E) = 240
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A card is drawn at random from a pack of cards. Find the probability that it is a/an: i. spade ii. ace
Solution: Total cards = 52 One card can be drawn from 52 in 52C1 = 52 ways. Hence, n(S) = 52.
i. Let A be the event that the card drawn is a spade. Number of spades = 13 One spade can be selected from 13 in 13C1 = 13 ways. Hence, n(A) = 13.
ii. Let B be the event that the card drawn is an ace. Number of aces = 4. One ace can be selected from 4 in 4C1 = 4 ways. Hence, n(B) = 4.
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There is an urn containing 3 white and 4 red balls. Two balls are drawn from this urn. Find the probability that: 1) Both balls are red. 2) Both balls are of different colours.
Solution: Number of balls in the urn = 7 Two balls can be selected from these in 7C2 = 21 ways. n(S) = 21
i. Let R be the event that both balls drawn are red. Number of red balls = 4 Two red balls can be selected from these in 4C2 = 6 ways. Hence, n(R) = 6.
ii. Let B be the event that both balls are of different colours. Total number of red balls = 4 One red ball can be selected from these in 4C1 = 4 ways. Total number of white balls = 3 One white ball can be selected from these in 3C1 = 3 ways. Thus, one red and one white ball can be selected in 4 Thus, balls of different colours can be selected in 12 ways.
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Four boys and eight girls are seated in a row. What is the probability that no two boys sit next to each other?
Solution: 12 people can be arranged among themselves in 12! ways.
Girls can be seated in 8! ways. __ G __ G __ G __ G __ G __ G __ G __ G __ There are nine places next to the girls in which the boys can be seated.
n(E) = 8!
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We know that for two sets A and B,
n(A
B) = n(A) + n(B) – n(A
B)
If A and B are events of a sample space S, then dividing the equation by n(S), we have
P(A
B) = P(A) + P(B) – P(A
B)
If A and B are mutually exclusive events,
n(A
B) = 0
P(A
B) = 0
P(A
B) = P(A) + P(B)
If A and B are exhaustive
events, n(A
B) = n(S)
P(A) + P(B) = 1
An event and its complement are always exhaustive events.
P(A) + P(A’) = 1
P(A) = 1 – P(A’)
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Three dice are thrown simultaneously. What is the probability that the sum of the results of the three throws is more than 5?
Solution: As three dice are thrown simultaneously, n(S) = 63 = 216. Rather than calculating the number of ways in which the sum of the results of three throws is more than 5, it is simpler to calculate the number of ways in which the sum of the results of three throws is less than or equal to 5. Now, let L be the event that the throws give a sum of less than or equal to 5 and G be the event that the throws give a sum greater than 5. Clearly, n(L) + n(G) = n(S) Or in other words, L and G are exhaustive events.
L
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A card is drawn at random from a well shuffled pack of 52 cards.
X. The card drawn is black or a king. Y. card drawn is a club or a heart or a jack. Z. The card drawn is an ace or a diamond or a queen. Then which of the following is correct? [IIFT 2009]
(1) P(X) > P(Y) > P(Z) (2) P(X) >= P(Y) = P(Z) (3) P(X) = P(Y) > P(Z) (4) P(X) = P(Y) = P(Z)
Solution: X is the event that the card drawn is black or a king. Y is the event that the card drawn is a club or a heart or a jack. Z is the event that the card drawn is an ace or a diamond or a queen. Consider X. A black card can be drawn in 26 ways. A king can be drawn in 4 ways. A black king can be drawn in 2 ways.
Consider Y. A club can be drawn in 13 ways. A heart can be drawn in 13 ways. A jack can be drawn in 4 ways. A Jack of hearts can be drawn in 1 way and a jack of club can be drawn in 1 way.
Consider Z. An ace can be drawn in 4 ways. A queen can be drawn in 4 ways and a diamond can be drawn in 13 ways. Queen of Diamonds can be drawn in 1 way and ace of diamonds can be drawn in 1 way.
Hence, option 3. |
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In a survey conducted among 500 students in a class, it was found that apart from 100 students, all the others watched at least one of the three news channels I, II and III. 75 watched only III, 270 watched II and 40 watched only II and III. Out of the 190 students who watched more than one of the three channels, 50 watched all the three and 70 watched I and II but not III. What is the probability that a student picked at random watched only one of the channels?
Solution:
The given information is summarised in the venn diagram. n(S) = 500 n(E) = 25 + 110 + 75 = 210
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A shop sells 10 lamps out of which 3 are defective. Sachin buys four lamps. Find the probability that all of Sachin’s lamps work. Also find the probability that at least two of the lamps that he buys work.
Solution: n(S) = 10C4 = 210 7 of the 10 lamps are not defective.
n(L) = 7C4 = 35
We need the probability that at least two of his lamps work. The event that less than two of his lamps work, and the event that at least two of his lamps work, are exhaustive. So we calculate the probability that less than two of his lamps work and subtract it from 1. The probability that none of Sachin’s lamps work = 0 as there are only 3 defective lamps and he buys 4. If K is the probability that only one of Sachin’s lamps works, n(K) = 7C1
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Find the probability that a random arrangement of the all letters of the word PARAGRAPH starts and ends with A.
Solution: The word PARAGRAPH has 9 letters, out of which A is repeated thrice, R is repeated twice and P is repeated twice.
For the number of arrangements that start and end with A, consider the following situation; A __ __ __ __ __ __ __ A The letters between the two As are P, A, R, G, R, P and H in some order.
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Amit, Sumit and seven other people sit around a circular table. Find the probability that Amit and Sumit are not next to each other.
Solution: Nine people can be seated around a circular table in 8! ways.
To find the number of arrangements with Amit and Sumit sitting together, we consider them as one entity. Then there are 8 people who can be seated in 7! ways. Amit and Sumit can sit in 2 ways.
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A manufacturer claims that only 2% items are defective in a shipment of 200 items sent by him. A random sample of two items is drawn from the shipment of 200 items. What is the probability that both the items drawn are defective? [FMS 2009]
Solution: 2 items out of 200 items are selected. Hence, the Sample Space = 200C2 Since only 2% of the items are defective in a shipment of 200 items,
We need to find the probability that both items drawn are defective. 2 defective items can be drawn from 4 defective items in 4C2 ways.
Hence, option 2. |
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Historical sales data of a retail store indicate that 40 percent of all customers that enter the store make a purchase. Determine the probability that exactly two of the next three customers will make a purchase. [JMET 2009] (1) 0.388 (2) 0.667 (3) 0.400 (4) 0.288
Solution: The probability that a customer who entered the store makes a purchase is 0.4. The activity of any single customer making a
purchase is independent of the event of any other customer making a
purchase. The probability that 2 customers make a purchase and one does
not is (0.4)2 The two of the three people can be selected in 3C2 i.e. 3 ways. Required probability = 3 Hence, option 4. |
Odds in favour and odds against
Odds in Favour 
Odds Against 
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The odds in favour of a player winning a chess game are 2/3. Find the probability that he loses the game and the probability that he wins the game.
Solution:
= favourable cases + unfavourable cases = 2 + 3 = 5 Let A be the event that the player wins the game. Then A’ represents the event that he loses the game.
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The chance of India winning a cricket match against Australia is 1/6. What is the minimum number of matches India should play against Australia so that there is a fair chance of winning at least one match? [XAT 2010]
(1) 3(2) 4 (3) 5(4) 6 (5) None of the above
Solution: A fair chance of winning is more than 50% chance of winning.
We need to find a fair chance winning atleast one match. 
By trial and error, n = 4 Hence, option 2. |
Let us consider the following two cases.
1. We draw a card from a pack of 52 cards, replace it, shuffle the cards and draw another card.
2. We draw a card from a pack of 52 cards, keep it aside and draw another card.
Let us find the probability that the second card drawn is a spade.
Let the event that the first card drawn is a spade be called A, and let the event that the second card drawn is a spade be called B.
As we replace the first card drawn in case 1, the probability that the second card drawn is a spade does not depend on what the first card was.
However, in the second case, we do not replace the first card drawn. So, the probability that the second card drawn is a spade cannot be determined unless we know whether the first card was a spade or not. In other words, event B depends on event A.
Therefore, the probability that the second card drawn is a spade under the assumption that the first card was a spade, denoted by P(B/A) = 12/51.
Similarly, the probability that the second card drawn is a spade, under the assumption that the first card was not a spade, denoted by
P(B/A’) = 13/51.
P(B/A) and P(B/A’) are called conditional probabilities.
P(A
B) represents the probability that both the events A and B
occur and is given by:
P(A
B) = P(A).P(B/A), where P(A) ≠ 0.
Similarly,
P(A
B) = P(B).P(A/B), where P(B) ≠ 0.
If events A and B are independent,
P(B/A) = P(B) and P(A/B) = P(A)
P(A
B) = P(A) 
P(B)
The concept of conditional probability can be better understood from the following example.
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Two cards are drawn at random from a pack of 52 cards. What is the probability that one card is an ace and the other is a diamond?
Solution: Let E be the event that one card is an ace and the other is a diamond.
Now two mutually exclusive cases arise. 1. The ace is a diamond.
Now there are 12 diamonds left in 51 cards.
2. The ace is a non-diamond.
Now there are 13 diamonds left in 51 cards.
Also, n(S) = 52C2 As the cases are mutually exclusive,
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Two cards are drawn from a pack of cards one after the other without replacement. What is the probability that the first card picked up belongs to the suit of spades and the second one belongs to the suit of diamonds?
Solution: Let A be the event that the first card drawn is a spade and B be the event that the second card drawn is a diamond. Number of spades = 13
Now, number of cards left = 51 Number of diamonds = 13
= P(B/A) = 13/51 Thus, probability that the first card belongs to spades and the next to diamonds
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Bayes’ Theorem
P(A ∩ B ) = P(A)
P(B/A)... (i)
Similarly,
P(A ∩ B ) = P(A)
P(A/B)... (ii)
Equating (i) and (ii) we get
P(A)
P(B/A) = P(B)
P(A/B)
This is called the Bayes’ Theorem.
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An urn contains 3 red balls and 2 blue balls. Two balls are picked from the urn one after the other without replacing the first ball back in the urn. Find the probability that the second ball is blue.
Solution: The second ball has to be blue. The first ball can be red or blue. So, we have two cases.
Case 1: The first ball is red. The probability that first ball is red is P(R1) is 3/5. Now, there are 4 balls left—two red and two blue. The probability that the second ball is blue is P(B2/R1) = 2/4 = 1/2 Thus, the probability that the first ball is red and the second ball is blue
Case 2: The first ball is blue. The probability that the first ball is blue P(B1) is 2/5. Now, there are 4 balls left—three red and one blue. The probability that the second ball is blue P(B2/B1 )= 1/4. Thus, the probability that the first ball as well as second ball is blue
Thus, the second ball is blue if either case 1 or case 2 occurs. Thus, probability that the second ball is blue is given by
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There are three identical boxes 1, 2 and 3. Box 1 contains 3 red balls and 2 blue balls. Box 2 contains 2 red and 5 white balls and box 3 contains 2 blue, 1 red and 1 blue balls. One ball is drawn from one of the boxes at random. Find the probability that it is red.
Solution: The ball is drawn from box 1 or box 2 or box 3. The probability that a ball drawn from box 1 is red is 3/5. The probability that a ball drawn from box 2 is red is 2/7. The probability that a ball drawn from box 3 is red is 1/4. Probability of selecting any one of the three boxes = 1/3.
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The chance of Geeta winning a competition is 1/9 and that of Seema winning it is 2/15. Find the probability that one of Geeta and Seema wins it.
Solution: The probability that Seema or Geeta wins the competition = 1 – the probability that both of them do not win it. Probability that Geeta does not win the competition
Probability that Seema does not win the competition
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There are three similar boxes, containing (i) 6 black & 4 white balls; (ii) 3 black & 7 white balls and (iii) 5 black & 5 white balls, respectively. If you choose one of the three boxes at random and from that particular box pick up a ball at random, and find that to be black, what is the probability that the ball was picked up from the second box? [JMET 2010]
Solution: We know that the ball picked up was black. There are 6 + 3 + 5 = 14 black balls in all.
Hence, option 1. |
REMEMBER:
- The sample space (S) is the set of all possible outcomes of an event.
- Any subset of S is called an event.
- Probability is the chance of occurrence of an event.
- If P(E) = 0, the event is an impossible event.
- If P(E) = 1, the event is a certain event.
- P(A
B) = P(A) + P(B) – P(A
B)
- If A and B are mutually exclusive events, P(A
B)
= P(A) + P(B)
- If A and B are exhaustive events, P(A) + P(B) = 1
=P(A)=1 - P(A’)
- Odds in Favour
- Odds Against
- If A and B are dependent events,
P(A
B) = P(A).P(B/A), where P(A) ≠ 0 ,P (A
B) = P(B).P(A/B), where P(B) ≠ 0
- If events A and B are independent,
P(A
B) = P(A)
P(B)
