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Probability

Key Concepts

Solved Examples: 75

Practice Questions: 7

Solved Examples

Question 1: A coin is tossed once. Find the probability of getting a Head.

The sample space is:
S = {H, T}
Total number of outcomes = 2
Favourable outcome = {H}
Number of favourable outcomes = 1
Therefore, P(Head) = 1/2
So probability of getting a head is 1/2

Question 2: A standard die is rolled once. Find the probability of getting a number greater than 4.

The sample space is:
S = {1, 2, 3, 4, 5, 6}
Favourable outcomes are: {5, 6}
Number of favourable outcomes = 2
Total number of outcomes = 6
Therefore,
P(number greater than 4) = 2/6 = 1/3

Question 3: A number is selected at random from the set {1, 2, 3, 4, 5, 6, 7, 8}. Find the probability of selecting an odd number.

Sample space:
S = {1, 2, 3, 4, 5, 6, 7, 8}
Favourable outcomes:
{1, 3, 5, 7}
Number of favourable outcomes = 4
Total number of outcomes = 8
Therefore,
P(odd number) = 4/8 = 1/2

Question 4: One letter is selected at random from the word PROBABILITY. Find the probability of selecting the letter B.

The word PROBABILITY contains 11 letters:
P, R, O, B, A, B, I, L, I, T, Y
The letter B occurs 2 times.
Total number of possible outcomes = 11
Favourable outcomes = 2
Therefore,
P(B) = 2/11

Question 5: A number is selected at random from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Find the probability that the selected number is prime.

The sample space contains 10 outcomes.
The prime numbers are: {2, 3, 5, 7}
Number of favourable outcomes = 4
Total number of outcomes = 10
Therefore,
P(prime number) = 4/10 = 2/5

Question 6: A standard die is rolled once. Find the probability of the event of getting an even number.

The sample space is:
S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting an even number.
Favourable outcomes:
E = {2, 4, 6}
Number of favourable outcomes = 3
Total number of outcomes = 6
Therefore, P(E) = 3/6 = 1/2

Question 7: A bag contains 5 red balls, 3 blue balls and 2 green balls. One ball is selected at random. Find the probability of the event that the selected ball is red.

Total number of balls: 5 + 3 + 2 = 10
Let R be the event of selecting a red ball.
Favourable outcomes = 5
Total outcomes = 10
Therefore, P(R) = 5/10 = 1/2

Question 8: A number is selected at random from the numbers 1 to 12. Find the probability of the event that the selected number is a multiple of 3.

Sample space:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Let E be the event of selecting a multiple of 3.
Favourable outcomes:
E = {3, 6, 9, 12}
Number of favourable outcomes = 4
Total number of outcomes = 12
Therefore, P(E) = 4/12 = 1/3

Question 9: One letter is selected at random from the letters of the word MATHEMATICS. Find the probability of selecting a vowel.

The word MATHEMATICS contains 11 letters.
The vowels are: A, A, E, A, I
Number of favourable outcomes = 5
Total number of outcomes = 11
Therefore, P(vowel) = 5/11

Question 10: A number is selected at random from the set {1, 2, 3, …, 20}. Find the probability of the event that the selected number is greater than 15.

The favourable outcomes are:
{16, 17, 18, 19, 20}
Number of favourable outcomes = 5
Total number of outcomes = 20
Therefore, P(E) = 5/20 = 1/4

Question 11: A spinner is divided into 8 equal sections numbered 1 to 8. What is the probability of landing on a number that is a perfect square?

The perfect squares among 1 to 8 are 1 and 4.
Favourable outcomes = 2
Total outcomes = 8
P(perfect square) = 2/8 = 1/4

Question 12: A two-digit number is selected at random. What is the probability that the number is divisible by 10?

The two-digit multiples of 10 are:
10, 20, 30, 40, 50, 60, 70, 80, 90
Favourable outcomes = 9
Total two-digit numbers = 90
P(divisible by 10) = 9/90 = 1/10

Question 13: A standard die is rolled once. Find the probability of not getting a 6.

The probability of getting a 6 is: P(6) = 1/6
The event “not getting a 6” is its complement.
P(not getting 6) = 1 − P(6)
= 1 − 1/6
= 5/6

Question 14: A bag contains 7 red balls and 5 blue balls. One ball is selected at random. Find the probability of not selecting a red ball.

Total balls = 7 + 5 = 12
P(red) = 7/12
Therefore,
P(not red) = 1 − 7/12
= 5/12

Question 15: A number is selected at random from 1 to 20. Find the probability that the number is not prime.

There are 8 prime numbers from 1 to 20:
{2, 3, 5, 7, 11, 13, 17, 19}
Therefore, P(prime) = 8/20 = 2/5
Using the complementary event:
P(not prime) = 1 − 2/5 = 3/5

Question 16: One card is drawn at random from a standard deck of 52 playing cards. Find the probability of not drawing an Ace.

There are 4 Aces in a standard deck.
P(Ace) = 4/52 = 1/13
Therefore,
P(not Ace) = 1 − 1/13
= 12/13

Question 17: A fair coin is tossed three times. Find the probability of getting at least one Head.

Instead of counting all outcomes containing at least one Head, consider the complementary event: getting no Heads.
Total possible outcomes = 2³ = 8
The sample space is:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
The only outcome with no Heads is: TTT
Therefore, P(no Head) = 1/8
Hence,
P(at least one Head) = 1 − 1/8
= 7/8

Question 18: A standard die is rolled twice. Find the probability of getting at least one 6.

For each roll, there are 6 possible outcomes. Therefore, the total number of possible ordered outcomes is:
6 × 6 = 36
The complement of “at least one 6” is “no 6”.
For each die, there are 5 outcomes that are not 6:
{1, 2, 3, 4, 5}
Therefore, the number of outcomes with no 6 is:
5 × 5 = 25
So, P(no 6) = 25/36
Using the complementary event:
P(at least one 6) = 1 − 25/36 = 11/36

Question 19: A coin is tossed 50 times, and Heads occurs 28 times. Find the experimental probability of getting Heads.

Experimental probability is:
Experimental probability = Number of times the event occurs ÷ Total number of trials
Here,
Number of Heads = 28
Total tosses = 50
Therefore,
P(Heads) = 28/50 = 14/25

Question 20: A die is rolled 60 times. The number 5 appears 12 times. Find the experimental probability of getting 5.

Number of times 5 appears = 12
Total rolls = 60
Therefore, P(5) = 12/60 = 1/5

Question 21: A bag contains balls of different colours. A ball is drawn, recorded and replaced 80 times. A red ball is obtained 26 times. Find the experimental probability of drawing a red ball.

Number of times a red ball is obtained = 26
Total trials = 80
Therefore,
P(red) = 26/80 = 13/40

Question 22: A fair coin is tossed 100 times and gives 47 Heads. Find the experimental probability of Heads and compare it with the theoretical probability.

Experimental probability:
P(Heads) = 47/100
For a fair coin, the theoretical probability of Heads is:
P(Heads) = 1/2 = 50/100
Therefore,
Experimental probability = 47/100
Theoretical probability = 1/2
The experimental probability is close to, but not exactly equal to, the theoretical probability.

Question 23: A student tosses a coin 200 times and obtains 108 Heads. Based on this experimental probability, approximately how many Heads would the student expect in 500 tosses?

Experimental probability of Heads:
P(Heads) = 108/200 = 27/50
For 500 tosses, the estimated number of Heads is:
500 × 27/50 = 270

Question 24: A fair coin is tossed three times. Find the probability of getting at least one Head.

The sample space is:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total outcomes = 8
“At least one Head” means one or more Heads.
Favourable outcomes:
{HHH, HHT, HTH, HTT, THH, THT, TTH}
Number of favourable outcomes = 7
Therefore, P(at least one Head) = 7/8

Question 25: A fair coin is tossed three times. Find the probability of getting at most one Head.

“At most one Head” means zero or one Head.
The outcomes are: TTT, HTT, THT, TTH
Favourable outcomes = 4
Total outcomes = 8
Therefore,
P(at most one Head) = 4/8 = 1/2

Question 26: A standard die is rolled three times. Find the probability of getting at least two 6s.

“At least two” means two or more.
The possible patterns are:
Exactly two 6s
Exactly three 6s
Number of outcomes with exactly two 6s:
3 × 5 = 15
Number of outcomes with exactly three 6s: 1
Total favourable outcomes = 15 + 1 = 16
Total possible outcomes: 6³ = 216
Therefore,
P(at least two 6s) = 16/216 = 2/27

Question 27: A die is rolled three times. Find the probability of getting at most two even numbers.

A die has 3 even outcomes: 2, 4 and 6.
Therefore, P(even) = 3/6 = 1/2
“At most two even numbers” means zero, one or two even numbers.
The only excluded case is getting three even numbers.
P(three even numbers) = (1/2)³ = 1/8
Therefore, P(at most two even numbers)
= 1 − 1/8
= 7/8

Question 28: A fair coin is tossed four times. Find the probability of getting at least two Heads.

“At least two Heads” means two, three or four Heads.
Total outcomes = 2⁴ = 16
Number of outcomes with:
2 Heads = 6
3 Heads = 4
4 Heads = 1
Total favourable outcomes:
6 + 4 + 1 = 11
Therefore, P(at least two Heads) = 11/16

Question 29: A fair coin is tossed four times. Find the probability of getting at most two Tails.

“At most two Tails” means zero, one or two Tails.
Total outcomes = 2⁴ = 16
Number of outcomes with:
0 Tails = 1
1 Tail = 4
2 Tails = 6
Total favourable outcomes:
1 + 4 + 6 = 11
Therefore,
P(at most two Tails) = 11/16

Question 30: A number is selected at random from 1 to 10. Find the probability that the selected number is at least one of the even numbers.

The integers from 1 to 10 are:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Total number of possible outcomes = 10
The even integers are:
{2, 4, 6, 8, 10}
Favourable outcomes = 5
Therefore,
P(even number) = 5/10 = 1/2

Question 31: A die is rolled four times. Find the probability of getting at most three 6s.

“At most three 6s” means zero, one, two or three 6s.
The only excluded case is getting four 6s.
Total outcomes = 6⁴ = 1296
Favourable outcomes:
1296 − 1 = 1295
Therefore,
P(at most three 6s) = 1295/1296

Question 32: A student answers four True/False questions by guessing. Find the probability that the student gets at least three answers correct.

Each question has 2 possible outcomes: correct or incorrect.
Total possible answer patterns 2⁴ = 16
“At least three correct” means:
Exactly 3 correct
Exactly 4 correct
Exactly 3 correct = 4 outcomes
Exactly 4 correct = 1 outcome
Total favourable outcomes = 5
Therefore,
P(at least three correct) = 5/16

Question 33: A box contains 5 items, of which 2 are defective and 3 are good. Two items are selected without replacement. Find the probability of selecting at most one defective item.

“At most one defective” means zero or one defective item.
Total ways of selecting 2 items from 5: 10
Ways of selecting two good items: 3
Ways of selecting one defective and one good item: 2 × 3 = 6
Favourable selections: 3 + 6 = 9
Therefore,
P(at most one defective item) = 9/10

Question 34: A standard die is rolled once. Find the probability of getting an odd number or an even number.

Odd outcomes = {1, 3, 5}
Even outcomes = {2, 4, 6}
The two events are mutually exclusive because a number cannot be both odd and even.
P(odd) = 3/6 = 1/2
P(even) = 3/6 = 1/2
Therefore,
P(odd or even) = P(odd) + P(even)
= 1/2 + 1/2
= 1

Question 35: One card is drawn from a standard deck of 52 cards. Find the probability of drawing a King or a Queen.

There are 4 Kings and 4 Queens.
A card cannot be both a King and a Queen, so the events are mutually exclusive.
Favourable outcomes = 4 + 4 = 8
Total outcomes = 52
Therefore,
P(King or Queen) = 8/52
= 2/13

Question 36: A spinner has 8 equal sections: 3 red, 2 blue and 3 green. Find the probability of landing on red or blue.

Red and blue are mutually exclusive outcomes because one spin cannot land on both colours.
P(red) = 3/8
P(blue) = 2/8
Therefore,
P(red or blue) = 3/8 + 2/8
= 5/8

Question 37: One letter is selected at random from the word PROBABILITY. Find the probability of selecting a P or a T.

The word PROBABILITY has 11 letters.
There is 1 P and 1 T.
A letter cannot be both P and T, so the events are mutually exclusive.
Favourable outcomes = 1 + 1 = 2
Therefore,
P(P or T) = 2/11

Question 38: A standard die is rolled once. Find the probability of getting 1 or 6.

P(1) = 1/6
P(6) = 1/6
A die cannot show 1 and 6 at the same time, so the events are mutually exclusive.
Therefore,
P(1 or 6) = 1/6 + 1/6
= 1/3

Question 39: One card is drawn from a standard deck of 52 cards. Find the probability of drawing a Heart or a Spade.

There are 13 Hearts and 13 Spades.
A card cannot belong to both suits, so the events are mutually exclusive.
Favourable outcomes = 13 + 13 = 26
Therefore,
P(Heart or Spade) = 26/52 = 1/2

Question 40: A number is selected at random from the integers 1 to 10. Find the probability that the number is less than 5 or greater than 7.

Numbers less than 5: {1, 2, 3, 4}
Numbers greater than 7: {8, 9, 10}
The two events cannot occur together, so they are mutually exclusive.
Favourable outcomes = 4 + 3 = 7
Total outcomes = 10
Therefore,
P(less than 5 or greater than 7) = 7/10

Question 41: Two standard dice are rolled together. Find the probability that the sum is 8.

For two dice, the total number of possible ordered outcomes is: 6 × 6 = 36
The outcomes with a sum of 8 are:
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
Favourable outcomes = 5
Therefore,
P(sum = 8) = 5/36

Question 42: One card is drawn from a standard deck of 52 cards. Find the probability of drawing a red card or an Ace.

There are 26 red cards and 4 Aces.
However, 2 Aces are already included among the red cards: the Ace of Hearts and Ace of Diamonds.
Therefore, favourable cards:
26 + 4 − 2 = 28
Total cards = 52
Therefore,
P(red or Ace) = 28/52
= 7/13

Question 43: Two standard dice are rolled. Find the probability that at least one die shows 5.

Total possible outcomes: 6 × 6 = 36
It is easier to consider the complementary event: neither die shows 5.
Each die then has 5 possible outcomes: 5 × 5 = 25
Therefore,
P(neither die shows 5) = 25/36
Hence, P(at least one 5) = 1 − 25/36
= 11/36

Question 44: An integer is selected at random from 1 to 30. Find the probability that the number is a multiple of 3 or 5.

Multiples of 3:
{3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
There are 10.
Multiples of 5:
{5, 10, 15, 20, 25, 30}
There are 6.
Multiples of both 3 and 5: {15, 30}
There are 2.
Therefore, favourable outcomes:
10 + 6 − 2 = 14
Total outcomes = 30
Therefore,
P(multiple of 3 or 5) = 14/30 = 7/15

Question 45: One card is drawn from a standard deck of 52 cards. Find the probability that the card is neither a King nor a Queen.

There are 4 Kings and 4 Queens.
These are mutually exclusive, so:
Number of Kings or Queens = 4 + 4 = 8
Therefore, cards that are neither Kings nor Queens:
52 − 8 = 44
Hence,
P(neither King nor Queen) = 44/52
= 11/13

Question 46: Two standard dice are rolled. Find the probability that the sum is even or both dice show the same number.

Total outcomes = 36
An even sum occurs when both numbers have the same parity.
There are:
3 × 3 = 9 odd-odd outcomes
and
3 × 3 = 9 even-even outcomes.
Therefore, even-sum outcomes = 18.
There are 6 outcomes where both dice show the same number:
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
All these are already included among the even-sum outcomes.
Therefore, favourable outcomes: 18 + 6 − 6 = 18
Hence,
P(even sum or double) = 18/36
= 1/2

Question 47: Three standard dice are rolled. Find the probability that exactly two dice show the same number.

Total possible ordered outcomes:
6 × 6 × 6 = 216
For exactly two dice to show the same number:
Choose the repeated number = 6 ways.
Choose the different number = 5 ways.
Choose the position of the different number = 3 ways.
Therefore,
Favourable outcomes:
6 × 5 × 3 = 90
Hence,
P(exactly two dice show the same number)
= 90/216
= 5/12

Question 48: A bag contains 4 red balls and 3 blue balls. Two balls are drawn without replacement. Find the probability that both balls are red.

Total balls = 7
Probability of drawing a red ball first:
P(red first) = 4/7
After one red ball is removed, 3 red balls remain among 6 balls.
P(red second | red first) = 3/6
Therefore,
P(both red) = 4/7 × 3/6
= 12/42
= 2/7

Question 49: A box contains 4 white, 6 black and 5 yellow balls. Two balls are drawn one after another without replacement. What is the probability that the first ball is white and the second ball is black?

Probability of first ball being white = 4/15
After drawing a white ball, 14 balls remain, including 6 black balls.
Probability of second ball being black = 6/14 = 3/7
Therefore,
Probability = 4/15 × 3/7 = 4/35

Question 50: A fair die is rolled twice. What is the probability that the sum of the two numbers is 7?

There are 36 equally likely outcomes.
The favourable outcomes are:
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
Favourable outcomes = 6
Probability = 6/36 = 1/6

What is Probability?

Probability is the measure of how likely an event is to occur. It is expressed as a number between 0 and 1, where 0 means that an event is impossible and 1 means that an event is certain to occur. Events that are more likely to happen have higher probabilities, while events that are less likely to happen have lower probabilities. For equally likely outcomes, probability can be calculated by comparing the number of favourable outcomes with the total number of possible outcomes.

Outcomes and Sample Space

An outcome is a possible result of a random experiment. The sample space is the complete set of all possible outcomes.
For example, when a coin is tossed once, the possible outcomes are Head and Tail.
Sample Space: S = {H, T}
When a standard die is rolled once, the sample space is:
S = {1, 2, 3, 4, 5, 6}
Each possible result in the sample space is called an outcome.

Theoretical Probability

When all possible outcomes are equally likely, the probability of an event is found by comparing the number of favourable outcomes with the total number of possible outcomes.
Formula:
P(E) = Number of favourable outcomes / Total number of possible outcomes
or, P(E) = n(E) / n(S)
where E is the event and S is the sample space.
Example:
A standard die is rolled once. What is the probability of getting an even number?
Favourable outcomes = {2, 4, 6} = 3
Total outcomes = 6
Therefore, P(E) = 3/6 = 1/2

Probability of an Event

An event is a collection of one or more outcomes from a sample space. An event may contain a single outcome or several outcomes.
For example, when a die is rolled, getting a number greater than 4 is an event.
Favourable outcomes = {5, 6}
Therefore,
P(number greater than 4) = 2/6 = 1/3
The probability of an event always lies between 0 and 1:
0 ≤ P(E) ≤ 1
A probability of 0 means that an event is impossible, while a probability of 1 means that an event is certain.

Complementary Events

The complement of an event consists of all outcomes in the sample space that are not part of that event.
If E is an event, its complement is written as E′.
Formula: P(E′) = 1 − P(E)
Example:
A die is rolled once. What is the probability of not getting a 6?
P(getting a 6) = 1/6
Therefore,
P(not getting a 6) = 1 − 1/6 = 5/6
Thus, the probabilities of an event and its complement always add up to 1.

Experimental Probability

Experimental probability is based on the results obtained by actually performing an experiment.
Formula:
Experimental Probability = Number of times the event occurs / Total number of trials
Example:
A coin is tossed 50 times and lands on Heads 28 times.
Experimental probability of Heads:
P(H) = 28/50 = 14/25
Experimental probability may differ from theoretical probability because it is based on actual results. With a larger number of trials, experimental probability generally tends to get closer to the theoretical probability.

Probability of Mutually Exclusive Events

Two events are mutually exclusive if they cannot occur at the same time.
For mutually exclusive events A and B:
Formula: P(A or B) = P(A) + P(B)
Example:
A standard die is rolled once. What is the probability of getting either a 2 or a 5?
The two events cannot occur together.
P(2) = 1/6
P(5) = 1/6
Therefore,
P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3

Interpreting Probability Statements

Probability questions often use phrases such as at least, at most, greater than, and less than. These phrases must be interpreted carefully when identifying the favourable outcomes.
At least 4 means 4 or more.
At most 4 means 4 or less.
Greater than 4 means values above 4.
Less than 4 means values below 4.
Example:
A standard die is rolled once. What is the probability of getting at least 4?
Favourable outcomes = {4, 5, 6} = 3
Therefore,
P(at least 4) = 3/6 = 1/2.

Summary of Probability

Concept	Key Formula / Rule
Outcome	A possible result of a random experiment.
Sample Space	The complete set of all possible outcomes of a random experiment.
Event	A collection of one or more outcomes from the sample space.
Theoretical Probability	P(E) = Number of favourable outcomes ÷ Total number of possible outcomes
Probability Range	0 ≤ P(E) ≤ 1
Impossible Event	An event that cannot occur has probability 0.
Certain Event	An event that is certain to occur has probability 1.
Complementary Event	P(E′) = 1 − P(E)
Event and Its Complement	P(E) + P(E′) = 1
Experimental Probability	Experimental Probability = Number of times the event occurs ÷ Total number of trials
Mutually Exclusive Events	Events that cannot occur at the same time.
Probability of Mutually Exclusive Events	P(A or B) = P(A) + P(B)
At Least	“At least n” means n or greater.
At Most	“At most n” means n or less.
Greater Than	“Greater than n” excludes n and includes values above n.
Less Than	“Less than n” excludes n and includes values below n.

Common Mistakes

Students often confuse an outcome with an event. An outcome is a single possible result, while an event can contain one or more outcomes.
Forgetting to include all possible outcomes in the sample space. Before calculating probability, carefully list the complete set of possible outcomes.
Using the wrong denominator in theoretical probability. The denominator must represent the total number of equally likely possible outcomes.
Confusing favourable outcomes with total outcomes. The numerator represents only the outcomes that satisfy the required event.
Forgetting that probability must lie between 0 and 1. A probability cannot be negative or greater than 1.
Confusing an impossible event with a certain event. An impossible event has probability 0, while a certain event has probability 1.
Forgetting to include the boundary value in phrases such as “at least” and “at most”. “At least 5” includes 5, while “at most 5” also includes 5.
Confusing “greater than” with “greater than or equal to”. “Greater than 4” excludes 4, whereas “at least 4” includes 4.
Confusing theoretical probability with experimental probability. Theoretical probability is based on equally likely possible outcomes, while experimental probability is based on actual results from trials.
Adding probabilities of events that are not mutually exclusive. The simple rule P(A or B) = P(A) + P(B) applies when the events cannot occur together.
Forgetting to use the complement when it makes a problem simpler. Remember that P(E′) = 1 − P(E).
Assuming that experimental probability must exactly equal theoretical probability. Actual results can differ from theoretical probability, especially when the number of trials is small.

Practice Questions

Question 1: A bag contains 5 red balls and 7 blue balls. Two balls are drawn successively without replacement. What is the probability that both balls are blue?

Question 2: A number is chosen at random from 1 to 30. What is the probability that it is a perfect square?

Question 3: A card is drawn from a standard deck of 52 cards. What is the probability that the card is neither a King nor a Queen?

Question 4: A fair coin is tossed three times. What is the probability of getting exactly two heads?

Question 5: A fair die is rolled three times. What is the probability that the sum of the numbers obtained is greater than 12 but less than 16?

Question 6: A number is selected at random from 1 to 100. What is the probability that it is divisible by 4 or 6 but not by 12?

Question 7: A fair coin is tossed five times. What is the probability that the number of heads is greater than the number of tails?