Question 1

Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = ___________.
(ii) The probability of an event that cannot happen is __________. Such an event is called ________.
(iii) The probability of an event that is certain to happen is _________. Such an event is called _________.
(iv) The sum of the probabilities of all the elementary events of an experiment is __________.
(v) The probability of an event is greater than or equal to ___ and less than or equal to __________.

Solution :

(i) Probability of an event E + Probability of the event ‘not E’ = 1 .

(ii) The probability of an event that cannot happen is 0. Such an event is called an impossible event.

(iii) The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event .

(iv) The sum of the probabilities of all the elementary events of an experiment is 1.

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2

Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to a Solution – a true-false question. The solution is right or wrong.
(iv) A baby is born. It is a boy or a girl.

Solution :

(i) The probability of a car starting depends on various factors such as its maintenance, fuel level, and weather conditions. If the car is well-maintained and has gas, it's very likely to start. If it's a very old car with a bad battery, it's very likely to not start. Therefore, the outcomes aren't equally likely.

(ii) Not Equally Likely. The likelihood of a player making a shot depends on their skill level, the distance from the hoop, and whether they are being guarded. Hence, the two outcomes are not equally likely.

(iii) Not Equally Likely. The probability of getting the right answer depends on the person's knowledge.

If a person knows the correct answer, the probability of them getting it right is close to 1. If they have no idea and are just guessing, the probability of getting it right is 0.5. Since the probability can vary, the outcomes are not equally likely.

(iv) Equally Likely. The birth of a boy or a girl is considered a random event with two possible outcomes.

The probability of having a boy is approximately 0.5, and the probability of having a girl is also approximately 0.5. Therefore, the two outcomes are considered equally likely.

Question 3

Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution :

Tossing a coin is considered a fair way to decide which team gets the ball because it provides two equally likely outcomes: heads or tails.

Each outcome has a probability of 1/2 or 50%, which means neither team has an inherent advantage.

This makes it an unbiased method of random selection.

Equal Probability: A standard, unweighted coin has a symmetrical design, ensuring that it is just as likely to land on one side as the other. This equal chance is the core of its fairness.

Question 4

Which of the following cannot be the probability of an event?

(A) 2/3, (B) -1.5, (C) 15%, (D) 0.7

Solution :

A probability of an event must always be a value between 0 and 1, inclusive. This can be expressed as:

0 ≤ Probability ≤ 1

Let's examine each option:

(A) 2/3 : This value is approximately 0.67, which is between 0 and 1. So, it can be a probability.

(B) -1.5: This value is less than 0. The probability of an event can never be negative.

(C) 15%: This can be written as 0.15, which is between 0 and 1. So, it can be a probability.

(D) 0.7: This value is between 0 and 1. So, it can be a probability.

Therefore, -1.5 cannot be the probability of an event.

Question 5

If P(E) = 0.05, what is the probability of ‘not E’?

Solution :

The probability of an event happening plus the probability of the event not happening is always equal to 1. This is because there are only two possible outcomes: the event occurs or it doesn't.

This relationship is expressed by the formula:

P(E) + P(not E) = 1

To find the probability of 'not E', you can rearrange the formula:

P(not E) = 1 − P(E)

Given that P(E) = 0.05:

P(not E) = 1 − 0.05 = 0.95

The probability of 'not E' is 0.95.

Question 6

A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out.

(i) an orange flavoured candy?
(ii) a lemon flavoured candy?

Solution :

(i) The probability of taking out an orange flavoured candy is 0.

Since the bag contains only lemon flavoured candies, it is impossible to draw an orange one. An event that is impossible has a probability of 0.

(ii) The probability of taking out a lemon flavoured candy is 1.

Because the bag contains only lemon flavoured candies, drawing a lemon flavoured candy is a certain event. An event that is certain to happen has a probability of 1.

Question 7

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution :

Let the event wherein 2 students having the same birthday be E

The relationship is:

P(E) + P(not E) = 1

Given that P(not E) = 0.992,

P(E) + 0.992 = 1

P(E) = 1 - 0.992

P(E) = 0.008

The probability that the two students have the same birthday is 0.008.

Question 8

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(i) red ?
(ii) not red ?

Solution :

The total number of balls in the bag is the sum of the red and black balls:

Total balls = 3 (red) + 5 (black) = 8

(i) Probability of drawing a red ball

The number of favorable outcomes is the number of red balls, which is 3.

The total number of possible outcomes is the total number of balls, which is 8.

P(red) = Number of favourable outcomes / Number of possible outcomes ​

P(red) = $ {3 \over 8 }$

(ii) Probability of the ball being not red

The event "not red" is the same as drawing a black ball, since those are the only other balls in the bag.

The number of favorable outcomes for "not red" is the number of black balls, which is 5.

P(not red ) = Number of favourable outcomes / Number of possible outcomes ​

P( not red) = $ {5 \over 8 }$

Question 9

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

(i) red ?
(ii) white ?
(iii) not green ?

Solution :

The total number of marbles in the box is the sum of all the marbles:

Total marbles = 5 (red) + 8 (white) + 4 (green) = 17 marbles 

(i) Probability of a red marblel

The number of favorable outcomes is the number of red marbles, which is 5.

The total number of possible outcomes is the total number of marbles, which is 17.

P(red) = Number of favourable outcomes / Number of possible outcomes ​

P(red) = $ {5 \over 17 }$

(ii) Probability of a white marble

The number of favorable outcomes is the number of white marbles, which is 8.

P( white ) = Number of favourable outcomes / Number of possible outcomes ​

P( white) = $ {8 \over 17 }$

(iii) Probability of a marble not being green

Number of marbles not green = (Number of red marbles) + (Number of white marbles) = 5 + 8 = 13.

P( not green ) = Number of favourable outcomes / Number of possible outcomes ​

P( not green ) = $ {13 \over 17 }$

Question 10

A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a 50 p coin ?
(ii) will not be a ₹ 5 coin ?

Solution :

The total number of coins in the piggy bank is the sum of all the different coins:

Total coins = 100(50p) + 50(₹1) + 20(₹2) + 10(₹5) = 180 coins. 

(i) Probability of being a 50p coin

The number of favorable outcomes is the number of 50p coins, which is 100.

The total number of possible outcomes is the total number of coins, which is 180.

P(50p coin) = Number of favourable outcomes / Number of possible outcomes ​

P(50p coin) = $ {100 \over 180}$ = $ {5 \over 9 }$

(ii) Probability of not being a ₹5 coin

The coins that are not ₹5 are the 50p, ₹1, and ₹2 coins..

Number of favorable coins = 100 + 50 + 20 = 170

P( not a ₹5 coin ) = Number of favourable outcomes / Number of possible outcomes ​

P( not a ₹5 coin) = $ {170 \over 180 }$ = $ {17 \over 18 }$

Question 11

Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution :

There are 5 male fish and 8 female fish, so the total number of fish is 5 + 8 = 13.

Number of favorable outcomes: The number of male fish is 5.

P(male fish) = Number of favourable outcomes / Number of possible outcomes ​

P(male fish) = $ {5 \over 13}$

Hence, probability of taking out a male fish = $ {5 \over 13}$

Question 12

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at

(i) 8 ?
(ii) an odd number ?
(iii)a number greater than 2 ?
(iv)a number less than 9 ?

Solution :

The total number of possible outcomes on the spinner is 8, as the arrow can come to rest at any of the numbers from 1 to 8.

(i) Probability of it pointing at 8

There is only one number 8 on the spinner.

P(pointing at 8) = Number of favourable outcomes / Number of possible outcomes ​

P(pointing at 8) = $ {1 \over 8 }$

(ii) Probability of an odd number

The odd numbers on the spinner are 1, 3, 5, and 7. There are 4 odd numbers.

P(odd number ) = Number of favourable outcomes / Number of possible outcomes ​

P( odd number) = $ {4 \over 8 }$ = $ {1 \over 2 }$

(iii) Probability of a number greater than 2

The numbers on the spinner that are greater than 2 are 3, 4, 5, 6, 7, and 8. There are 6 such numbers.

P( number greater than 2 ) = Number of favourable outcomes / Number of possible outcomes ​

P( number greater than 2 ) = $ {6 \over 8 }$ = $ {3 \over 4 }$

(iv) Probability of a number less than 9

All the numbers on the spinner (1, 2, 3, 4, 5, 6, 7, 8) are less than 9. There are 8 such numbers.

P( number less than 9 ) = Number of favourable outcomes / Number of possible outcomes ​

P( number less than 9 ) = $ {8 \over 8 }$ = $ 1 $

Question 13

A die is thrown once. Find the probability of getting

(i) a prime number
(ii) a number lying between 2 and 6
(iii) an odd number

Solution :

The total number of possible outcomes when a standard die is thrown is 6 (the numbers 1, 2, 3, 4, 5, 6).

(i) A prime number

The prime numbers on a die are numbers greater than 1 that can only be divided by 1 and themselves. The prime numbers are: 2, 3, 5.

P(prime number) = Number of favourable outcomes / Number of possible outcomes ​

P(prime number) = $ {3 \over 6 }$ = $ {1 \over 2 }$

(ii) A number lying between 2 and 6

The numbers that are greater than 2 and less than 6 are: 3, 4, 5.

P(number between 2 and 6 ) = Number of favourable outcomes / Number of possible outcomes ​

P( number between 2 and 6) = $ {3 \over 6 }$ = $ {1 \over 2 }$

(iii) An odd number

The odd numbers on a die are: 1, 3, 5.

P( odd number ) = Number of favourable outcomes / Number of possible outcomes ​

P(odd number ) = $ {3 \over 6 }$ = $ {1 \over 2 }$

Question 14

One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds

Solution :

There are 52 cards in a standard, well-shuffled deck. The total number of possible outcomes is 52 for each draw.

(i) A king of red colour

The red suits are hearts and diamonds. There is one king in each of these suits, so there are 2 red kings (the King of Hearts and the King of Diamonds).

P(king of red colour) = Number of favourable outcomes / Number of possible outcomes ​

P(king of red colour) = $ {2 \over 52 }$ = $ {1 \over 26 }$

(ii) A face card

Face cards are Jacks, Queens, and Kings. There are 3 face cards in each of the 4 suits, so there are a total of 3 × 4 = 12 face cards.

P(face card ) = Number of favourable outcomes / Number of possible outcomes ​

P( face card) = $ {12 \over 52 }$ = $ {3 \over 13 }$

(iii) A red face card

The red suits (hearts and diamonds) each have 3 face cards (Jack, Queen, King). Therefore, there are a total of 3 × 2 = 6 red face cards.

P( red face card ) = Number of favourable outcomes / Number of possible outcomes ​

P(red face card ) = $ {6 \over 52 }$ = $ {3 \over 26 }$

(iv) The jack of hearts

There is only one Jack of Hearts in the entire deck.

P( jack of hearts ) = Number of favourable outcomes / Number of possible outcomes ​

P(jack of hearts ) = $ {1 \over 52 }$

(v) A spade

There are 4 suits in a deck, and each suit has 13 cards. Therefore, there are 13 spade cards..

P( spade ) = Number of favourable outcomes / Number of possible outcomes ​

P(spade ) = $ {13 \over 52 }$ = $ {1 \over 4 }$

(vi) The queen of diamonds

There is only one Queen of Diamonds in the entire deck.

P( queen of diamonds ) = Number of favourable outcomes / Number of possible outcomes ​

P(queen of diamonds ) = $ {1 \over 52 }$

Question 15

Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?
(ii)If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Solution :

Initially, there are 5 cards in total: the ten, jack, queen, king, and ace of diamonds.

(i) What is the probability that the card is the queen?

There is only one queen in the set of 5 cards.

P(queen) = Number of favourable outcomes / Number of possible outcomes ​

P(queen) = $ {1 \over 5 }$

(ii) If the queen is drawn and put aside

After the queen is drawn, the total number of cards remaining is 5 −1 = 4. The remaining cards are the ten, jack, king, and ace of diamonds.

(a) What is the probability that the second card picked up is an ace?

There is one ace left among the 4 remaining cards.

P(ace) = Number of favourable outcomes / Number of possible outcomes ​

P(ace) = $ {1 \over 4 }$

(b) What is the probability that the second card picked up is a queen?

Since the queen was already drawn and put aside, there are now zero queens left in the deck.

P(queen) = Number of favourable outcomes / Number of possible outcomes ​

P(queen) = $ {0 \over 4 }$ = 0

Question 16

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution :

Total number of pens:.

Total pens = (Good pens) + (Defective pens)

Total pens = 132 + 12 = 144

The number of good pens is the number of favorable outcomes, which is 132.

P(good pen) = Number of favourable outcomes / Number of possible outcomes ​

P(good pen) = $ {132 \over 144 }$ = $ {11 \over 12 }$

Question 17

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?.
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?

Solution :

(i) Probability of drawing a defective bulb

Total number of bulbs: 20

Number of defective bulbs: 4

P(defective bulb) = Number of favourable outcomes / Number of possible outcomes ​

P(defective bulb) = $ {4 \over 20 }$ = $ {1 \over 5 }$

(ii) Probability of the second bulb not being defective

A non-defective bulb has already been drawn and not replaced. This changes the total number of bulbs and the number of non-defective bulbs.

Total bulbs remaining: 20 − 1 = 19

The original number of non-defective bulbs was 20 − 4 = 16.

Since one was drawn, the remaining number is 16 − 1 = 15.

P(not defective ) = Number of favourable outcomes / Number of possible outcomes ​

P( not defective) = $ {15 \over 19 }$

Question 18

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number
(ii)a perfect square number
(iii) a number divisible by 5.

Solution :

(i) A two-digit number

The number of two-digit numbers is the total number of discs minus the single-digit numbers.

Number of two-digit numbers = 90 − 9 = 81.

P(two-digit number) = Number of favourable outcomes / Number of possible outcomes ​

P(two-digit number) = $ {81 \over 90 }$ = $ {9 \over 10 }$

(ii) A perfect square number

A perfect square number is the product of an integer with itself.There are 9 perfect square numbers.

Perfect square numbers between 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81

P(perfect square ) = Number of favourable outcomes / Number of possible outcomes ​

P( perfect square) = $ {9 \over 90 }$ = $ {1 \over 10 }$

(iii) A number divisible by 5

The numbers between 1 and 90 that are divisible by 5 are the multiples of 5: 5, 10, 15, ..., 90.

There are 18 numbers divisible by 5.

P(number divisible by 5) = Number of favourable outcomes / Number of possible outcomes ​

P(number divisible by 5) = $ {18 \over 90 }$ = $ {1 \over 5 }$

Question 19

A child has a die whose six faces show the letters as given below:

[A] , [B], [C], [D], [E], [A]
The die is thrown once. What is the probability of getting
(i) A?
(ii) D?

Solution :

The die has six faces, so there are a total of 6 possible outcomes.

(i) Probability of getting A

The letter 'A' appears on two of the six faces.

P(A) = Number of favourable outcomes / Number of possible outcomes ​

P(A) = $ {2 \over 6 }$ = $ {1 \over 3 }$

(ii) Probability of getting D

The letter 'D' appears on only one of the six faces..

P(D) = Number of favourable outcomes / Number of possible outcomes ​

P(D) = $ {1 \over 6 }$

Question 20

Suppose you drop a die at random on the rectangular region shown. What is the probability that it will land inside the circle with diameter 1 m?
The die is thrown once. What is the probability of getting
(i) Length of rectangular region = 3 m
(ii) Breadth of rectangular region = 2 m

Solution :

(i) Calculate the Total Area of the rectangular region

The total area is the area of the rectangular region = length × breadth

3 m × 2 m = 6 $m^2 $

Calculate the Favorable Area of the circle

The diameter of the circle is 1 m, so the radius is half of that: radius (r) = 0.5 m

Area = π×$r^2 $

Area of circle = π×($0.5^2 $) = 0.25π$m^2 $ ​

Calculate the Probability

The probability that it will land inside the circle

P(E) = Number of favourable outcomes / Number of possible outcomes ​

P(E) = Area of circular region / Area of the rectangular region ​

P(E) = $ {0.25π \over 6 }$ = $ {π \over 24 }$

Question 21

A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it ?
(ii) (ii) She will not buy it ?

Solution :

No. of possible outcomes = 144

(i) What is the probability that she will buy it?

Nuri will buy the pen if it is good. The probability is the ratio of good pens to the total number of pens.

let's determine the number of good pens: = 144 − 20 = 124

P(buy) = Number of favourable outcomes / Number of possible outcomes ​

P(buy) = $ {124 \over 144 }$ = $ {31 \over 36 }$

(ii) What is the probability that she will not buy it?

Nuri will not buy the pen if it is defective. The probability is the ratio of defective pens to the total number of pens.

No. of Defective pens = 20

P(not buy) = Number of favourable outcomes / Number of possible outcomes ​

P(not buy) = $ {20 \over 144 }$ = $ {5 \over 36 }$

Question 22

Complete the following table :

Event : 'Sum on two dice'	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36						5/ 36				1/36

(ii) A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1 /11 . Do you agree with this argument? Justify your answer.

Solution :

Here is the completed table showing the probabilities for the sum of two dice.

If 2 dice are thrown, the possible events are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(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)

The denominator for each probability is 36, as there are 6×6=36 possible outcomes when rolling two standard dice.

The number of ways to achieve each sum is as follows:

For getting the sum as 2, the possible events (or outcomes) = E (sum 2) = (1, 1) = 1 way

So, P(sum 2) = 1/36

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1, 2), (2, 1) = 2 ways

So, P(sum 3) = 2/36

For getting the sum as 4, the possible events (or outcomes) = E (sum 4) = (1, 3), (2, 2), (3, 1) = 3 ways

So, P(sum 4) = 3/36

For getting the sum as 5, the possible events (or outcomes) = E (sum 5) = (1, 4), (2, 3), (3, 2), (4, 1) = 4 ways

So, P(sum 5) = 4/36

For getting the sum as 6, the possible events (or outcomes) = E (sum 6) = 1, 5), (2, 4), (3, 3), (4, 2), (5, 1) = 5 ways

So, P(sum 6) = 5/36

For getting the sum as 7, the possible events (or outcomes) = E (sum 7) = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6 ways

So, P(sum 7) = 6/36

For getting the sum as 8, the possible events (or outcomes) = E (sum 8) = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5 ways

So, P(sum 8) = 5/36

For getting the sum as 9, the possible events (or outcomes) = E (sum 9) = (3, 6), (4, 5), (5, 4), (6, 3) = 4 ways

So, P(sum 9) = 4/36

For getting the sum as 10, the possible events (or outcomes) = E (sum 10) = (4, 6), (5, 5), (6, 4) = 3 ways

So, P(sum 10) = 3/36

For getting the sum as 11, the possible events (or outcomes) = E (sum 11) = (5, 6), (6, 5) = 2 ways

So, P(sum 11) = 2/36

For getting the sum as 12, the possible events (or outcomes) = E (sum 12) = (6, 6) = 1 way

So, P(sum 12) = 1/36

Probability Table

Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	1/36	2/36	3/36	4/36	5/36	6/ 36	5/ 36	4/36	3/36	2/36	1/36

(ii) I don't agree with the student's argument. The reasoning is flawed because it assumes that all 11 outcomes (the sums of two dice) are equally likely, which they are not

The fundamental principle of probability states that for an event to have a probability of 1/n , there must be n equally likely outcomes. While there are 11 possible sums (from 2 to 12) when rolling two dice, the number of ways to achieve each sum is different.

Question 23

A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hani wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution :

A fair coin is tossed three times. Each toss has two possible outcomes (Heads or Tails). The total number of possible outcomes is calculated by multiplying the outcomes of each toss:

Total possible outcomes are = {HHH, TTT, HTH, HHT, THH, THT, TTH, HTT}

2 (Toss 1) × 2 (Toss 2) × 2 (Toss 3) = 8 possible outcomes

Identify the winning outcomes:

if all three tosses give the same result. The winning outcomes are:

{ HHH , TTT }

This means there are 2 winning outcomes.

We can find the number of losing outcomes by subtracting the winning outcomes from the total outcomes:

8 (Total outcomes) − 2 (Winning outcomes) = 6 losing outcomes

P( lose the game) = Number of favourable outcomes / Number of possible outcomes ​

P( lose the game) = $ {6 \over 8 }$

Question 24

A die is thrown twice. What is the probability that

(i) 5 will not come up either time ?

(ii) 5 will come up at least once ?

Solution :

If 2 dice are thrown, the possible events are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(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)

The denominator for each probability is 36, as there are 6 × 6 = 36 possible outcomes when rolling two standard dice.

(i) Probability that 5 will not come up either time

When a die is thrown, there are 6 possible outcomes: {1, 2, 3, 4, 5, 6}. The number of outcomes

where a 5 does not come up is 5: {1, 2, 3, 4, 6}.

Favourable outcomes for 5 to not come up either time are

{(1,1) (1,2) (1,3) (1,4) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,6)
(6,1) (6,2) (6,3) (6,4) (6,6)}

No. of favourable outcomes = 25

P(no 5 either time) = Number of favourable outcomes / Number of possible outcomes ​

P(no 5 either time) = $ {25 \over 36 }$

(ii) Probability that 5 will come up at least once

The event "5 will come up at least once" is the complement of the event "5 will not come up either time." This means that the sum of their probabilities is 1.

We can use the complement rule:

P(at least one 5) = 1− P(no 5 either time)

Using the result from part (i):

P(at least one 5)= $ 1 − {25 \over 36 }$

P(at least one 5)= $ {11 \over 36 }$

Question 25

Which of the following arguments are correct and which are not correct? Give reasons for your answer.

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2.

Solution :

(i) The above statement is incorrect.

The outcome "one of each" can happen in two distinct ways: a head on the first coin and a tail on the second (HT), or a tail on the first and a head on the second (TH). Therefore, the correct and complete list of equally likely outcomes is:

HH (Head, Head)

HT (Head, Tail)

TH (Tail, Head)

TT (Tail, Tail)

There are four equally likely outcomes, not three. The probability of each of these outcomes is 1/4

The probability of getting "one of each" (HT or TH) is 2/4 = 1/2

while the probabilities of "two heads" and "two tails" are each 1/4

∴ This statement is incorrect.

(ii) This argument is correct.

A standard die has six faces with the numbers {1, 2, 3, 4, 5, 6}.

The odd numbers are {1, 3, 5}, which is a total of three outcomes.

The even numbers are {2, 4, 6}, which is a total of three outcomes.

The probability of getting an odd number is the number of odd outcomes divided by the total number of outcomes:

= $ {3 \over 6 }$ = $ {1 \over 2 }$

On a die, "odd" has 3 outcomes and "even" has 3 outcomes. Since they’re equal, the probability really is 1/2 .

∴ This statement is is correct.