Question1
Complete the following statements:
(i) Probability of an event E + Probability of the event â€˜not Eâ€™ = ________.
(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) 1
(ii) 0, impossible event
(iii) 1, sure or certain event
(iv) 1
(v) 0, 1.
(i) Probability of an event E + Probability of the event â€˜not Eâ€™ = ________.
(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) 1
(ii) 0, impossible event
(iii) 1, sure or certain event
(iv) 1
(v) 0, 1.
Question2
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 answer a truefalse question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) If a driver attempts to start a car, the two possible outcomes are either starting a car or not starting a car. Thus it is an experiment with equally likely outcomes.
(ii) If a player attempts to shoot a basketball, the two possible outcomes are either shooting or missing the shot. Hence experiment has equally likely outcomes.
(iii) If a trail is made to answer a truefalse question. The two possible outcomes are either the answer is right or wrong. Thus this is an experiment with equally likely outcomes.
(iv) When a baby is born the possible outcomes are likely a boy or a girl. Hence experiment has equally likely outcomes.
Hence, all events have two possible outcomes so both outcomes are equally likely.
(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 answer a truefalse question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) If a driver attempts to start a car, the two possible outcomes are either starting a car or not starting a car. Thus it is an experiment with equally likely outcomes.
(ii) If a player attempts to shoot a basketball, the two possible outcomes are either shooting or missing the shot. Hence experiment has equally likely outcomes.
(iii) If a trail is made to answer a truefalse question. The two possible outcomes are either the answer is right or wrong. Thus this is an experiment with equally likely outcomes.
(iv) When a baby is born the possible outcomes are likely a boy or a girl. Hence experiment has equally likely outcomes.
Hence, all events have two possible outcomes so both outcomes are equally likely.
Question3
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:
When we toss a coin, the two possible outcomes are head and tail which is equally likely. Thus the result of tossing a coin is considered as a fair way of deciding which team should get the ball at the beginning of a football game.
Solution:
When we toss a coin, the two possible outcomes are head and tail which is equally likely. Thus the result of tossing a coin is considered as a fair way of deciding which team should get the ball at the beginning of a football game.
Question4
Which of the following cannot be the probability of an event?
(A) (B) â€“1.5 (C) 15% (D) 0.7.
Solution:
B cannot be the probability of an event.
(A) (B) â€“1.5 (C) 15% (D) 0.7.
Solution:
B cannot be the probability of an event.
Question5
If P(E) = 0.05, what is the probability of â€˜not Eâ€™?
Solution:
We know that P(E) + P() =1, here P() denotes that the probability is not E.
Hence,
P(E) + P() = 1
0.05 + P() = 1
P() = 1 â€“ 0.05
= 0.95.
Solution:
We know that P(E) + P() =1, here P() denotes that the probability is not E.
Hence,
P(E) + P() = 1
0.05 + P() = 1
P() = 1 â€“ 0.05
= 0.95.
Question6
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) Since the bag contains only lemon flavoured candies the probability that Malini takes out an orange flavoured candy is zero.
(ii) As the bag contains only lemon flavoured candies the probability that Malini takes out a lemon flavoured candy is one.
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since the bag contains only lemon flavoured candies the probability that Malini takes out an orange flavoured candy is zero.
(ii) As the bag contains only lemon flavoured candies the probability that Malini takes out a lemon flavoured candy is one.
Question7
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 E be the event of two students having the same birthday, be the event of two students not having the same birthday. Then
P(E) + P() = 1
But P() = 0.992 thus
P(E) + 0.992 = 1
Ãž P(E) = 1 â€“ 0.992
= 0.008.
Solution:
Let E be the event of two students having the same birthday, be the event of two students not having the same birthday. Then
P(E) + P() = 1
But P() = 0.992 thus
P(E) + 0.992 = 1
Ãž P(E) = 1 â€“ 0.992
= 0.008.
Question8
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:
Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 5 = 8 (i.e.) n(S) = 8
Let A be the event of drawing red ball and B be the event of drawing black ball.
Number of red balls (i.e.) n(A) = 3
Number of black balls (i.e.) n(B) = 5
Probability that the ball is red, P(A) = [P(A) = ]
Probability that the ball is not red, P(B) = [P(B) = ]
âˆ´ The probability that the ball drawn is (i) red =
(ii) not red = .
Solution:
Number of red balls = 3
Number of black balls = 5
Total number of balls = 3 + 5 = 8 (i.e.) n(S) = 8
Let A be the event of drawing red ball and B be the event of drawing black ball.
Number of red balls (i.e.) n(A) = 3
Number of black balls (i.e.) n(B) = 5
Probability that the ball is red, P(A) = [P(A) = ]
Probability that the ball is not red, P(B) = [P(B) = ]
âˆ´ The probability that the ball drawn is (i) red =
(ii) not red = .
Question9
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:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles (i.e.) = n(S) = 5 + 8 + 4 = 17
Let A be the event of taking a red, B be the event of taking a white and C be the event of taking not a green marbles respectively.
Probability that the marble taken out is red marble (i.e.) P(A)= [P(A) = ]
Probability that the marble taken out is white marble (i.e.) P(B)= [P(B) = ]
(i.e.) the marble taken out is other than green hence 5 + 8 = 13
Probability that the marble taken out is not green marble (i.e.) P(C)= [P(C) = ]
âˆ´ The probability that the marble taken out would be (i) red = (ii) white = (iii) notgreen = .
Solution:
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles (i.e.) = n(S) = 5 + 8 + 4 = 17
Let A be the event of taking a red, B be the event of taking a white and C be the event of taking not a green marbles respectively.
Probability that the marble taken out is red marble (i.e.) P(A)= [P(A) = ]
Probability that the marble taken out is white marble (i.e.) P(B)= [P(B) = ]
(i.e.) the marble taken out is other than green hence 5 + 8 = 13
Probability that the marble taken out is not green marble (i.e.) P(C)= [P(C) = ]
âˆ´ The probability that the marble taken out would be (i) red = (ii) white = (iii) notgreen = .
Question10
Solution:
Number of 50p coins = 100
Number of `1 coins = 50
Number of `2 coins = 20
Number of `5 coins = 10
Total number of coins = 100 + 50 + 20 + 10 = 180
Probability of a event =
Let A denote the event that the coin fallen out is 50p coin,
B denote the event that the coin fallen out will not be `5[(i.e)., the coin is other than `5]
P(A) = =
=
=
P(B) = =
=
âˆ´ The probability that the coin is (i) 50p coin =
(ii) not a `5 coin = .
Number of 50p coins = 100
Number of `1 coins = 50
Number of `2 coins = 20
Number of `5 coins = 10
Total number of coins = 100 + 50 + 20 + 10 = 180
Probability of a event =
Let A denote the event that the coin fallen out is 50p coin,
B denote the event that the coin fallen out will not be `5[(i.e)., the coin is other than `5]
P(A) = =
=
=
P(B) = =
=
âˆ´ The probability that the coin is (i) 50p coin =
(ii) not a `5 coin = .
Question11
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 (see Figure). What is the probability that the fish taken out is a male fish?
Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13
Probability of a event =
Let A be the event of taking a male fish
P(A) =
= .
Solution:
Number of male fish = 5
Number of female fish = 8
Total number of fish = 5 + 8 = 13
Probability of a event =
Let A be the event of taking a male fish
P(A) =
= .
Question12
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 (see Figure), 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:
Total Number of outcomes when an arrow is spanned = 8
(i) The probability of getting 8 =
(ii) Number of odd number = 4 (i.e.) 1, 3, 5, 7
The probability of getting a odd number = =
(iii) Numbers greater than 2 = 3, 4, 5, 6, 7, 8
The probability of getting a number greater than 2 = =
(iv) Numbers less than 9 = 1, 2, 3, â€¦, 8
The probability of getting a number less than 9 = = 1.
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Solution:
Total Number of outcomes when an arrow is spanned = 8
(i) The probability of getting 8 =
(ii) Number of odd number = 4 (i.e.) 1, 3, 5, 7
The probability of getting a odd number = =
(iii) Numbers greater than 2 = 3, 4, 5, 6, 7, 8
The probability of getting a number greater than 2 = =
(iv) Numbers less than 9 = 1, 2, 3, â€¦, 8
The probability of getting a number less than 9 = = 1.
Question13
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:
Sample space = {1, 2, 3, 4, 5, 6}
n(s) = 6
Let A denote the event of getting a prime number, B denote the event of getting a number lying between 2 and 6, C denote the probability of getting an odd number.
A = {2, 3, 5}, B = {3, 4, 5}, C = {1, 3, 5}
Probability of a event =
P(A) = = =
P(B) = = =
P(C) = = =
Thus the probability of getting (i) a prime number =
(ii) a number lying between 2 and 6 =
(iii) an odd number = .
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Sample space = {1, 2, 3, 4, 5, 6}
n(s) = 6
Let A denote the event of getting a prime number, B denote the event of getting a number lying between 2 and 6, C denote the probability of getting an odd number.
A = {2, 3, 5}, B = {3, 4, 5}, C = {1, 3, 5}
Probability of a event =
P(A) = = =
P(B) = = =
P(C) = = =
Thus the probability of getting (i) a prime number =
(ii) a number lying between 2 and 6 =
(iii) an odd number = .
Question14
One card is drawn from a wellshuffled 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:
Total number of cards in a wellshuffled deck = 52
Let A be the event of getting a king of red colour, B be the event of getting a face card, C be the event of getting a red face card, D be the event of getting the jack of hearts, E be the event of getting a spade and F be the event of getting the queen of diamonds.
Number of king of red colour n(A) = 2
Number of face card n(B) = 12
Number of red face card n(C) = 6
Number of jack of hearts n(D) = 1
Number of spade n(E) = 13
Number of queen of diamonds n(F) = 1
Probability of a event =
(i) Probability of getting a king of red colour P(A)=
(ii) Probability of getting a face card P(B) =
(iii) Probability of getting a red face card P(C) =
(iv) Probability of getting a jack of hearts P(D) =
(v) Probability of getting a spade P(E) =
(vi) Probability of getting a queen of diamonds P(F) =.
(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:
Total number of cards in a wellshuffled deck = 52
Let A be the event of getting a king of red colour, B be the event of getting a face card, C be the event of getting a red face card, D be the event of getting the jack of hearts, E be the event of getting a spade and F be the event of getting the queen of diamonds.
Number of king of red colour n(A) = 2
Number of face card n(B) = 12
Number of red face card n(C) = 6
Number of jack of hearts n(D) = 1
Number of spade n(E) = 13
Number of queen of diamonds n(F) = 1
Probability of a event =
(i) Probability of getting a king of red colour P(A)=
(ii) Probability of getting a face card P(B) =
(iii) Probability of getting a red face card P(C) =
(iv) Probability of getting a jack of hearts P(D) =
(v) Probability of getting a spade P(E) =
(vi) Probability of getting a queen of diamonds P(F) =.
Question15
Five cards, the ten, jack, queen, king and ace of diamonds, are wellshuffled 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:
Number of ten = 1
Number of Jack = 1
Number of queen = 1
Number of king = 1
Number of ace of diamonds = 1
Probability of a event =
(i) Probability that the card is queen =
(ii) If the queen is drawn, then the number of cards available is 4
a) Probability that the second card is an ace =
b) Probability that the second card is an queen = 0 [since it has only one queen card
which is already removed the
number of queen card = 0].
(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:
Number of ten = 1
Number of Jack = 1
Number of queen = 1
Number of king = 1
Number of ace of diamonds = 1
Probability of a event =
(i) Probability that the card is queen =
(ii) If the queen is drawn, then the number of cards available is 4
a) Probability that the second card is an ace =
b) Probability that the second card is an queen = 0 [since it has only one queen card
which is already removed the
number of queen card = 0].
Question16
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:
Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
Let A denote the event of taking a good pen
Probability of a event =
P(A) =
= = .
Solution:
Number of defective pens = 12
Number of good pens = 132
Total number of pens = 12 + 132 = 144
Let A denote the event of taking a good pen
Probability of a event =
P(A) =
= = .
Question17
(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:
Total number of bulbs n(S) = 20
Number of defective bulbs = 4
Number of good bulbs = 16
(i) Probability of a event =
Let A be the event of drawing a defective bulb
P(A) =
= =
(ii) If suppose the bulb drawn is not defective and is not replaced (i.e.) one good bulb is taken. The number of bulbs will be 19.
Number of good bulbs = 16 â€“ 1
= 15
âˆ´ The probability of the bulbs not 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:
Total number of bulbs n(S) = 20
Number of defective bulbs = 4
Number of good bulbs = 16
(i) Probability of a event =
Let A be the event of drawing a defective bulb
P(A) =
= =
(ii) If suppose the bulb drawn is not defective and is not replaced (i.e.) one good bulb is taken. The number of bulbs will be 19.
Number of good bulbs = 16 â€“ 1
= 15
âˆ´ The probability of the bulbs not defective = .
Question18
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 twodigit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
Number of discs in a box = 90
Two digit number = {10, 11, 12, â€¦, 90}
Number of two digit number = 90 â€“ 9 = 81
Perfect square number = {1, 4, 9, 16, 25, 36, 49, 64, 81}
Number of perfect square number = 9
Numbers divisible by 5 = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}
Let A denote a two â€“ digit number, B denote a perfect square, and C denote a number divisible by 5.
(i) P(A) =
= =
(ii) P(B) =
= =
(iii) P(C) =
= =
âˆ´ The probability that the disc drawn from the box
(i) a twoâ€“digit number =
(ii) a perfect square number =
(iii) a number divisible by 5 = .
(i) a twodigit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
Number of discs in a box = 90
Two digit number = {10, 11, 12, â€¦, 90}
Number of two digit number = 90 â€“ 9 = 81
Perfect square number = {1, 4, 9, 16, 25, 36, 49, 64, 81}
Number of perfect square number = 9
Numbers divisible by 5 = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90}
Let A denote a two â€“ digit number, B denote a perfect square, and C denote a number divisible by 5.
(i) P(A) =
= =
(ii) P(B) =
= =
(iii) P(C) =
= =
âˆ´ The probability that the disc drawn from the box
(i) a twoâ€“digit number =
(ii) a perfect square number =
(iii) a number divisible by 5 = .
Question19
A child has a die whose six faces show the letters as given below:
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Letters displayed on the die = {A, B, C, D, E, A}
Number of all possible outcomes = 6
Number of Aâ€™s = 2
Number of Dâ€™s = 1
Probability of an event =
Let A denote the event of getting Aâ€™s and B denote the event of getting Dâ€™s then
P(A) =
= =
P(B) =
=
âˆ´ The probability of getting (i) A = (ii) D = .
The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Letters displayed on the die = {A, B, C, D, E, A}
Number of all possible outcomes = 6
Number of Aâ€™s = 2
Number of Dâ€™s = 1
Probability of an event =
Let A denote the event of getting Aâ€™s and B denote the event of getting Dâ€™s then
P(A) =
= =
P(B) =
=
âˆ´ The probability of getting (i) A = (ii) D = .
Question20
Suppose you drop a die at random on the rectangular region shown in the figure. What is the probability that it will land inside the circle with diameter 1 m?
Solution:
Area of the rectangular region = l Ã— b
= 3 Ã— 2
= 6 m^{2}
Area of the circle = Ï€ r^{2}
Diameter of a circle = 1 m
Radius of a circle =
Area of the circle = Ï€ Ã— Ã—
= Ï€ Ã— Ã—
= m^{2}
Probability that the land inside is the circle of =
= = .
Solution:
Area of the rectangular region = l Ã— b
= 3 Ã— 2
= 6 m^{2}
Area of the circle = Ï€ r^{2}
Diameter of a circle = 1 m
Radius of a circle =
Area of the circle = Ï€ Ã— Ã—
= Ï€ Ã— Ã—
= m^{2}
Probability that the land inside is the circle of =
= = .
Question21
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) She will not buy it?
Solution:
Total number of ball pens, n(s) = 144
Number of defective pens = 20
Number of good pens = 144 â€“ 20 = 124
Let A be the event of Nuri buying the pen and B be the event of Nuri not buying the pen (since Nuri will buy a pen only if it is good and will not buy the pen if it is defective)
Probability of buying a pen = = [P(A) = ]
Probability of not buying a pen = 1 â€“ == [P(B) = ]
(i) She will buy it?
(ii) She will not buy it?
Solution:
Total number of ball pens, n(s) = 144
Number of defective pens = 20
Number of good pens = 144 â€“ 20 = 124
Let A be the event of Nuri buying the pen and B be the event of Nuri not buying the pen (since Nuri will buy a pen only if it is good and will not buy the pen if it is defective)
Probability of buying a pen = = [P(A) = ]
Probability of not buying a pen = 1 â€“ == [P(B) = ]
âˆ´ Probability Nuri will (i) buy a pen =
(ii) not buy a pen = .
Question22
(i) Complete the following table:
(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. Do you agree with this argument? Justify your answer.
Solution:
(i)
(ii) No, I donâ€™t agree with this argument.
Since there are 11 possible outcomes and all are not equally likely.
Event: 'Sum on 2 dice' 
2  3  4  5  6  7  8  9  10  11  12 
Probability 
(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. Do you agree with this argument? Justify your answer.
Solution:
(i)
3 
4 
5 
6 
7 
9 
10 
11 








(ii) No, I donâ€™t agree with this argument.
Since there are 11 possible outcomes and all are not equally likely.
Question23
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif 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:
When a one rupee coin is tossed 3 times
All possible outcomes = {(H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T),(H, T, T), (T, T, T), (H, H, H)}
n(S) = 8
If Hanif wins all the tosses must give the same result (i.e.) three heads or three tails.
Hanif would win if the outcomes is (T, T, T) and (H, H, H).
Probability of an event =
Let A be the event of winning then is the event of losing the game.
âˆ´ P(A) =
= =
The sum of all the probabilities is one
P(A) + P( = 1
â‡’ P( = 1 
= .
Solution:
When a one rupee coin is tossed 3 times
All possible outcomes = {(H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T),(H, T, T), (T, T, T), (H, H, H)}
n(S) = 8
If Hanif wins all the tosses must give the same result (i.e.) three heads or three tails.
Hanif would win if the outcomes is (T, T, T) and (H, H, H).
Probability of an event =
Let A be the event of winning then is the event of losing the game.
âˆ´ P(A) =
= =
The sum of all the probabilities is one
P(A) + P( = 1
â‡’ P( = 1 
= .
Question24
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?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
The possible outcomes of throwing two dice = {(1, 1), (1, 2), â€¦, (1, 6)
(2, 1), (2, 2), â€¦, (2, 6)
(6, 1), (6, 2), â€¦, (6, 6)}
n(s) = 36
(i) Let A denote the event such that five will not come up either time
A = {(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)}
n(A) = 25
The probability of an event =
P(A) = =
(ii) Let denote the event such that five will come up at least once. (i.e.) it is other than (i), thus
P(A)+ P() = 1 [sum of all the probabilities]
â‡’ P() = 1 P(A)
= 1 
=
âˆ´ The probability (i) of not getting 5 either time =
(ii) 5 will come up atleast once = .
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].
Solution:
The possible outcomes of throwing two dice = {(1, 1), (1, 2), â€¦, (1, 6)
(2, 1), (2, 2), â€¦, (2, 6)
(6, 1), (6, 2), â€¦, (6, 6)}
n(s) = 36
(i) Let A denote the event such that five will not come up either time
A = {(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)}
n(A) = 25
The probability of an event =
P(A) = =
(ii) Let denote the event such that five will come up at least once. (i.e.) it is other than (i), thus
P(A)+ P() = 1 [sum of all the probabilities]
â‡’ P() = 1 P(A)
= 1 
=
âˆ´ The probability (i) of not getting 5 either time =
(ii) 5 will come up atleast once = .
Question25
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 .
(ii) If a die is thrown, there are two possible outcomesâ€”an odd number or an even number. Therefore, the probability of getting an odd number is .
Solution:
If two are tossed simultaneously three possible outcomes â€“ two heads, two tails or one of each.
All possible outcomes = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Then each of the outcomes (i.e) two heads, two tails or one of each doesnâ€™t have the probability as . Because
Probability of an event =
The probability of getting two heads =
The probability of getting two tails =
(i) It is not correct. If we want to get the probability of them we shall classify the outcomes like this but they are not "equally likely". Because "one of each" can result in two ways from a head on first coin and tail on the second coin or from a tail on the first coin and head on the second coin. This makes it twicely as likely as two heads(or two tails).
(ii) Correct. The two outcomes considered in this question are equally likely.
(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 .
(ii) If a die is thrown, there are two possible outcomesâ€”an odd number or an even number. Therefore, the probability of getting an odd number is .
Solution:
If two are tossed simultaneously three possible outcomes â€“ two heads, two tails or one of each.
All possible outcomes = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Then each of the outcomes (i.e) two heads, two tails or one of each doesnâ€™t have the probability as . Because
Probability of an event =
The probability of getting two heads =
The probability of getting two tails =
(i) It is not correct. If we want to get the probability of them we shall classify the outcomes like this but they are not "equally likely". Because "one of each" can result in two ways from a head on first coin and tail on the second coin or from a tail on the first coin and head on the second coin. This makes it twicely as likely as two heads(or two tails).
(ii) Correct. The two outcomes considered in this question are equally likely.