NCERT
10th Mathematics
Exercise 14.1
Topic: Probability
"With our thorough approach to 'Probability ,' as presented in the NCERT textbooks, discover the joy of learning." We simplify complex ideas into manageable steps so students can proceed through the content confidently."
Q. 1. Complete the following statements:
(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 impossible event.
(iii) The probability of an event that is certain to happen is 1. Such an
event is called 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 .
Q. 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 answer a true-false question. The answer is
right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) A driver attempts to start a car. The car starts or does not start.
Ans. As it depends on various factors such as the car being old, it may or may not
have fuel, the factors for both the conditions are not same. Therefore this is not equally likely event.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Ans. It is not an equally likely event, as it depends on the player's ability.
(iii) A trial is made to answer a true-false question. The answer is right or
wrong.
Ans. As the answer may be either right or wrong, it is an equally likely event.
(iv) A baby is born. It is a boy or a girl.
Ans. As the answer may be either boy or girl, it is an equally likely event.
Q. 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:
Ans. When we toss a coin, there are only two possible outcomes: heads or tails.
Each outcome is equally likely, making the result entirely unpredictable.
Q. 4. Which of the following cannot be the probability of an event?
(A) 2/3 (B) –1.5 (C) 15% (D) 0.7
Solution:
Ans. The probability of any event E must always be between 0 and 1, inclusive.
Since − 1.5 falls outside this range, it cannot represent the probability of any event. Therefore answer is B
Q. 5. If P(E) = 0.05, what is the probability of ‘not E’?
Ans.
1) We know that,
P(Not E) = 1 - P(E)
P(Not E) = 1 - 0.05P(Not E) = 0.95
2) So, P(Not E) = 0.95. i.e.
The probability of ‘not E’ is 0.95.
Q. 6. A bag contains lemon-flavored 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) What is the probability that she takes out an orange-flavored candy?
Ans: The bag contains only lemon-flavored candies. This means that every time
Malini takes a candy, it will be lemon-flavored. Therefore, the event of Malini picking an 'orange-flavored candy' is impossible, and the probability of this happening is zero.
Therefore P(an orange flavoured candy) = 0.
(ii) What is the probability that she takes out a lemon-flavored candy?
Ans: Since the bag contains only lemon-flavored candies, Malini is certain to pick a
lemon-flavored candy. Therefore, this is a sure event, and the probability of it happening is 1.
Therefore P(a lemon flavoured candy) = 1.
Q. 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:
1) Let the event E represent the situation where two students do not have the
same birthday.
2) Therefore, P(E) = 0.992.
3) We know that:
P(Not E) = 1 - P(E)
P(Not E) = 1 - 0.992
P(Not E) = 0.008
4) Therefore, the probability that the 2 students have the same birthday is
0.008.
Q. 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:
(i) What is the probability that the ball drawn is red?
1) Bag containd 3 red balls, and 5 black balls.
2) Therefore, total number of balls in a bag is 3 + 5 = 8.
3) Let the event E represents drawing a red ball.
4) Now, let's find the probability of drawing a red ball:
probability of drawing a red ball
= (Number of favarable out comes)/(Total number of outcomes)
probability of drawing a red ball= (Number of red balls)/(Total number of balls)
probability of drawing a red ball = 3/8
P(E) = 3/8.
5) Therefore, the probability that the ball drawn is red is 3/8.
(ii) What is the probability that the ball drawn is not red?
6) We know that:
P(Not E) = 1 - P(E)
P(Not E) = 1 - (3/8)
P(Not E) = 5/8
7) Therefore, the probability that the ball drawn is not red is 5/8.
Q. 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:
(i) What is the probability that the marble taken out will be red?
1) Box containd 5 red marbles, 8 white marbles and 4 green marbles.
2) Therefore, total number of marbles in a box is 5 + 8 + 4 = 17.
3) Now, let's find the probability of drawing a red marble:
probability of drawing a red marble
= (Number of favarable out comes)/(Total number of outcomes)
probability of drawing a red marble= (Number of red marbles)/(Total number of marbles)
probability of drawing a red marbles = 5/17
P(drawing a red marbles) = 5/17.
4) Therefore, the probability that the marble taken out is red is 5/17.
(ii) What is the probability that the marble taken out will be white?
5) Now, let's find the probability of drawing a white marble:
probability of drawing a white marble
= (Number of favarable out comes)/(Total number of outcomes)
probability of drawing a white marble= (Number of red marbles)/(Total number of marbles)
probability of drawing a white marbles = 8/17
P(drawing a white marbles) = 8/17.
6) Therefore, the probability that the marble taken out is white is 8/17.
(iii) What is the probability that the marble taken out will not be green?
7) Now, let's find the probability of drawing a green marble:
probability of drawing a green marble
= (Number of favarable out comes)/(Total number of outcomes)
probability of drawing a green marble= (Number of red marbles)/(Total number of marbles)
probability of drawing a green marbles = 4/17
P(drawing a green marbles) = 4/17.
8) Therefore
P(not drawing a green marbles) = 1 - P(drawing a green marbles)P(not drawing a green marbles) = 1 - 4/17
P(not drawing a green marbles) = 13/17
9) Therefore,
the probability that the marble taken out is not green is 13/17.
Q. 10. A piggy bank contains a hundred 50p coins, fifty Re 1 coins, twenty Rs
2 coins and ten Rs 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 Rs 5 coin?
Solution:
(i) What is the probability that the coin will be a 50 p coin?
1) A piggy bank contains 100 50p coins, 50 Re 1 coins, 20 Rs 2 coins and 10
Rs 5 coins.
2) Therefore, total number of coins in a piggy bank is 100+50+20+10 = 180.
3) Now, let's find the probability of falling a 50p coin:
probability of falling a 50p coin
= (Number of favarable out comes)/(Total number of outcomes)
probability of falling a 50p coin= (Number of 50p coins)/(Total number of coins)
probability of falling a 50p coin = 100/180 = 5/9
P(falling a 50p coin) = 5/9.
4) Therefore, the probability of falling a 50p coin is 5/9.
(ii) what is the probability that the coin will not be a Rs 5 coin?
5) Now, let's find the probability of not falling a Rs 5 coin:
probability of falling a Rs 5 coin
= (Number of favarable out comes)/(Total number of outcomes)
probability of falling a Rs 5 coin= (Number of Rs 5 coins)/(Total number of coins)
probability of falling a Rs 5 coin = 10/180 = 1/18
P(not falling a Rs 5 coin) = 1 - P(falling a Rs 5 coin).
P(not falling a Rs 5 coin) = 1 - 1/18.
P(not falling a Rs 5 coin) = 17/18.
6) Therefore, the probability of not falling a Rs 5 coin is 17/18.
Q. 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 (see the following fig.). What is the probability that the fish taken out is a male fish?
Solution:
1) A tank contains 5 male fish, and 8 female fish.
2) Therefore, total number of fishes in tank is 5 + 8 = 13.
3) Now, let's find the probability of male fish:
probability of male fish
= (Number of favarable out comes)/(Total number of outcomes)
probability of male fish= (Number of male fishes)/(Total number of fishes)
probability of male fish = 5/13
P(male fish) = 5/13.
4) Therefore, the probability that the fish taken out is a male fish is 5/13.
Q. 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 (see the following fig), 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:
(i) What is the probability that it will point at 8?
1) Here, total number of outcomes is 8.
2) Now, let's find the probability of getting 8:
probability of getting 8
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting 8= (Number of getting 8)/(Total number of outcomes)
probability of getting 8 = 1/8.
3) Therefore, the probability that it will point at 8 is 1/8.
(ii) What is the probability it will point at an odd number?
4) Here, total number of outcomes is 8, and there are 4 odd numbers.
5) Now, let's find the probability of getting odd numbers:
probability of getting odd numbers
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting odd numbers= (Number of getting odd numbers)/(Total number of outcomes)
probability of getting odd numbers = 4/8 = 1/2.
6) Therefore, the probability that it will point at an odd number is 1/2.
(iii) What is the probability that it will point at a number greater than 2?
7) Here, total number of outcomes is 8, and a numbers > 2 are 6.
(i.e. 3, 4, 5, 6, 7, and 8)
8) Now, let's find the probability of getting a number > 2:
probability of getting a number > 2
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a number > 2= (Number of getting a number > 2)/(Total number of outcomes)
probability of getting a number > 2 = 6/8 = 3/4.
9) Therefore,
the probability that it will point at a number greater than 2 is 3/4.
(iv) What is the probability that it will point at a number < 9?
10) Here, total number of outcomes is 8, and a numbers < 9 are 8.
(i.e. 1, 2, 3, 4, 5, 6, 7, and 8)
11) Now, let's find the probability of getting a numbers < 9:
probability of getting a numbers < 9
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a numbers < 9= (Number of getting a numbers < 9)/(Total number of outcomes)
probability of getting a numbers < 9 = 8/8 = 1.
12) Therefore, the probability that it will point at a number less than 9 is 1.
Q. 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:
(i) Find the probability of getting a prime number.
1) Here, total number of outcomes is 6. (i.e. 1, 2, 3, 4, 5, 6) and there are 3
primes (i.e. 2, 3, and 5)
2) Now, let's find the probability of getting a prime number:
probability of getting a prime number
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a prime number= (Number of getting a prime number)/(Total number of outcomes)
probability of getting a prime number = 3/6 = 1/2.
3) Therefore, the probability of getting a prime number is 1/2.
(ii) What is the probability of getting a number between 2 and 6.
4) Here, total number of outcomes is 6. (i.e. 1, 2, 3, 4, 5, 6) and there are 3
numbers lying between 2 and 6 (i.e. 3, 4, and 5)
5) Now, let's find the probability of getting a number lying between 2 and 6:
probability of getting odd numbers
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a number lying between 2 and 6= (Number of getting a number lying between 2 and 6)/(Total number
of outcomes)
probability of getting a number lying between 2 and 6 = 3/6 = 1/2.
6) Therefore,
the probability of getting a number lying between 2 and 6 is 1/2.
(iii) What is the probability of an odd number.
7) Here, total number of outcomes is 6. (i.e. 1, 2, 3, 4, 5, 6) and there are 3
odd numbers (i.e. 1, 3, 5)
8) Now, let's find the probability of getting odd numbers:
probability of getting odd numbers
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting odd numbers= (Number of getting odd numbers)/(Total number of outcomes)
probability of getting odd numbers = 3/6 = 1/2.
9) Therefore, the probability of getting an odd number is 1/2.
Q. 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:
(i) Find the probability of getting a king of red color:
1) Here, total number of cards in a well-shuffled deck is 52 and there are 2
king cards of red colour. (i.e. diamond and heart)
2) Now, let's find the probability of getting a king of red colour:
probability of getting a king of red colour
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a king of red colour= (Number of getting a king of red colour)/(Total number of
outcomes)
probability of getting a king of red colour = 2/52 = 1/26.
3) Therefore, the probability of getting a king of red colour is 1/26.
(ii) Find the probability of getting a face card:
4) Here, total number of cards in a well-shuffled deck is 52 and there are
12 face cards. (i.e. 4 kings, 4 queens, 4 jacks)
5) Now, let's find the probability of getting a face card:probability of getting a face card= (Number of favarable out comes)/(Total number of outcomes)probability of getting a face card= (Number of getting a face card)/(Total number of outcomes)probability of getting a face card = 12/52 = 3/13.6) Therefore, the probability of getting a face card is 3/13.
(iii) Find the probability of getting a red-face card:
7) Here, total number of cards in a well-shuffled deck is 52 and there are6 red-face cards. (i.e. 2 kings, 2 queens, 2 jacks)8) Now, let's find the probability of getting a red-face card:probability of getting a red-face card= (Number of favarable out comes)/(Total number of outcomes)probability of getting a red-face card= (Number of getting a red-face card)/(Total number of outcomes)
probability of getting a red-face card = 6/52 = 3/26.
9) Therefore, the probability of getting a red-face card is 3/26.
(iv) Find the probability of getting the Jack of Hearts:
10) Here, total number of cards in a well-shuffled deck is 52 and there is1 Jack of Heart card. (i.e. 1 jack of Heart)11) Now, let's find the probability of getting Jack of Heart card:probability of getting Jack of Heart card= (Number of favarable out comes)/(Total number of outcomes)probability of getting Jack of Heart card= (Number of getting Jack of Heart card)/(Total number of outcomes)
probability of getting Jack of Heart card = 1/52.
12) Therefore, the probability of getting Jack of Heart card is 1/52.
(v) Find the probability of getting a spade:
13) Here, total number of cards in a well-shuffled deck is 52 and there is13 spade cards. (i.e. 13 spade cards)14) Now, let's find the probability of getting a spade card:probability of getting a spade card= (Number of favarable out comes)/(Total number of outcomes)probability of getting a spade card= (Number of getting a spade card)/(Total number of outcomes)
probability of getting a spade card = 13/52 = 1/4.
15) Therefore, the probability of getting a spade card is 1/4.
(vi) Find the probability of getting the queen of diamonds:
13) Here, total number of cards in a well-shuffled deck is 52 and there is1 queen of diamond card. (i.e. 1 queen of diamond card)14) Now, let's find the probability of getting the queen of diamonds:probability of getting the queen of diamonds= (Number of favarable out comes)/(Total number of outcomes)probability of getting the queen of diamonds= (Number of getting the queen of diamonds)/(Total number of
outcomes)
probability of getting the queen of diamonds = 1/52 = 1/52.
15) Therefore, the probability of getting the queen of diamonds is 1/52.
Q. 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:
(i) What is the probability that the card is the queen?
1) Here, total number of cards in a well-shuffled is 5 and there is 1
queen card of diamond. (i.e. queen card of diamond)
2) Now, let's find the probability that the card is the queen:
probability that the card is the queen
= (Number of favarable out comes)/(Total number of outcomes)
probability that the card is the queen
= (Number of queen card)/(Total number of
outcomes)
probability that the card is the queen = 1/5 = 1/5.
3) Therefore, the probability that the card is the queen is 1/5.
(ii) (a) If the queen is drawn and put aside, what is the probability that the
second card picked up is an ace?
4) Here, the queen is drawn and kept aside, therefore total number of
remaining cards will be 4 and there is 1 ace card.
5) Now, let's find the probability that the second card picked up is an ace:probability that the second card picked up is an ace= (Number of favarable out comes)/(Total number of outcomes)probability that the second card picked up is an ace= (Number of getting ace card)/(Total number of outcomes)probability that the second card picked up is an ace = 1/4.6) Therefore,
the probability that the second card picked up is an ace is 1/4.
(ii) (b) If the queen is drawn and put aside, what is the probability that the
second card picked up is a queen?
7) Here, the queen is drawn and kept aside, therefore total number of
remaining cards will be 4 and as queen is already drawn, there is no queen card.
8) Now, let's find the probability that the second card picked up is queen:probability that the second card picked up is queen= (Number of favarable out comes)/(Total number of outcomes)probability that the second card picked up is queen= (Number of getting queen card)/(Total number of outcomes)probability that the second card picked up is queen = 0/4 = 0.9) Therefore, the probability that the second card picked up is queen is 0.
Q. 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:
1) Here, total number of pens = 12 + 132 = 144, and there are 132 good pens.
2) Now, let's find the probability that thepen taken out is a good one:
probability of getting good pen
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting good pen
= (Number of good pens)/(Total number of outcomes)
probability of getting good pen = 132/144 = 11/12.
3) Therefore, the probability getting good pen is 11/12.
Q. 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) 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?
1) Here, total number of bulbs = 20, and there are 4 defective bulbs.
2) Now, let's find the probability of getting a defective bulb:
probability of getting a defective bulb
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a defective bulb
= (Number of defective bulb)/(Total number of outcomes)
probability of getting a defective bulb = 4/20 = 1/5.
3) Therefore, the probability of getting a defective bulb is 1/5.
(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?
4) As the bulb drawn in (i) is not defective, so total number of bulbs = 19, and
there will be 16 - 1 = 15 non-defective bulbs.
5) Now, let's find the probability of getting non-defective bulb:
probability of getting non-defective bulb
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting non-defective bulb
= (Number of non-defective bulb)/(Total number of outcomes)
probability of getting non-defective bulb = 15/19 = 15/19.
6) Therefore, the probability of getting non-defective bulb is 15/19.
Q. 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) Find the probability that it bears a two-digit number.
1) The total number of discs is 90 and the number of discs with two-digit
numbers are 90 − 9 = 81.
2) Now, let's find the probability of getting two-digit number:
probability of getting two-digit number
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting two-digit number= (Number of two-digit number)/(Total number of outcomes)
probability of getting two-digit number = 81/90 = 9/10.
3) Therefore, the probability of getting two-digit number is 9/10.
(ii) Find the probability that it bears a perfect square number.
4) The total number of discs is 90. The perfect square numbers are 1, 4, 9, 16,
25, 36, 49, 64, 81, making a total of 9 perfect squares.
5) Now, let's find the probability of getting perfect square number:
probability of getting perfect square number
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting perfect square number= (Number perfect squares)/(Total number of
outcomes)
probability of getting perfect square number = 9/90 = 1/10.
6) Therefore, the probability of getting perfect square number is 1/10.
(iii) Find the probability that it bears a number divisible by 5.
7) The total number of discs is 90. The numbers divisible by 5 are 5, 10, 15,
20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90, making a total of 18 numbers.
8) Now, let's find the probability of getting a number divisible by 5:
probability of getting a number divisible by 5
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting a number divisible by 5= (Number of number divisible by 5)/(Total number of
outcomes)
probability of getting a number divisible by 5 = 18/90 = 1/5.
9) Therefore, the probability of getting a number divisible by 5 is 1/5.
Q. 19. 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:
(i) What is the probability of getting A?
1) The total number of outcomes on die is 6 and total number of faces having
A on it is 2.
2) Now, let's find the probability of getting A:
probability of getting A
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting A= (Number of A)/(Total number of outcomes)
probability of getting A = 2/6 = 1/3.
3) Therefore, the probability of getting A is 1/3.
(ii) What is the probability of getting D?
1) The total number of outcomes on die is 6 and total number of faces havingD on it is 1.2) Now, let's find the probability of getting D:probability of getting D= (Number of favarable out comes)/(Total number of outcomes)probability of getting D= (Number of D)/(Total number of outcomes)probability of getting D = 1/6.3) Therefore, the probability of getting D is 1/6.
Q. 20. Suppose you drop a die at random on the rectangular region shown in
the following Fig. What is the probability that it will land inside the circle with a diameter of 1m?
Solution:
1) Area of rectangle = l x b = 3 x 2 = 6 m2.
2) Diameter of the circle =1 m, so its radius = (1/2) m.
3) Therefore the area of circle = π r2 = π (1/2)2 = π/4.
4) Now, let's find the probability of die landing inside the circle:
probability of getting die landing inside the circle
= (Area of the circle)/(Area of the rectangle)
probability of getting die landing inside the circle = (π/4)/6 = π/24
5) Therefore, the probability of getting die landing inside the circle is π/24.
Q. 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) She will not buy it?
Solution:
(i) What is the probability that Nuri will buy a pen?
1) The total number of pens is 144 and the number of
good pens = 144 - 20 = 124.
2) Now, let's find the probability that Nuri will buy a pen:
probability of getting good pen
= (Number of favarable out comes)/(Total number of outcomes)
probability of getting good pen= (Number of good pens)/(Total number of outcomes)
probability of getting good pen = 124/144 = 31/36.
3) Therefore, the probability of Nuri buys pen is 31/36.
(ii) What is the probability that Nuri will not buy a pen?
4) Now, let's find the probability that Nuri will not buy a pen:
probability that Nuri will not buy a pen
= 1 - (probability that Nuri will buy a pen)
= 1 - (31/36)
= ((36 - 31)/36)
= 5/36
5) Therefore, the probability that Nuri will not buy a pen is 5/36.
Q. 22. Two dice, one blue, and one grey, are thrown at the same time.
(i) Write down all the possible outcomes and 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 1/11.
Do you agree with this argument? Justify your answer.
Solution:
(i) Write down all the possible outcomes and complete the following
table:
1) There are 36 possible outcomes when two dice are rolled simultaneously.
2) The only outcome for a sum of 2 is (1, 1),
3) Therefore, there's just 1 possible outcome. P(E) = 1/36.
Therefore, the probability of getting a sum of 2 is 1/36.
4) The outcomes for a sum of 3 are (1, 2) and (2, 1).
5) Therefore, there are only 2 possible outcomes. P(E) = 2/36.
Therefore, the probability of getting a sum of 3 is 2/36.
6) The outcomes for a sum of 4 are (1, 3), (2, 2) and (3, 1).
7) Therefore, there are only 3 possible outcomes. P(E) = 3/36.
Therefore, the probability of getting a sum of 4 is 3/36.
8) The outcomes for a sum of 5 are (1, 4), (2, 3), (3, 2) and (4, 1).
9) Therefore, there are only 4 possible outcomes. P(E) = 4/36.
Therefore, the probability of getting a sum of 5 is 4/36.
10) The outcomes for a sum of 6 are (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1).
11) Therefore, there are only 5 possible outcomes. P(E) = 5/36.
Therefore, the probability of getting a sum of 6 is 5/36.
12) The outcomes for a sum of 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1).
13) Therefore, there are only 6 possible outcomes. P(E) = 6/36.
Therefore, the probability of getting a sum of 7 is 6/36.
14) The outcomes for a sum of 8 are (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).
15) Therefore, there are only 5 possible outcomes. P(E) = 5/36.
Therefore, the probability of getting a sum of 8 is 5/36.
16) The outcomes for a sum of 9 are (3, 6), (4, 5), (5, 4), and (6, 3).
17) Therefore, there are only 4 possible outcomes. P(E) = 4/36.
Therefore, the probability of getting a sum of 9 is 4/36.
18) The outcomes for a sum of 10 are (4, 6), (5, 5), and (6, 4).
19) Therefore, there are only 3 possible outcomes. P(E) = 3/36.
Therefore, the probability of getting a sum of 10 is 3/36.
20) The outcomes for a sum of 11 are (5, 6), and (6, 5).
21) Therefore, there are only 2 possible outcomes. P(E) = 2/36.
Therefore, the probability of getting a sum of 11 is 2/36.
22) The outcomes for 12 are (6, 6).
23) Therefore, there is only 1 possible outcome. P(E) = 1/36.
Therefore, the probability of getting a sum of 12 is 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.
1) The probability of each of these sums is not 1/11 because the sums are not
equally likely to occur.
Q. 23. 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:
1) The possible outcomes are {HHH, HHT, HTH, THH, TTT, TTH, THT, HTT}
2) Therefore the total number of outcomes is 8.
3) The number of favorable outcomes is 2. i.e. {HHH, TTT}.
4) Now, let's find the probability that Hanif will win the game:
probability that Hanif will win the game
= (Number of favarable out comes)/(Total number of outcomes)
probability that Hanif will win the game = 2/8 = 1/4.5) Therefore, Hanif's probability of winning the game is 1/4.
6) Now, let's find the probability that Hanif will lose the game:
probability that Hanif will lose the game
= 1 - (probability that Hanif will win the game)
= 1 - (1/4)
= ((4 - 1)/4)
= 3/4
7) Therefore, the probability that Hanif will lose the game is 3/4.
Q. 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?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
(i) What is the probability that 5 will not come up either time?
1) When a die is rolled twice, there are 36 possible outcomes.
2) The total number of outcomes when 5 comes up on either time are (1, 5), (2, 5),
(3, 5), (4, 5), (5, 5), (6, 5), (5,1), (5, 2), (5, 3), (5, 4) and (5, 6).
3) Thus, there are 11 favorable outcomes where 5 comes up.
4) Therefore, p(5 will be up either times) = 11/36.
5) Therefore,
p(5 will not come up either time) = 1 - p(5 will be up either times).
p(5 will not come up either time) = 1 - 11/36.
p(5 will not come up either time) = (36 - 11)/36.
p(5 will not come up either time) = 25/36.
Therefore, the probability that 5 will not come up either time is 25/36.
(ii) What is the probability that 5 will come up at least once?
6) The number of outcomes where 5 comes up at least once is 11.
7) Therefore, p(5 will come up at least once) = 11/36.
The probability that 5 will come up either time is 11/36.
Q. 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) 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.
1) This argument is incorrect.
2) When two coins are tossed, the possible outcomes are {HH, HT, TH, TT}
3) Therefore, the probability of getting two heads is 1/4.
4) Similarly, the probability of getting two tails is 1/4.
5) The possible outcomes for getting one head and one tail are {HT, TH}
6) Therefore, the probability of getting one of each is 4/2 = 1/2.
7) Therefore, this argument is not correct.
(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.
1) This argument is correct.
2) When a die is thrown, possible outcomes are {1, 2, 3, 4, 5, and 6}
3) Here, odd numbers are 1, 3, 5, and even numbers are 2, 4, 6.
4) Thus, the probability of getting an odd number is 3/6 = 1/2.
5) Therefore, this argument is correct.
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