NCERT Class 10th Math Chapter 14 Exercise 14.1 Question Answer and can take online test here.. CCL Chapter helps you in it.. NCERT Solution for Class 10 Maths Chapter 14 Probability NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.
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NCERT Class 10 Math Chapter 14 Probability Exercise 14.1 Question Answer with Online test of Objective type Solution.
Class 10 Math Chapter 14 Exercise 14.1 Question Answer
1. Complete the following statements :
(i) Probability of an event E + Probability of the event not E = ____________
Ans. 1
(ii) The probability of an event that cannot happen is __________ . Such an event is called __________ .
Ans. zero, Impossible event
(iii) The probability of an event that is certain to happen is __________ . Such an event is called __________ .
Ans. one, Sure event.
(iv) The sum of the probabilities of all the elementary events of an experiment is __________ .
Ans. 1
(v) The probability of an event is greater than or equal to and less than or equal to ____________ .
Ans. 0, 1
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.
Ans. This experiment has not equally likely outcomes.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
Ans. This experiment has not equally likely outcomes.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
Ans. This experiment has equally likely outcomes because either true or false will come in trial.
(iv) A baby is born. It is a boy or a girl.
Ans. This experiment has equally likely outcomes because either a boy child or a girl child be born.
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?
Ans. Because tossing a coin contains two outcomes which are equally likely. In this either first team or the opponent team wins.
4. Which of the following cannot be the probability of an event?
(A) 2/3
(B) –1.5
(C) 15%
(D) 0.7
Ans. (B) –1.5
This is because Probability of any event lies in 0≤ p(x) ≤ 1.
5. If P(E) = 0.05, what is the probability of ‘not E’?
Ans. Probability of not E = 1 – P(E)
= 1 – 0.05
= 0.95
Therefore, Probability of not E = 0.95.
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?
Ans. Probability of an orange flavoured candy = 0.
Because it doesn’t contain any orange flavoured candy.
(ii) a lemon flavoured candy?
Ans. Probability of a lemon flavoured candy = 1.
because all candies are lemon flavoured and one candy that is drawn is also of that flavour.
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?
Ans.
Probability that these 2 students having same birthday = 1 – Probability of 2 students not having same birthday
Probability of 2 students not having same birthday = 0.992
Therefore,
Probability that these 2 students having same birthday = 1 – 0.992
= 0.008
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 ?
Ans.
Probability that a ball drawn is red = favorable outcomes / Total outcome = total red balls/ total balls
Total no of balls = 8
Total no of red balls = 3
Therefore,
Probability that a ball drawn is red = 3/8
(ii) not red?
Ans.
Probability of the ball drawn is not red = 1 – Probability of ball drawn is red
= 1 – 3/8
= 5/8
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 ?
Ans.
Probability that the marble taken out is red = total no of red marbles/ total no of marbles
Total no of marbles = 17
total no of red marbles = 5
Therefore,
Probability that the marble taken out is red = 5/17
(ii) white ?
Ans.
Probability that the marble taken out is white = total no of white marbles/ total no of marbles
Total no of marbles = 17
total no of white marbles = 8
Therefore,
Probability that the marble taken out is white = 8/17
(iii) not green?
Ans.
Probability that the marble taken out is not green = total no of not green marbles/ total no of marbles
Total no of marbles = 17
total no of not green marbles = 13
Therefore,
Probability that the marble taken out is not green = 13/17
Or
Probability that the marble taken out is not green = 1 – probability that marble taken out is green
Probability that marble is green = total green marble/ total marbles = 4/17
Therefore,
Probability that the marble taken out is not green = 1 – 4/17 = 13/17
10. A piggy bank contains hundred 50p coins, fifty Rs.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 ?
Ans. Probability that the coin fall out will be a 50 p coin = total 50 p coins / total coins
total 50 paise coins = 100
total coins = 180 ( 100 + 50 + 20)
Therefore,
Probability that the coin fall out will be a 50 p coin = 100/180 = 10/18 = 5/9
(ii) will not be a Rs.5 coin?
Ans. Probability that the coin fall will not be a Rs. 5 coin = total coins which are not of Rs. 5 / total coins
total coins which are not of Rs. 5 = 170
total coins = 180
Therefore,
Probability that the coin fall will not be a Rs. 5 coin = 170/180 = 17/18
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?
Ans.
Probability that the fish taken out is a male fish = total male fishes / total fishes in the tank
total male fishes in the tank = 5
total fishes in the tank = 13 ( 8 male + 5 female )
Therefore,
Probability that the fish taken out is a male fish = 5/13
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 ?
Ans. Probability of getting 8 = total outcomes of 8 / total outcomes
total outcomes of 8 = 1
total outcomes = 8
Therefore,
Probability of getting 8 = 1/8
(ii) an odd number?
Ans. Probability of getting an odd number = total odd outcomes / total outcomes
total odd outcomes = 4
total outcomes = 8
Therefore,
Probability of getting an odd number = 4/8 = ½
(iii) a number greater than 2?
Ans. Probability of getting a number greater than 2 = total favorable outcomes / total outcomes
total favorable outcomes or total outcomes greater than 2 = 6
total outcomes = 8
Therefore,
Probability of getting a number greater than 2 = 6/8 = 3/4
(iv) a number less than 9?
Ans. Probability of getting a no less than 9 = 1
because all the given numbers are less than 9.
13. A die is thrown once. Find the probability of getting
(i) a prime number;
Ans.
Probability of getting a prime number in a die = total prime no in a die/ total no in a die
total no in a die = 6
total prime no in a die = 3 ( 2, 3, 5)
Therefore,
Probability of getting a prime number in a die = 3/6
= 1/2
(ii) a number lying between 2 and 6;
Ans.
Probability of a number lying between 2 and 6 in a die = total no lying between 2 and 6 in a die/ total no in a die
total no in a die = 6
total no lying between 2 and 6 = 3 ( 3, 4, 5)
Therefore,
Probability of a number lying between 2 and 6 in a die = 3/6 = 1/2
(iii) an odd number.
Ans.
Probability of getting an odd number = total odd no of a die/total no of a die
Total no of a die = 6
total odd no of a die = 3 ( 1, 3, 5)
Therefore,
Probability of getting an odd number = 3/6 = 1/2
14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
Ans. Probability of getting a king of red colour = total king of red colour/ total cards in a pack
total king of red colour = 2
total cards in a pack = 52
Therefore,
Probability of getting a king of red color = 2/52 = 1/26
(ii) a face card
Ans.
Probability of getting a face card = total face cards/total cards
total face cards = 12
total cards in a pack = 52
Therefore,
Probability of getting a face card = 12/52 = 3/13
(iii) a red face card
Ans.
Probability of getting a red face card = total red face card/ total cards
total red face cards = 6
total cards in a pack = 52
Therefore,
Probability of getting a red face card = 6/52 = 3/26
(iv) the jack of hearts
Ans.
Probability of getting the jack of hearts = total cards of the jack of hearts /total cards
total cards of the jack of hearts = 1
total cards in a pack = 52
Therefore,
Probability of getting the jack of hearts = 1/52
(v) a spade
Ans. Probability of getting a spade card = total spade cards / total cards in a pack
total spade cards = 13
total cards in a pack = 52
Therefore,
Probability of getting a spade card = 13/52 = 1/4
(vi) the queen of diamonds
Ans.
Probability of getting the queen of diamonds = total cards of the queen of diamonds / total cards in a pack
total cards of the queen of diamonds = 1
total cards in a pack = 52
Therefore,
Probability of getting the queen of diamonds = 1/52
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?
Ans.
Probability that the card is queen = total queen in given cards / total given cards
total queen in given cards = 1
total given cards = 5
Therefore,
Probability that the card is queen = 1/5
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is
(a) an ace?
Ans. Now total 4 cards remained.
Probability of getting an ace = total given ace/total given cards
total given ace = 1
total given cards = 4
Therefore,
Probability of getting an ace = 1/4
(b) a queen?
Ans. Now total 4 cards remained.
Probability of getting queen = zero . Because there is no zero in remined 4 cards.
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.
Ans.
Probability of getting a good pen = total good pens / total pens
total good pens = 132
total pens = 12 + 132 = 144
Therefore,
Probability of getting a good pen = 132/144 = 11/12
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?
Ans.
Probability of getting a defective bulb = total defective bulb / total bulbs
total defective bulbs = 4
total bulbs = 20
Therefore,
Probability of getting a defective bulb = 4/20 = 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 ?
Ans. The bulb drawn in (i) is not defective and is not replaced.
Therefore, total bulbs at present are 19 in which 15 good one and 4 defective.
Now, Probability that a bulb drawn in not defective = 1 – probability of getting a defective bulbs
Probability of getting a defective bulb = 4/19
Therefore,
Probability that a bulb drawn is not defective = 1 – 4/19
= 15/19
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
Ans.
Probability that a disc drawn is a two digit no = total two digit numbers/total given numbers
total two digits given discs = 81
total given discs = 90
therefore,
Probability that a disc drawn is a two digit no = 81/90 = 9/10
(ii) a perfect square number
Ans.
Probability that a disc drawn is a perfect square no = total perfect square no discs/ total discs
total perfect square no = 9
total discs = 90
Therefore,
Probability that a disc drawn is a perfect square no = 9/90 = 1/10
(iii) a number divisible by 5.
Ans.
Probability that a disc drawn is a no divisible by 5 = total disc divisible by 5/ total given discs
total discs divisible by 5 = 18 ( 5, 10, …. 85, 90)
total given discs = 90
Therefore,
Probability that a disc drawn is a no divisible by 5 = 18/90 = 1/5
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?
Ans. P(A) = favorable outcomes of A/ total outcomes
favorable outcomes of A = 2
total outcomes = 6
Therefore,
P(A) = 2/6 = 1/4
(ii) D?
Ans. P(D) = total no of D/ total outcomes
total no of D = 1
total outcomes = 6
Therefore,
P(D) = 1/6
20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with diameter 1m?
Ans. Given that diameter of circle is 1m. Therefore radius of circle is ½m.
Probability of getting a drop of die inside the circle = area of circle/area of total land
Area of circle = πr2 = π(½)2 = π/4 m2
Area of total land = 3 × 2 = 6m2
Therefore,
Probability of getting a drop of die inside the circle = π/4/6 = π/24
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 ?
Ans.
Probability that Nuri will buy a ball pen = total good ball pens / total ball pens
total good ball pens = 144 – 20 = 124
total ball pens = 144
Therefore,
Probability that Nuri will buy it = 124/144 = 31/36
(ii) She will not buy it ?
Ans.
Probability that Nuri will not by a ball pen = 1 – Probability that Nuri will by a good ball pen
= 1 – 31/36
= 5/36
22. Refer to Example 13. example 13- Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) Complete the following table:
| Event ‘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 |
Ans. total outcomes are as :
| (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) |
total outcomes with sum 3 = 2 { (1,2), (2,1)}
Therefore, probability of getting sum 3 = 2/36
total outcomes with sum 4 = 3 {(1,3), (3,1), (2,2)}
Therefore, probability of getting sum 4 = 3/36
total outcomes with sum 5 = 4 {(4,1), (3,2), (2,3), (1,4)}
Therefore, probability of getting sum 5 = 4/36
total outcomes with sum 6 = 5 {(5,1), (4,2), (3,3), (2,4), (1,5)}
Therefore, probability of getting sum 6 = 5/36
total outcomes with sum 7 = 6 {(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)}
Therefore, probability of getting sum 7 = 6/36
total outcomes with sum 8 = 5 {(6,2), (5,3), (4,4), (3,5), (2,6)}
Therefore, probability of getting sum 8 = 5/36
total outcomes with sum 9 = 4 {(6,3), (5,4), (4,5), (3,6)}
Therefore, probability of getting sum 9 = 4/36
total outcomes with sum 10 = 3 {(6,4), (4,6), (5,5)}
Therefore, probability of getting sum 10 = 3/36
total outcomes with sum 11 = 2 {(5,6), (6,5)}
Therefore, probability of getting sum 11 = 2/36
total outcomes with sum 12 = 1 {(6,6)}
Therefore, probability of getting sum 12= 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.
Ans. No. The eleven sums are not equally likely.
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.
Ans. Given that Hanif will loses if he didn’t get three heads or three tails
Probability that Hanif will lose the game = 1- Probability that he wins
Probability that he wins = total no of outcomes of three heads or three tails/ total outcomes
total no of outcomes of three heads or three tails = 2 ( HHH, TTT)
Total outcomes = 23 = 8 (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)
Therefore,
Probability that he wins = 2/8
and
Probability that Hanif will lose the game = 1 – 2/8 = 6/8 = 3/4
24. A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
Ans.
Probability that 5 will not come up either time = 1 – Probability that 5 will come on either time
Therefore,
Probability that 5 will come on either time = total outcome with no 5/total outcomes
total outcomes with no 5 = 11 { (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (1, 5), (2, 5), (3,5), (4,5) (6,5) }
total outcomes = 62 =36
Probability that 5 will come on either time = 11/36
Probability that 5 will not come up either time = 1 – 11/36
= 25/36
(ii) 5 will come up at least once?
Ans. Probability that 5 will come at least = 11/36
Done in (i) part.
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
Ans. Wrong
We can classify the outcomes like this but they are not then equally likely. Reason is that 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 tales)
(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 .
Ans. Correct
The two outcomes considered in the question are equally likely.