(i) 1 (ii) 0, impossible event (iii) 1, sure event or certain event (iv) 1 (v) 0,1 The above question is from ncert class 10-chapter 15 probability exercise 15.1. See here for video explanation of the exercise 15.1 👇
(i) 1
(ii) 0, impossible event
(iii) 1, sure event or certain event
(iv) 1
(v) 0,1
The above question is from ncert class 10-chapter 15 probability exercise 15.1.
See here for video explanation of the exercise 15.1 👇
(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same. (ii) It is not an equally likely event, as it depends on the player's ability and there is no information given about that. (iii) IRead more
(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same.
(ii) It is not an equally likely event, as it depends on the player’s ability and there is no information given about that.
(iii) It is an equally likely event.
(iv) tis an equally likely event.
This question is from ncert chapter 15 probability exercise 15.1
See this for better understanding 😀👇
Total number of pens = 144 Total number of defective pens = 20 Total number of good pens = 144 - 20 = 124 (i) Probability of getting a good pen = 124/144 = 31/36 P (Nuri buys a pen) = 31/36 (ii) P (Nuri will not buy a pen) = 1 - 31/36 = (36 - 31)/36 = 5/36
Total number of pens = 144
Total number of defective pens = 20
Total number of good pens = 144 – 20 = 124
(i) Probability of getting a good pen = 124/144 = 31/36
P (Nuri buys a pen) = 31/36
(ii) P (Nuri will not buy a pen) = 1 – 31/36 = (36 – 31)/36 = 5/36
(i) It can be observed that, To get the sum as 2, possible outcomes =(1, 1) To get the sum as 3, possible outcomes = (2,1) and (1, 2) To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2) To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2) To get the sum as 6, possibleRead more
(i) It can be observed that,
To get the sum as 2, possible outcomes =(1, 1)
To get the sum as 3, possible outcomes = (2,1) and (1, 2)
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4). (4, 2), (3, 3)
To get the sum as 7, possible outcomes (6, 1). (1, 6), (2, 5), (5, 2), (3, 4), (4, 3)
To get the sum as 8, possible outcomes = (6, 2), (2, 6),. (3, 5), (5, 3), (4, 4)
To get the sum as 9, possible outcomes (3, 6), (6, 3), (4, 5), (5, 4)
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
To get the sum as 11, possible outcomes = (5, 6). (6, 5)
To get the sum as 12, possible outcomes [6. 6)
Total number of outcomes = 6 × 6 36 (i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2). (5, 3). (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) Hence, total number of favourable cases = 11 P (5 will come up either time) = 11/36 P (5 will not come up eitherRead more
Total number of outcomes = 6 × 6 36
(i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2). (5, 3). (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Hence, total number of favourable cases = 11
P (5 will come up either time) = 11/36
P (5 will not come up either time) = 1 – 11/36 = 25/36
(ii) Total number of cases, when 5 can come at least once = 11
P (5 will come at least once) = 11/36
(i) Incorrect When two coins are tossed, the possible outcomes are (H, H). (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways -(H, T). (T, H). Therefore, the probability of getting two heads is 1/4, the probability of getting two tails is 1/4, and the pRead more
(i) Incorrect
When two coins are tossed, the possible outcomes are (H, H). (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways -(H, T). (T, H).
Therefore, the probability of getting two heads is 1/4, the probability of getting two tails is 1/4, and the probability of getting one of each is 1/2.
It can be observed that for each outcome, the probability is not 1/3.
(ii) Correct
When a dice is thrown, the possible outcomes are 1,2, 3, 4, 5, and 6. Out of these, 1,3, 5 are odd and 2,4,6 are even numbers.
Therefore, the probability of getting an odd number is 1/2.
Total number of possible outcomes when two dice are thrown = 6 x 6= 36 (i) Total times when the sum is even = 18 P (getting an even number) = 18/36 = 1/2 (ii) Total times when the sum is 6 = 4 P (getting sum as 6) = 4/36 = 1/9 (iii) Total times when the sum is at least 6 i.e., greater than 5) = 15 PRead more
Total number of possible outcomes when two dice are thrown = 6 x 6= 36
(i) Total times when the sum is even = 18
P (getting an even number) = 18/36 = 1/2
(ii) Total times when the sum is 6 = 4
P (getting sum as 6) = 4/36 = 1/9
(iii) Total times when the sum is at least 6 i.e., greater than 5) = 15
P (getting sum at least 6) = 15/36 = 5/12
Total number of marbles = 24 Let the total number of green marbles be x. Then, total number of blue marbles 24 - x P (getting a given marble) = x/24 According to the condition given in the question, x/24 = 2/3 ⇒ x = 16 Therefore, total number of green marbles in the jar = 16 Hence, total number of bRead more
Total number of marbles = 24
Let the total number of green marbles be x.
Then, total number of blue marbles 24 – x
P (getting a given marble) = x/24
According to the condition given in the question,
x/24 = 2/3
⇒ x = 16
Therefore, total number of green marbles in the jar = 16
Hence, total number of blue marbles = 24 – x = 24 – 16 = 8
Total number of possible outcomes on the dice = 6 (i)Total number of faces having A on it = 2 P (getting A) = 2/6 = 1/3 (ii) Total number of faces having D on it = 1 P (getting D) = 1/6 See here for video explanation of exercise 15.1✌😄
Total number of possible outcomes on the dice = 6
(i)Total number of faces having A on it = 2
P (getting A) = 2/6 = 1/3
(ii) Total number of faces having D on it = 1
P (getting D) = 1/6
See here for video explanation of exercise 15.1✌😄
Area of rectangle = l x b = 3 x 2 = 6m² Area of circle (of diameter 1 m) = πr² = π(1/2)² = π/4 P (die will land inside the circle) = (π/4)/6 = π/24 Explanation of Exercise 15.1 👇😄
Area of rectangle = l x b = 3 x 2 = 6m²
Area of circle (of diameter 1 m) = πr² = π(1/2)² = π/4
P (die will land inside the circle) = (π/4)/6 = π/24
Complete the following statements:
(i) 1 (ii) 0, impossible event (iii) 1, sure event or certain event (iv) 1 (v) 0,1 The above question is from ncert class 10-chapter 15 probability exercise 15.1. See here for video explanation of the exercise 15.1 👇
(i) 1
(ii) 0, impossible event
(iii) 1, sure event or certain event
(iv) 1
(v) 0,1
The above question is from ncert class 10-chapter 15 probability exercise 15.1.
See lessSee here for video explanation of the exercise 15.1 👇
Which of the following experiments have equally likely outcomes? Explain.
(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same. (ii) It is not an equally likely event, as it depends on the player's ability and there is no information given about that. (iii) IRead more
(i) It is not an equally likely event, as it depends on various factors such as whether the car will start or not. And factors for both the conditions are not the same.
(ii) It is not an equally likely event, as it depends on the player’s ability and there is no information given about that.
(iii) It is an equally likely event.
(iv) tis an equally likely event.
This question is from ncert chapter 15 probability exercise 15.1
See lessSee this for better understanding 😀👇
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
Total number of pens = 144 Total number of defective pens = 20 Total number of good pens = 144 - 20 = 124 (i) Probability of getting a good pen = 124/144 = 31/36 P (Nuri buys a pen) = 31/36 (ii) P (Nuri will not buy a pen) = 1 - 31/36 = (36 - 31)/36 = 5/36
Total number of pens = 144
See lessTotal number of defective pens = 20
Total number of good pens = 144 – 20 = 124
(i) Probability of getting a good pen = 124/144 = 31/36
P (Nuri buys a pen) = 31/36
(ii) P (Nuri will not buy a pen) = 1 – 31/36 = (36 – 31)/36 = 5/36
Refer to 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) It can be observed that, To get the sum as 2, possible outcomes =(1, 1) To get the sum as 3, possible outcomes = (2,1) and (1, 2) To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2) To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2) To get the sum as 6, possibleRead more
(i) It can be observed that,
See lessTo get the sum as 2, possible outcomes =(1, 1)
To get the sum as 3, possible outcomes = (2,1) and (1, 2)
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4). (4, 2), (3, 3)
To get the sum as 7, possible outcomes (6, 1). (1, 6), (2, 5), (5, 2), (3, 4), (4, 3)
To get the sum as 8, possible outcomes = (6, 2), (2, 6),. (3, 5), (5, 3), (4, 4)
To get the sum as 9, possible outcomes (3, 6), (6, 3), (4, 5), (5, 4)
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
To get the sum as 11, possible outcomes = (5, 6). (6, 5)
To get the sum as 12, possible outcomes [6. 6)
A die is thrown twice. What is the probability that
Total number of outcomes = 6 × 6 36 (i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2). (5, 3). (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) Hence, total number of favourable cases = 11 P (5 will come up either time) = 11/36 P (5 will not come up eitherRead more
Total number of outcomes = 6 × 6 36
See less(i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2). (5, 3). (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Hence, total number of favourable cases = 11
P (5 will come up either time) = 11/36
P (5 will not come up either time) = 1 – 11/36 = 25/36
(ii) Total number of cases, when 5 can come at least once = 11
P (5 will come at least once) = 11/36
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) Incorrect When two coins are tossed, the possible outcomes are (H, H). (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways -(H, T). (T, H). Therefore, the probability of getting two heads is 1/4, the probability of getting two tails is 1/4, and the pRead more
(i) Incorrect
See lessWhen two coins are tossed, the possible outcomes are (H, H). (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways -(H, T). (T, H).
Therefore, the probability of getting two heads is 1/4, the probability of getting two tails is 1/4, and the probability of getting one of each is 1/2.
It can be observed that for each outcome, the probability is not 1/3.
(ii) Correct
When a dice is thrown, the possible outcomes are 1,2, 3, 4, 5, and 6. Out of these, 1,3, 5 are odd and 2,4,6 are even numbers.
Therefore, the probability of getting an odd number is 1/2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
Total number of possible outcomes when two dice are thrown = 6 x 6= 36 (i) Total times when the sum is even = 18 P (getting an even number) = 18/36 = 1/2 (ii) Total times when the sum is 6 = 4 P (getting sum as 6) = 4/36 = 1/9 (iii) Total times when the sum is at least 6 i.e., greater than 5) = 15 PRead more
Total number of possible outcomes when two dice are thrown = 6 x 6= 36
See less(i) Total times when the sum is even = 18
P (getting an even number) = 18/36 = 1/2
(ii) Total times when the sum is 6 = 4
P (getting sum as 6) = 4/36 = 1/9
(iii) Total times when the sum is at least 6 i.e., greater than 5) = 15
P (getting sum at least 6) = 15/36 = 5/12
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3.
Total number of marbles = 24 Let the total number of green marbles be x. Then, total number of blue marbles 24 - x P (getting a given marble) = x/24 According to the condition given in the question, x/24 = 2/3 ⇒ x = 16 Therefore, total number of green marbles in the jar = 16 Hence, total number of bRead more
Total number of marbles = 24
See lessLet the total number of green marbles be x.
Then, total number of blue marbles 24 – x
P (getting a given marble) = x/24
According to the condition given in the question,
x/24 = 2/3
⇒ x = 16
Therefore, total number of green marbles in the jar = 16
Hence, total number of blue marbles = 24 – x = 24 – 16 = 8
A child has a die whose six faces show the letters as given below:
Total number of possible outcomes on the dice = 6 (i)Total number of faces having A on it = 2 P (getting A) = 2/6 = 1/3 (ii) Total number of faces having D on it = 1 P (getting D) = 1/6 See here for video explanation of exercise 15.1✌😄
Total number of possible outcomes on the dice = 6
(i)Total number of faces having A on it = 2
P (getting A) = 2/6 = 1/3
(ii) Total number of faces having D on it = 1
P (getting D) = 1/6
See here for video explanation of exercise 15.1✌😄
See lessSuppose you drop a die at random on the rectangular region shown in Figure.
Area of rectangle = l x b = 3 x 2 = 6m² Area of circle (of diameter 1 m) = πr² = π(1/2)² = π/4 P (die will land inside the circle) = (π/4)/6 = π/24 Explanation of Exercise 15.1 👇😄
Area of rectangle = l x b = 3 x 2 = 6m²
Area of circle (of diameter 1 m) = πr² = π(1/2)² = π/4
P (die will land inside the circle) = (π/4)/6 = π/24
Explanation of Exercise 15.1 👇😄
See less