(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
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