(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
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
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