ML Aggarwal Solutions Class 8 Mathematics Solutions for Data Handling Exercise 19.3 in Chapter 19 - Data Handling

Question 9 Data Handling Exercise 19.3

A box contains 17 cards numbered 1, 2, 3…….., 17 and are mixed thoroughly. A card is drawn at random from the box. Find the probability that the number on that card is

(i) odd

(ii) even

(iii) prime

(iv) divisible by 3

(v) divisible by 2 and 3 both

Answer:

Given A box contains 17 cards numbered 1 to 17 So, total number of outcomes = 17

(i) Card bearing odd number (1, 3, 5, 7, 9, 11, 13, 15, 17) = 9 Therefore, Probability P(E) = 9 / 17

(ii) Even number (2, 4, 6, 8, 10, 12, 14, 16) = 8 Therefore, Probability P(E) = 8 / 17

(iii) Prime numbers (2, 3, 5, 7, 11, 13, 17) = 7 Therefore,

Probability P(E) = 7 / 17

(iv) Numbers divisible by 3 3, 6, 9, 12, 15 = 5 Therefore, Probability P(E) = 5 / 17

(v) Numbers divisible by 2 and 3 both 6, 12 = 2 Therefore, Probability P(E) = 2 / 17

Video transcript

"hello everyone i am suman jaleen math tutor from leader learning welcome to question and answer session given question is a box contains 17 cards numbered 1 2 3 and so on up to 17 and are mixed thoroughly a card is drawn at random from the box find the probability that the number on that card card is five different conditions was given we need to calculate we need to identify the probability of these conditions okay so first of all like yeah we need to understand we need to understand the card numbers that is from 1 to 17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 okay so out of this the odd numbers are 1 3 5 7 9 11 13 15 and 17 so there are eight eight odd numbers okay so there are not eight there are nine odd numbers okay so the probability is 9 out of 17 and the even numbers so the probability of getting a card with even number is 2 4 6 8 10 12 14 and 16 okay so these are these are eight numbers so 8 out of 17 is the probability and the third question is a prime number prime number is nothing but which is having two factors only one and the same number okay one and the same number so here the prime numbers are 2 3 5 7 11 13 17. these are the prime numbers there are totally seven prime numbers so the probability is 7 out of 17 okay so and the fourth condition is divisible by 3 okay the numbers here divisible by 3 are 3 6 9 12 15 15 that's it okay so there are five outcomes so 5 out of 17 is the answer and the last question is divisible by both 2 and 3 so the numbers which are divisible by both 2 and 3 are first of all 6 12 6 12 that's it only two numbers so the probability is 2 out of 17 okay so this is the probability and the answer hope you understand if you have further questions reach us at comment section and subscribe to the channel leader learning for further queries thank you "

Related Questions

List the outcomes you can see in these experiments

A die is rolled once. Find the probability of getting(i) an even number(ii) a multiple of 3(iii) not...

Two coins are tossed together. Find the probability of getting(i) two tails(ii) atleast one tail(iii...

Three coins are tossed together. Find the probability of getting(i) atleast two heads(ii) atleast on...

Two dice are rolled simultaneaously. Find the probability of getting(i) the sum as 7(ii) the sum as ...

A box contains 600 screws, one tenth are rusted. One screw is taken out atrandom from the box. Find ...

Exercises

Data Handling Exercise 19.1

Data Handling Exercise 19.2

Data Handling Exercise 19.3

Chapters