# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 19 - Probability

Page / Exercise

## Chapter 15 - Probability Ex. 19

Solution 1

(i) The probability of an impossible event is zero.

(ii) The probability of a sure event is one.

(iii) For any event E, P(E) + P(not E)= one .

(iv) The probability of a possible but not a sure event lies between zero and one.

(v) The sum of probabilities of all the outcomes of an experiment is one.

Solution 2

Solution 3

Solution 4

In a throw of a dice, all possible outcomes are 1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

(i)Let E be event of getting even number

Then, the favorable outcomes are 2, 4, 6

Number of favorable outcomes= 3

P(getting a even number)= P(E) =

(ii)Let R be the number less than 5

Then, the favorable outcomes are 1, 2, 3, 4

Number of favorable outcomes = 4

P(getting a number less than 5)= P(R) =

(iii)Let M be the event of getting a number greater than 2

Then, the favorable outcomes are 3, 4, 5, 6

Number of favorableoutcomes = 4

P(getting a number greater than 2)= P(M)

(iv)Let N be the number lying between 3 and 6

Then the favorable outcomes are 4, 5

Number of favorable outcomes = 2

P(getting a number 3 and 6)= P(N) =

(v)Let G be event of getting a number other than 3

Then the favorable outcomes are 1, 2, 4, 5, 6

Number of favorable outcomes = 5

P(getting a number other than 5)=P(G) =

(vi)Let T be event of getting a number 5

Then the favorable outcome is 5

Number of favorable outcomes = 1

P(getting a number 5)=P(T)

Solution 5

Solution 6

Solution 8

Solution 9

Total number of tickets sold = 250

Number of prizes = 5

Let E be the event getting a prize

Number of favorable outcomes = 5

P(getting a prize) =

Solution 10

Solution 11

Spinning arrow may come to rest at one of the 12 numbers

total number of outcomes = 12

(i)Probability that it will point at 6 =

(ii)Even numbers are 2, 4, 6, 8, 10 and 12. There are 6 numbers.

Probability that it points at even numbers =

(iii)The prime numbers are 2,3 5, 7 and 11. There are 5 prime numbers.

Probability that it points at prime number =

(iv)There are 2 numbers divisible by 5. These are 5 and 10.

Probability that a number is a multiple of 5 =

Solution 12

Solution 13

Solution 15

Solution 17

Solution 18

Solution 20

Solution 22

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Total number of tickets = 100

(i)Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100

Total number of even number = 50

P(getting a even number) =

(ii)Numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Total number of numbers less than 16 is 14

P(getting a number less than 16) =

(iii)Numbers which are perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100

Total number of perfect squares = 9

P(getting a perfect square) =

(iv)Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Total number of prime numbers =12

P(getting a prime number less 40) =

Solution 31

Solution 37

Solution 39

Solution 40

Solution 42

Solution 43

Solution 44

There are 26 red cards containing a 2 queensand 2 more black queens are there in a pack of cards

P(getting a red card or a queen)

P(getting neither a red card nor a queen) =

Solution 45

Solution 46

Total number of cards = 52

(i)There are 13 cards of spade (including 1 ace) and 3 more ace cards are there in a pack of cards

P(getting a card of spades or an ace) =

(ii)There are 2 red kings in a pack of cards

P(getting a red king) =

(iii)There are 4 kings and 4 queens in a pack of cards

P(getting either a king or a queen) =

(iv)P(getting neither a king nor a queen) =

### STUDY RESOURCES

REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India.