Selina Solutions Concise Maths Class 10 Chapter 25 Probability Exercise 25(B)

Events like tossing a coin once and several times, throwing a dice or two dice simultaneously and picking cards from a pack are the major topics covered in this exercise. The Selina Solutions for Class 10 Maths is the best resource a student can have for doing quick reference and preparations for exams. The Concise Selina Solutions for Class 10 Maths Chapter 25 Probability Exercise 25(B) solutions are available in a PDF in the below link.

Selina Solutions Concise Maths Class 10 Chapter 25 Probability Exercise 25(B) Download PDF

Access Selina Solutions Concise Maths Class 10 Chapter 25 Probability Exercise 25(B)

1. Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:

(i) an even number

(ii) a multiple of 3

(iii) an even number and a multiple of 3

(iv) an even number or a multiple of 3

Solution:

We know that, there are totally 9 cards from which one card is drawn.

Total number of elementary events = n(S) = 9

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(i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10

So, favorable number of events = n(E) = 5

Hence, probability of selecting a card with an even number = n(E)/ n(S) = 5/9

(ii) From numbers 2 to 10, there are 3 numbers which are multiples of 3 i.e. 3, 6, 9

So, favorable number of events = n(E) = 3

Hence, probability of selecting a card with a multiple of 3= n(E)/ n(S) = 3/9 = 1/3

(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6

So, favorable number of events = n(E) = 1

Hence, probability of selecting a card with a number which is an even number as well as multiple of 3

= n(E)/ n(S) = 1/9Â

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10

So, favorable number of events = n(E) = 7

Hence, probability of selecting a card with a number which is an even number or a multiple of 3

=Â n(E)/ n(S) = 7/9

2. Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:

(i) a multiple of 5

(ii) a multiple of 6

(iii) between 40 and 60

(iv) greater than 85

(v) less than 48

Solution:

We kwon that, there are 100 cards from which one card is drawn.

Total number of elementary events = n(S) = 100

(i)Â From numbers 1 to 100, there are 20 numbers which are multiple of 5 i.e.Â {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}Â

So, favorable number of events = n(E) = 20

Hence, probability of selecting a card with a multiple of 5 = n(E)/ n(S) = 20/ 100 = 1/5

(ii) From numbers 1 to 100, there are 16 numbers which are multiple of 6 i.e.Â {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}Â

So, favorable number of events = n(E) = 16

Hence, probability of selecting a card with a multiple of 6 = n(E)/ n(S) = 16/ 100 = 4/25

(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e.Â {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59}Â

So, favorable number of events = n(E) = 19

Hence, probability of selecting a card between 40 and 60 = n(E)/ n(S) = 19/100

(iv) From numbers 1 to 100, there are 15 numbers which are greater than 85 i.e.Â {86, 87, â€¦., 98, 99, 100}Â

So, favorable number of events = n(E) = 15

Hence, probability of selecting a card with a number greater than 85 = n(E)/ n(S) = 15/100 = 3/20

(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e.Â {1, 2, â€¦â€¦â€¦.., 46, 47}Â

So, favorable number of events = n(E) = 47

Hence, probability of selecting a card with a number less than 48 = n(E)/ n(S) = 47/100

3. From 25 identical cards, numbered 1, 2, 3, 4, 5, â€¦â€¦,Â 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:

(i) 3

(ii) 5

(iii) 3 and 5

(iv) 3 or 5

Solution:

We know that, there are 25 cards from which one card is drawn.

So, the total number of elementary events = n(S) = 25

(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e.Â {3, 6, 9, 12, 15, 18, 21, 24}Â

So, favorable number of events = n(E) = 8

Hence, probability of selecting a card with a multiple of 3 = n(E)/ n(S) = 8/25

(ii) From numbers 1 to 25, there are 5 numbers which are multiple of 5 i.e.Â {5, 10, 15, 20, 25}Â

So, favorable number of events = n(E) = 5

Hence, probability of selecting a card with a multiple of 5 = n(E)/ n(S) = 5/25 = 1/5

(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e.Â {15}Â

So, favorable number of events = n(E) = 1

Hence, probability of selecting a card with a multiple of 3 and 5 = n(E)/ n(S) = 1/25

(iv) From numbers 1 to 25, there are 12 numbers which are multiple of 3 or 5 i.e.Â {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}Â

So, favorable number of events = n(E) = 12

Hence, probability of selecting a card with a multiple of 3 or 5 = n(E)/ n(S) = 12/25

4. A die is thrown once. Find the probability of getting a number:

(i) less than 3

(ii) greater than or equal to 4

(iii) less than 8

(iv) greater than 6

Solution:

We know that,

In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}

So, n(S) = 6

(i) On a dice, numbers less than 3 = {1, 2}

So, n(E) = 2

Hence, probability of getting a number less than 3 =Â n(E)/ n(S) = 2/6 = 1/3

(ii) On a dice, numbers greater than or equal to 4 = {4, 5, 6}

So, n(E) = 3

Hence, probability of getting a number greater than or equal to 4 = n(E)/ n(S) = 3/6 = 1/2

(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6}

So, n(E) = 6

Hence, probability of getting a number less than 8 = n(E)/ n(S) = 6/6 = 1Â

(iv) On a dice, numbers greater than 6 = 0

So, n(E) = 0

Hence, probability of getting a number greater than 6 = n(E)/ n(S) = 0/6 = 0Â

5. A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?

Solution:

We know that,

Number of pages in the book = 85

Number of possible outcomes = n(S) = 85

Out of 85 pages, pages that sum up to 8 = {8, 17, 26, 35, 44, 53, 62, 71, 80}

So, pages that sum up to 8 = n(E) = 9

Hence, probability of choosing a page with the sum of digits on the page equals 8 = n(E)/ n(S) = 9/85

6. A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.

Solution:

In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}

So, n(S) = 6

For two dice, n(S) = 6Â x 6 = 36

Favorable cases where the sum is 10 or more with 5 on 1stÂ die = {(5, 5), (5, 6)}

Event of getting the sum is 10 or more with 5 on 1stÂ die = n(E) = 2

Hence, the probability of getting a sum of 10 or more with 5 on 1stÂ die = n(E)/ n(S) = 2/ 36 = 1/18

7. If two coins are tossed once, what is the probability of getting:

(iii)Â bothÂ heads or both tails.

Solution:

We know that, when two coins are tossed together possible number of outcomes = {HH, TH, HT, TT}

So, n(S) = 4

(i) E = event of getting both heads = {HH}

n(E) = 1

Hence, probability of getting both heads =Â n(E)/ n(S) = Â¼

(ii) E = event of getting at least one head = {HH, TH, HT}

n(E) = 3

Hence, probability of getting at least one head =Â n(E)/ n(S) = Â¾

(iii) E = event of getting both heads or both tails = {HH, TT}

n(E) = 2

Hence, probability of getting both heads or both tails =Â n(E)/ n(S) = 2/4 = Â½