NCERT solutions class 10 maths chapter 15 Probability are given here. NCERT textbooks introduce the important concepts of probability. In this chapter, students learn several important terms and concepts that form the main foundation of the vast concept of probability.Â NCERT class 10 solutionsÂ for chapter 15 probability given here can help the students to clear their doubts that can arise while solving questions and help them to develop problem-solving abilities. These solutions given here are simple and can any students can easily understand the steps involved. Check the class 10 probability chapter 15 NCERT solutions below. It is important to understand the different types of problems one may encounter on the topic of probability. Probability is an important chapter in 10th-grade mathematics and needs to be remembered as there can be important questions that can be asked in the examination.

### NCERT Solutions Class 10 Maths Chapter 15 Exercises

*1-Two dice are thrown simultaneously. What is the probability that the sum of the numbers is *

*a) *8?

*b) *2?

*c) *Prime number?

**Sol:Â Â Â Â Â **When two dice are thrown simultaneously, the maximum number of possibilities in the sample

space is (\(6\times 6\)=36).

**a)**Sum of numbers is 8 {(2,6),(3,5),(4,4),(6,2),(5,3)}

Therefore, P(sum 8) =Â \(\frac{5}{36}\)

** b)**Sum of numbers is 2 {(1,1)}

Therefore, P(sum 2) =Â \(\frac{1}{36}\)

**c)**Sum is a Prime number {(1,1),(1,2),(1,4),(2,1),(2,3),(3,2),(4,1),(1,6),(2,5),(3,4),(4,3),(5,2),(5,6),(6,1),(6,5)}

(Because, prime numbers within 12 are, 2, 3, 5, 7, 11)

Therefore, P(sum is a prime number) =Â \(\frac{15}{36}\)= \(\frac{5}{12}\)

*2- Two dice are thrown together. Find the probability that the product of the numbers on the top**of the dice is:*

**i) 8**

**ii) 12**

**iii) 5**

**Sol:Â Â Â Â **Â Â When two dice are thrown simultaneously, the maximum number of possibilities in the sample

space is (\(6\times 6\)=36).

i) Product of the numbers on the top is 8 {(2,4), (4,2)}

Therefore, P(product is 8) = \(\frac{2}{36}\)= \(\frac{1}{18}\)

ii) Product of on the top is 12 {(2,6), (3, 4),(4,3),(6,2)}

Therefore, P(product is 12) = \(\frac{4}{36}\)= \(\frac{1}{9}\)

iii) Product of numbers on the top is 5 {(1,5),(5,1)}

Therefore, P(product is 5) = \(\frac{2}{36}\)= \(\frac{1}{18}\)

*3- A bag contains 24 balls of which A are red, 2A are white and 3A are blue. A ball is selected at**random. What is the probability that it is:*

**i)Not blue?**

**ii)Red?**

**Sol:Â Â **Â Â Â Â Total number of balls in the bag = 24

number of red balls in the bag Â Â Â Â = A

number of white balls in the bag = 2A

number of blue balls in the bag Â Â Â = 3A

Total number of balls = \(A+2A+3A\)= 6A which is actually 24.

Therefore, 6A = 24 and A = 4.

number of red balls in the bagÂ Â Â Â = 4

number of white balls in the bag = 8

number of blue balls in the bagÂ Â Â = 12

i)Not blue?

P(not blue) = \(\frac{12}{24}\)= \(\frac{1}{2}\)

ii)Red?

P(Red) = \(\frac{4}{24}\)= \(\frac{1}{6}\)

*4- At a gala, cards bearing numbers 1 to 500, one number on one card are put in a box. Each player**selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 250, the player wins a prize. What is the probability that:*

*i)The first player wins a prize? *

*ii)The second player wins a prize if the first has won?*

**Sol:**Â Â Â Â Â Â Â Total number of cards â€“ 500

i) The first player wins a prize.

Perfect squares greater than 250 are 256, 289, 324, 361, 400, 441 and 484 (7 outcomes only).

P(first player winning) = \(\frac{7}{500}\)

ii) The second player wins a prize, if the first has won.

If the first has won, the number of perfect squares available will be 6 and

The total number of available outcomes becomes 499.

P(second player winning after first has won) = \(\frac{6}{499}\)

*5- Â All the jacks, queens and kings are removed from a deck of 52 cards. The remaining cards are**well shuffled and then one card is drawn at random. Giving ace a value of 1 and similar values for**other cards, find the probability that the card has a value:*

*i) 6*

*ii) Greater than 6 *

*iii) Less than 6*

**Sol:Â Â Â Â Â Â **Â Â When all jacks (4) queens (4) and kings (4) are removed then, sample space = 52 -12 = 40 cards.

i)P(6) = \(\frac{4}{40}\)= \(\frac{1}{10}\)

ii)P(>6) = \(\frac{16}{40}\)= \(\frac{2}{5}\)

iii)P(<6) = \(\frac{20}{40}\)= \(\frac{1}{2}\)

**Â **

*6- A box contains 10 red marbles, 5 blue marbles and 7 green marbles. One marble is taken out of at random. What is the probability that the marble taken out will be:*

*i) Neither green nor blue*

*ii) Blue*

*iii Not green*

**Sol:Â Â Â Â Â **Â Number of red marbles = 10

Number of blue marbles = 5

Number of green marbles = 7

Total number of marbles = \(10+5+7= 22\)

i)P(Neither green nor blue) = P(Blue) = \(\frac{10}{22}\)= \(\frac{5}{11}\)

ii)P(Not green) = \(\frac{15}{22}\)

*7- Two dice are thrown at the same time. Find the probability of getting:*

*i)Same number on both dice*

*ii)Different numbers on both dice*

**Sol:**Â Â Â Â Â Â When two dice are thrown. then sample space contains 36 outcomes(\(6\times 6\)).

i)P(Same number on both dice) = \(\frac{6}{36}\)= \(\frac{1}{6}\)

{The six outcomes are [(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)]}

ii)P(Different numbers on both dice) = \(\frac{30}{36}\)= \(\frac{5}{6}\)

*8- A lot consists of 50 mobile phones of which 43 good, 3 have only minor defects and 4 have major defects. Mani will buy a phone only if it is good. AndÂ the trader will Â buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is:**Â *

*i)Acceptable to Mani?*

*ii)Acceptable to Trader?*

**Sol: Â Â Â Â **

i)Â Phones are acceptable to Mani when it is good

P(Acceptable to Mani) = \(\frac{43}{50}\)

ii) Phones are acceptable to the trader only if it has no major defect

P(Acceptable to the trader) = \(\frac{46}{50}\)= \(\frac{23}{25}\)

*9- Box A contains 25 slips of which 20 are marked $1 and the rest are marked $5 each. Box B contains 50 slips of which 45 are marked $1 each and the rest are marked $13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than $5?*

**Sol:Â Â Â Â Â **Total number of slips = \(25+50\)= 75

Number of slips marked $1 = 65

Number of slips marked $5 = 5

Number of slips marked $13 = 5

P( Slip marked other than $5) = \(\frac{70}{75}\)= \(\frac{14}{15}\)

*Â *

*10- A carton of 30 bulbs contains 5 defective bulbs. One bulb is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?*

**Sol:Â Â Â Â Â Â **Total number of bulbs = 30

Number of non-defective bulbs =Â 30-5 = 25

P(Non-defective bulb) = \(\frac{25}{30}\)= \(\frac{5}{6}\)

If the bulb is defective and removed, the number of defective bulbs becomes 4.

Therefore, P(Defective bulb) = \(\frac{4}{29}\)

11-Â *In a game, the entry fee is $5. The game consists of tossing a coin 3 times. If one or two heads show, Shiva gets his entry fee back. If he throws 3 heads, he receives double the entry fee. Otherwise, he will lose. For tossing a coin three times, find the probability that he: (i) Loses the entry fee (ii) Gets double the entry fee (iii) Just gets his entry fee*

**Sol:Â Â Â Â Â Â Â **When a coin is tossed thrice, there are 8 possible outcomes.

{(HHH),(HHT),(HTT),(HTH),(THH),(TTT),(THT),(TTH)}

i)P(Loses his entry fee) = \(\frac{1}{8}\)

ii)P(Gets double the entry fee) = \(\frac{1}{8}\)

iii)P(Just gets his entry fee) = \(\frac{6}{8}\)= \(\frac{3}{4}\)

*12- The probability of getting a rotten egg from a lot of 400 eggs is 0.07. Find the number of rotten**eggs in the lot.*

Sol:Â Â Â Â Â Â Â Â Â Number of rotten eggs = \(400×0.07\)= 28

*13- A bag contains 3 red balls, 5 white balls and 7 black balls. What is the probability that a ball**drawn from the bag at random will be neither red nor white?*

Sol:Â Â Â Â Â Â Â Â Â Total number of balls = \(3+5+7\)= 15

P(Neither red nor white) = \(\frac{7}{15}\)

*14- If the probability of an event is 0.75, find the probability of its complimentary event.*

Sol:Â Â Â Â Â Â Â Â P(complimentary) = \(1-0.75\)= 0.25

*15- A card is selected from a deck of 52 playing cards. Find the probability of a black face card.*

Sol:Â Â Â Â Â Â Â Â Total number of cards = 52

Number of black face cards = 6

P( Black face card) = \(\frac{6}{52}\)= \(\frac{3}{26}\)

*16- Find the probability that a leap year selected at random will contain 53 Mondays.*

Sol:Â Â Â Â Â Â Â Â A leap year contains 366 days (52 weeks and 2 days). The possible outcomes of the two days are

(Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun), (Sun,Mon)

i.e., there areÂ 7 outcomes. Out of them, favourable outcomes = 2 {(Mon, Tue) and (Sun, Mon)}

P(53 Mondays) = \(\frac{2}{7}\)

*17- When a die is thrown once, find the probability of getting an odd number less than 5.*

Sol:Â Â Â Â Â Â Â Â When a die is thrown once, the possible outcomes are 1,2,3,4,5 and 6.

Out of them, odd numbers less than 5 are 1 and 3.

P(Odd no less than 5) = \(\frac{2}{6}\)= \(\frac{1}{3}\)

*18- A card is drawn from a deck of 52 cards. The event E is that the card is not an ace of Spades. Find**the number of outcomes favourable to E.*

Sol:Â Â Â Â Â Â Â Â Number of ace of spades = 1

Number of outcomes favourable to E = \(52-1\)= 51

*19- A girl calculates that the probability of her winning the lottery is 0.08. if 5000 tickets were sold,**how many tickets has she bought?*

Sol:Â Â Â Â Â Â Â Â Number of tickets bought = \(0.08×5000\)= 400

*20- One ticket is drawn at random from a bag containing tickets numbered 1 to 80. Find the**probability that the selected ticket has a number which is a multiple of 4.*

Sol:Â Â Â Â Â Â Â Â Numbers between 1 and 80 which are multiples of 4 = 20

P(Multiple of 4) = \(\frac{20}{80}\)= \(\frac{1}{4}\)

*21- Mani is asked to take a number from 1 to 50. Find the probability that it is a prime number.*

Sol:Â Â Â Â Â Â Â Â The prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41,43 and 47.

There are 15 favourable outcomes. Therefore, P(Prime number) = \(\frac{15}{50}\)= \(\frac{3}{10}\)

*22- A school has five houses A, B, C, D and E. A class has 50 students, 8 from house A, 16 from house**B, 10 from house C, 4 from house D and rest from house E. A single student is selected at**random to be the class monitor. Find the probability that the selected student is not from A, D**and E.*

Sol:Â Â Â Â Â Â Â Â Number of students in A = 8

Number of students in D = 4

Number of students in E = 50 â€“ (\(8+16+10+4\)) = 12

Number of students not from A,D and E = \(50-(8+4+12)\)= 26

P(Student not from A,D and E) = \(\frac{26}{50}\)= \(\frac{13}{25}\)

Thus, this explains theÂ NCERT solutionsÂ for class 10 maths chapter 15. The chapter is probability and is very beneficial in learning through. Besides probability there are other different topics that are included in the syllabus for the final exam. Some of the major types of problems that are asked in the subject of mathematics on the topic of probability are probabilities of dices, probability of sum of numbers, probability of product of numbers, probability of balls and probability of cards. Thus, these along with other types of problem sets are commonly asked as questions during the final exam. Different problems such as Card problems, marble problems and other problems are useful in differentiating between the methods in which the probability theorem is used in solving for problems.