Number of favourable outcomes when card is neither red nor queen = 28. (ii) a multiple of 5. A card is then drawn at random from the pack. (HH, HT, TH, TT} You could also express this as 0.058 or 5.8%. (i) a king of red colour. Question 34. Favourable outcomes for a card of red colour are 24 as 2 red queens have been removed. Question 5. Probability of getting a red face card =6/52=3/26, Question 77. 1, 2, 4, 5, 7, 8, , 10,11,13,14,16,17,19, 20,22,23,25, 26, 28 and 29). Total number of cards from 11 to 60 = 50. Is this correct? Another number y is selected at random from the numbers 1, 4, 9 and 16. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8 see Fig, and these are equally likely outcomes. A card is drawn at random from a well-shuffled deck of playing cards. ∴ P(E) = \(\frac{24}{52}\) = \(\frac{16}{13}\), Question 10. The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’. 3, 4). (ii) D? Two dice are thrown. On each spin, each sector has equal chance of selection by the arrow. A box contains 35 blue, 25 white and 40 red marbles. (iii) exactly one head. (ii) P (black or white ball) = \(\frac{8}{15}\) Number of cards bearing a prime number less than 20 are 7 (i.e. Find the probability of getting: The number of possible outcomes = 52 Kewal will buy shirt if a shirt is not having major defect. (1.1),(1,2), (1,3), (1,4),(1,5), (1,6) Solution: (2, 4) (2,5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3,5) (4, 1) Number of cards neither a jack nor an ace = 52 – 8 = 44, Question 43. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is 1/4. Possible outcomes of drawing one card from 48 cards is 48. Product less than 16 = (1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 3 × 1, 3 × 4, 4 × 1) Required probability =8/80=1/10, Question 74. Required probability =3/4, Question 31. = 1 – \(\frac{25}{36}\) = \(\frac{11}{36}\). Favourable outcomes : 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59 (iii) Let E be the event of getting an odd number. Card marked with numbers 1, 3, 5, …, 101 are placed in a bag and mixed thoroughly. Question 41. As blackjacks, black kings and black aces are removed from the pack, so, number of remaining cards in a pack = 46, Kerala Syllabus 9th Standard Physics Solutions Guide, Kerala Syllabus 9th Standard Biology Solutions Guide, MCQ Questions for Class 6 Science with Answers PDF Download Chapter Wise, MCQ Questions for Class 7 Science with Answers PDF Download Chapter Wise, NCERT Solutions for Class 6 Sanskrit Ruchira Bhag 1, MCQ Questions for Class 8 Science with Answers PDF Download Chapter Wise, NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes InText Questions, NCERT Solutions for Class 7 Maths Chapter 14 Symmetry InText Questions, NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers InText Questions, NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions InText Questions, NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions, NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers InText Questions, NCERT Solutions for Class 7 Maths Chapter 8 Comparing Quantities InText Questions, NCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles InText Questions, NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties InText Questions, NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles InText Questions, NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions. Number lying between 2 and 5 are 2 (i.e. ∴ Probability that arrow points at a number less than 9 = \(\frac{8}{8}\) = 1. A card is drawn at random from the box. Required probability=3/4, Question 46. Given: Number of blue marbles = 35; Number of white marbles = 25; A coin is tossed two times. Now, Ramesh will lose the game if he gets Number of cases, where product is less than 16 = 8 A card is drawn from a well shuffled deck of 52 cards. a = 2, b can take 1 value The die is thrown once. Favourable outcomes when sum of the numbers appearing on the dice is 6 or 7 are, i.e. ∴ Total number of possible outcomes = 17. (iv) a number less than 9? ∴ Favourable number of elementary events = 1 Find the probability that product of x and y is less than 16. Solution: ∴ Probability that arrow points at an odd number = \(\frac{4}{8}\) = \(\frac{1}{2}\), (iii) We have 6 numbers greater than 2, i.e., 3, 4, 5, 6, 7 and 8. So, one face card can be chosen in 12 ways. The remaining cards are well shuffled and then a card is drawn at random. Favourable outcomes: 6 red face cards (king, queen and jack of diamond and heart suits), Question 7. Total cards = 52. Multiple of 3 or 7 in between 1 to 20 are 3, 6, 9,12,15,18, 7,14 The probability of Sangeeta’s winning = P(S) = 0.62 ∴ Probability of getting an ace card = \(\frac{1}{4}\), (b) Since, there is no queen (as queen is put aside) (i) be an ace. There are four types of cards in the pack of 52 cards. Find the probability that the drawn card is, Solution: The probability of the first event happening is 13/52. A bag contains cards numbered from 1 to 49. Rahim tosses two different coins simultaneously. P (E) + P (not E) = 1 MCQ on Statistics Class 10 Question 1. ∴ Favourable number of elementary events Find the probability that the drawn card is neither a jack nor an ace Possible outcomes are 4, 9, 16, 25, 36, 49, i.e., 6. The number of outcomes favourable to the event Ē = 52 – 4 = 48. Total number of possible outcomes = 46 Solution: Required probability=23/49, Question 49. a = 4, b can take 3 value, i.e. Solution: A bag contains 5 red balls and some blue balls. All red face cards art removed from a pack of playing cards. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that … P (a prime number on each die) = \(\frac{9}{36}\) or \(\frac{1}{4}\) P(E) = P (number greater than 4) = \(\frac{2}{6}\) = \(\frac{1}{3}\) ∴ Favourable number of elementary events = 1 10. Number of favourable outcomes = 8 4. Five cards: the ten, jack, queen, king and ace of diamonds are shuffled with faces downwards. (1) P (red ball) = \(\frac{7}{15}\) Number of cards with a prime number are 11 (i.e. A box consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. {1, 6}, (2,5), (3, 4), (4,3), (5, 2), (6,1), Question 71. (iii) is a doublet of odd numbers. Therefore, total outcomes are 36. Answer with step by step detailed solutions to question from HashLearn's X ICSE: Math, Probability- "Froma well- shuffled deck of 52 cards, a card is drawn at random. Cards remaining after removing black face cards = red cards + black cards excluding face cards (i) As we know that there are two suits of red card, i.e., diamond and heart and each suit contains one king. P(R)=+ 2/5 + 3/10 = 1 No. It is given that there are 5 prizes on these 1000 tickets. Hence, P (Winning a prize) = \(\frac{5}{1000}\) = \(\frac{1}{200}\), Question 13. {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, Question 20. 2 red kings), Number of perfect square numbers from 11 to 60 = 4 (i.e. Number of outcomes when product is even = 27 [(1, 2), (1, 4) … (6,6)], Question 1. Total number of cases = 36 5 are 170 (100 + 50 + 20), Favourable outcomes for falling a coin of value 1 or 2 = 70(50 + 20), Probability of selecting extremely patient = 3/12=1/4, Probability of selecting extremely kind or honest = 3+6/12=9/12=3/4. Ace, jack, queen, king of clubs are removed from a deck of playing cards. If 2 more red balls are put in the bag, the probability of drawing a red ball will be 9/8 times the probability of drawing a red ball in the first case. Hence, required probability = \(\frac{28}{52}\) = \(\frac{7}{13}\), (iv) A card drawn will be red as well as king, if it is a red king. Possible outcomes when 3 coins are tossed are 8 (i.e. 5, 6). What is the probability of getting a number greater than 4? Thus, a leap year always has 52 Tuesdays. From a well-shuffled pack of playing cards, blackjacks, black kings and black aces are removed. Why? If one card is drawn at random from the box, find the probability that it bears, Solution: Two dice, one blue and one grey, are thrown at the same time. (v) a face card. Well-shuffling ensures equally likely outcomes. You can also download NCERT Solutions For Class 10 … Hence, required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\). ∴ Probability of getting a queen = \(\frac{0}{4}\), Question 8. Solution: n(E2) = 1 [(6 × 6)] The ace, jack, queen and king of clubs are removed from a pack of 52 playing cards. Solution: Question 4. If a card is drawn at random from the box, find the probability that the number on the drawn card is. Solution: Spade is of black colour] Find the probability of getting: Solution: A card is then drawn at random from the bag. Therefore, P(F) = \(\frac{4}{6}\) = \(\frac{2}{3}\), Question 6. If Saket has purchased one lottery ticket, what is the probability of winning a prize ? D : Card with composite number between 50 and 70. that the sum of numbers appearing on the two dice is 5. Example 1: A card is drawn at random from a pack of 52 playing cards. Favourable outcomes that the sum of the numbers appearing on two dice is 5 are(1,4), (2, 3), (3, 2), (4,1), i.e. Total possible outcomes when one card is drawn = 48 Number of toal possible outcomes when one card is drawn = 52 Two dice are rolled once. (ii) a face card. (i) There are four ace cards in a pack of 52 cards. Favourable outcomes for drawing a king are 4. Numbers from 1 to 80 which are perfect square are 1, 4, 9,16, 25, 36, 49, 64. Number of cards in the box = 65, 2014 Find the probability of getting a red face card. 1, 2, 3, 4, 5, 6). Hence, P(E) = \(\frac{5}{36}\), (ii) As you can see from figure, there is no outcome favourable to the event F, ‘the sum of two numbers is 13’. Question 16. After gaining knowledge of the ideas, solving probability questions in exams will get convenient for you. ∴ Non-prime numbers from 1 to 25 = 25 – 9 = 16. Total number of marbles = 100, Question 61. HHH, HHT, HTH, THH, TTH, TTT, THT, HTT). Very Short Answer Type Questions [1 Mark], Question 1. A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. Short Answer Type Questions I [2 Marks], Question 42. Find the probability of getting a: What is the probability that she gets at least one head? (vi) a red face card. Short Answer Type Questions I [2 Marks], Question 76. All the questions are prepared based on the latest exam pattern. (iv) the jack of hearts. ∴ P (product less than 16) = \(\frac{25}{36}\) A card is drawn at random from a well shuffled pack of 52 playing cards. A piggy bank contains hundred 50 p coins, fifty 1 coins, twenty 2 coins and ten 5 coins. Number of favourable cases whose product is 12 are given as {(2,6), (3,4), (4,3), (6,2)}, i.e. (b) a queen? Out of 52 cards, one card can be drawn in 52 ways. If one card is drawn at random from the box, find the probability that it bears, Question 62. Possible outcomes HHH, HHT, HTH, THH, TTT, TTH, THT, HTT Hence, required probability = \(\frac{12}{52}\) = \(\frac{3}{13}\), (vi) There are 6 red face cards 3 each from diamonds and hearts. A ball is drawn from the bag at random. Solution: Probability of getting a number greater than 4=2/6=1/3, Question 78. Total number of outcomes while rolling two dice = 36. The total number of elementary events associated with random experiment of throwing a die is 6. n(s) = 20, Multiples of 3 or 7, A: {3, 6, 9, 12, 15, 18, 7, 14), n(A) = 8 Question 35. In 10th grade worksheet on probability we will practice various types of problems based on definition of probability and the theoretical probability or classical probability. Number of kings = 4, number of queens = 4, number of aces = 4 Number of black queens in a pack of cards = 2 A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Cards marked with number 3,4,5,50 are placed in a box and mixed thoroughly. What is the probability that the sum of the two numbers appearing on the top of the dice is: (i) 8? ∴ Favourable number of elementary events = 3 (i.e., 1, 3, 5) Total number of outcomes = 49 If one disc is drawn at random from the box, find the probability that it bears a perfect square number. A box contains 5 red marbles, 8 white marbles and 4 green marbles. 2 + 2 = 4 cards are removed from a pack of 52 playing cards. Question - Probability. So, total number of elementary events = 52 In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice ? Here, the total number of possible outcomes = 5. Find the probability of getting exactly one head. Solution: (i) Since, there are 5 red marbles in the box. Find the probability that the selected ticket has a number which is a multiple of 5. Five cards – the ten, jack, queen, king and ace of diamonds, are well shuffled with their faces downwards. Ex.3 Match the following: (i) P() = (a) 0. If P(E) = 0.05, what is the probability of ‘not E’? A box contains 75 cards which are numbered from 1 to 75. A number x is selected at random from the numbers 1, 2, 3 and 4. So, total number of possible outcomes are 36. Remaining cards are well shuffled and a card is drawn from them at random. Find the probability of getting at least one Total number of cards in the bag = 51. Number of exactly two heads are HHT, HTH and THH. .’. Solution: A = Product of the numbers on the top of the dice is 6. Question 8. All kings, queens and aces are removed from a pack of 52 cards. Let R = getting a red ball NCERT Solutions for Class 10 Maths Probability - Exercise 15.1 Q1: The first question contains five fill in the blanks, and you need to have a proper understanding of the text to answer the same. i.e., P (Ē) = 0.992 (Given) Two different dice are thrown together. It is given … Find the probability that the selected card bears a perfect square number. of favourable outcomes = 25 When two dice are rolled, total number of cases = 36 Solution: Solution: Total number of defective bulbs in the box = 15 Hanif wins if he gets three heads or three tails, and loses otherwise. (i) face card. Find the probability that the fraction \(\frac{a}{b}\) > 1. Let there be x blue balls in the bag. Total number of outcomes = 8. is divisible by 9 and is a perfect square. (iv) a prime number less than 20. Total cases are {HH, HT, TH, TT}. What is the probability that, Question 7. x can be 1, 4, 9 or 16 andy can be 1, 2,3 or 4. Determine the probability that the chosen letter is a consonant. Solution: contains cards numbered from 1 to 49. Students who are preparing for their Class 10 exams must go through Important Questions for Class 10 Math Chapter 15 Probability. Solution: True, because the outcomes odd number’ and `even number’ are equally likely here. = \(\frac{25}{50}\) = \(\frac{1}{2}\), (ii) A perfect square number (ii) have a product less than 16 (iii) As you can see from figure, all the outcomes are favourable to the event G, ‘sum of two numbers ≤ 12. Total outcomes of drawing a card from 46 cards = 46. Three coins are tossed simultaneously. A coin is tossed two times. Solution: ∴ Favourable number of elementary events = 13 ∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\) Find the probability of getting a black queen. a = 2, b can take 1 value, Find the probability that the card drawn is, Question 29. The sample space is {HH, HT, TH, TT} (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)]. A bag contains cards numbered from 1 to 49. Find the probability that it bears, Question 81. Number from 1 to 25 which are divisible by 3 or 5 are 3, 6, 9,12,15,18, 21, 24, 5,10,20,25. Favourable outcomes = 1 + 2 + 3 + 4 + 5 = 15, 2015 Favourable outcomes for even number on both dice = 9, [i.e. Favourable cases = 6 (red face cards, i.e. ∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\) Are the outcomes 1, 2 and 3 equally likely to occur? Solution: ∴ P (no head) = \(\frac{1}{4}\), Question 9. Solution: P (a total of 9 or 11) = \(\frac{6}{36}\) or \(\frac{1}{6}\), Question 14. Find the probability of getting the following: Solution: The probability of selecting a blue ball at random from the same jar is 1/3. Favourable cases of not getting a diamond card are = 46 – 13 = 33. ∴ Favourable number of elementary events = 13 (6.1),(6,2),(6, 3),(6,4), (6, 5),(6, 6) 4 cards) Probability of card bears a one digit number=4/65; B : Number on the cards is divisible by 5. P(B)=13/65=1/5; C: Cards … Total number of cases of product of x and y = 16 Here we are providing Probability Class 10 Extra Questions Maths Chapter 15 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. Solution: Download free printable worksheets for CBSE Class 10 Probability with important topic wise questions, students must practice the NCERT Class 10 Probability worksheets, question banks, workbooks and exercises with solutions which will help them in revision of important concepts Class 10 Probability. (i) red. Number of favourable cases whose sum is 7 are 6, i.e. Favourable cases are (1,1), (1,4), (2,2), (3, 3), (4,1), (4,4), (5,5), (6, 6), Question 73. (i) an odd number. A bag contains 25 cards numbered from 1 to 25. Very Short Answer Type Questions [1 Mark], Question 15. (i) Let E be the event of getting a letter A. Concise Mathematics Class 10 ICSE Solutions 2018, Class 12 Biology Important Questions Chapter 15 Biodiversity and Conservation, CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions, Class 12 Biology Important Questions Chapter 14 Ecosystem, Class 12 Biology Important Questions Chapter 13 Organisms and Populations, CBSE Sample Papers for Class 10 Science Set 5 for Practice, CBSE Sample Papers for Class 10 Science Set 4 for Practice, CBSE Sample Papers for Class 10 Science Set 3 with Solutions, Class 12 Biology Important Questions Chapter 12 Biotechnology and its Applications, Class 12 Biology Important Questions Chapter 11 Biotechnology: Principles and Processes, CBSE Sample Papers for Class 10 Science Set 2 with Solutions, Concise Mathematics Class 10 ICSE Solutions. A card is drawn at random from a pack of 52 playing cards. If I toss a coin 3 times and get head each tir ne, then I should expect a tail to have a higher chance in the 4th toss. Therefore, Ē is the event of not having the same birthday.
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