CBSE Class 10 Maths HOTs Probability Set 02

Question. A card is drawn at random a well-shuffled pack of 52 cards. The probability that the card is not an ace is 
(a) \( \frac{1}{13} \)
(b) \( \frac{9}{13} \)
(c) \( \frac{4}{13} \)
(d) \( \frac{12}{13} \)
Answer: (d) \( \frac{12}{13} \)

Question. An experiment whose outcomes has to be among a set of events that are completely known but whose exact outcomes is unknown is a
(a) sample space
(b) elementary event
(c) random experiment
(d) None of the options
Answer: (c) random experiment

Question. The experiments which when repeated under identical conditions produce the same results or outcomes are known as
(a) random experiments
(b) probabilistic experiment
(c) elementary experiment
(d) deterministic experiment
Answer: (d) deterministic experiment

Question. For an event E, \( P(E) + P(\overline{E}) = q \), then
(a) \( 0 \le q < 1 \)
(b) \( 0 < q \le 1 \)
(c) \( 0 < q < 1 \)
(d) None of the options
Answer: (b) \( 0 < q \le 1 \)

Question. A man is known to speak truth 3 out of 4 times. He throws a die and a number other than six comes up. Find the probability that he reports it is a six.
(a) \( \frac{3}{4} \)
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{2} \)
(d) 1
Answer: (b) \( \frac{1}{4} \)

Question. One ticket is selected at random from 100 tickets numbered 00, 01, 02, ......, 99. Suppose x is the sum of digits and y is the product of digits, then probability that \( x = 9 \) and \( y = 0 \) is
(a) \( \frac{2}{17} \)
(b) \( \frac{3}{27} \)
(c) \( \frac{1}{50} \)
(d) \( \frac{1}{25} \)
Answer: (c) \( \frac{1}{50} \)

Question. Find the probability of a non-leap year having 53 Mondays.
Answer: \( \frac{1}{7} \)

Question. What name is given to each of the outcomes of a random experiment?
Answer: Elementary event

Question. Write the name of the experiment whose outcomes has to be among the set of events that are completely known but whose exact outcome is unknown.
Answer: Random experiment

Question. A man is known to speak truth 5 out of 7 times. He throws a die and a number other than six comes up. Find the probability that he reports it is a six.
Answer: \( \frac{2}{7} \)

Question. The probability of getting a bad egg in a lot of 800 eggs is 0.125. Find the number of bad eggs in the lot.
Answer: \( 0.125 \times 800 = 100 \) bad eggs.

Question. The probability of getting a bad pen in a lot of 400 pens is 0.25. Find the number of good pen in the lot.
Answer: Probability of a good pen \( = 1 - 0.25 = 0.75 \). Number of good pens \( = 0.75 \times 400 = 300 \).

Question. Archana calculates that probability of her winning the first prize in a lottery is 0.04. If 12000 tickets are sold, how many tickets has she bought?
Answer: \( 0.04 \times 12000 = 480 \) tickets.

Question. A letter is chosen at random from the letters of the word ‘ASSASSINATION’ Find the probability that the letter chosen is a vowel?
Answer: Total letters \( = 13 \). Vowels are {A, A, I, A, I, O} i.e., 6. Probability \( = \frac{6}{13} \).

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting the jack of hearts.
Answer: \( \frac{1}{52} \)

Question. Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of winning the match by Sangeeta is 0.62. What is the probability of winning the match by Reshma?
Answer: \( 1 - 0.62 = 0.38 \)

Question. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant. 
Answer: Total alphabets \( = 26 \). Consonants \( = 21 \). Probability \( = \frac{21}{26} \).

Question. 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. Find the probability that the arrow will point at any factor of 8. 
Answer: Factors of 8 are {1, 2, 4, 8} i.e., 4. Probability \( = \frac{4}{8} = \frac{1}{2} \).

Question. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.
Answer: Multiples of 5 are {5, 10, 15, 20, 25, 30, 35, 40} i.e., 8. Probability \( = \frac{8}{40} = \frac{1}{5} \).

Question. Three cards of spades are lost from a pack of 52 playing cards. The remaining cards were well shuffled and then a card was drawn at random from them. Find the probability that the drawn cards is of black colour. 
Answer: Total cards remaining \( = 52 - 3 = 49 \). Originally there are 26 black cards (13 spades + 13 clubs). After losing 3 spades, black cards remaining \( = 26 - 3 = 23 \). Probability \( = \frac{23}{49} \).

Question. Two different dice are tossed together. Find the probability
(i) that the number on each dice is even.
(ii) that the sum of numbers appearing on two dice is 5. 
Answer: Total outcomes \( = 36 \).
(i) Favourable outcomes: {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)} i.e., 9. Probability \( = \frac{9}{36} = \frac{1}{4} \).
(ii) Favourable outcomes: {(1,4), (4,1), (2,3), (3,2)} i.e., 4. Probability \( = \frac{4}{36} = \frac{1}{9} \).

Question. Find the probability that a leap year should have exactly 52 tuesday.
Answer: A leap year has 366 days \( = 52 \) weeks and 2 days. The remaining 2 days can be {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}. Total outcomes \( = 7 \). Exactly 52 Tuesdays means the 2 days should not include a Tuesday. Possible pairs are 5. Probability \( = \frac{5}{7} \).

Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is
(i) a card of spade or an ace
(ii) a red king
(iii) neither a king nor a queen
(iv) either a king or queen. 
Answer: Total cards \( = 52 \).
(i) Spades (13) + Aces (4) - Ace of Spades (1) \( = 16 \). Probability \( = \frac{16}{52} = \frac{4}{13} \).
(ii) Red kings \( = 2 \). Probability \( = \frac{2}{52} = \frac{1}{26} \).
(iii) Kings (4) + Queens (4) \( = 8 \). Neither \( = 52 - 8 = 44 \). Probability \( = \frac{44}{52} = \frac{11}{13} \).
(iv) Kings (4) + Queens (4) \( = 8 \). Probability \( = \frac{8}{52} = \frac{2}{13} \).

Question. Cards marked with numbers 3, 4, 5, ....., 50 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the drawn card is 
(i) divisible by 7
(ii) a number which is a perfect square.
Answer: Total cards \( = 50 - 3 + 1 = 48 \).
(i) Divisible by 7: {7, 14, 21, 28, 35, 42, 49} i.e., 7. Probability \( = \frac{7}{48} \).
(ii) Perfect squares: {4, 9, 16, 25, 36, 49} i.e., 6. Probability \( = \frac{6}{48} = \frac{1}{8} \).

Question. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of
(i) heart
(ii) queen
(iii) clubs.
Answer: Remaining cards \( = 52 - 3 = 49 \).
(i) Hearts \( = 13 \). Probability \( = \frac{13}{49} \).
(ii) Queens remaining \( = 4 - 1 = 3 \). Probability \( = \frac{3}{49} \).
(iii) Clubs remaining \( = 13 - 3 = 10 \). Probability \( = \frac{10}{49} \).

Question. Cards bearing numbers 1, 3, 5, ..........., 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing
(a) a prime number less than 15.
(b) a number divisible by 3 and 5.
Answer: Total cards are odd numbers from 1 to 35: \( \{1, 3, 5, ..., 35\} \). Total \( n = 18 \).
(a) Primes < 15: {3, 5, 7, 11, 13} i.e., 5. Probability \( = \frac{5}{18} \).
(b) Divisible by 3 and 5 (LCM 15): {15} i.e., 1. Probability \( = \frac{1}{18} \).

Question. Two dice are rolled once. Find the probability of getting such numbers on the two dice, whose product is 12.
Answer: Favourable outcomes: {(2,6), (6,2), (3,4), (4,3)} i.e., 4. Total \( = 36 \). Probability \( = \frac{4}{36} = \frac{1}{9} \).

Question. All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is 
(i) a black face card.
(ii) a red card.
Answer: Total removed \( = 4 \) (K) \( + 4 \) (Q) \( + 4 \) (A) \( = 12 \). Remaining \( = 52 - 12 = 40 \).
(i) Face cards are K, Q, J. Since K and Q are removed, only Jacks are left. Black Jacks \( = 2 \). Probability \( = \frac{2}{40} = \frac{1}{20} \).
(ii) Original red cards \( = 26 \). Removed red cards \( = 2 \) (K) \( + 2 \) (Q) \( + 2 \) (A) \( = 6 \). Remaining red cards \( = 26 - 6 = 20 \). Probability \( = \frac{20}{40} = \frac{1}{2} \).

Question. A bag contains 12 balls out of which x are white.
(i) If one ball is drawn at random, what is the probability that it will be a white ball?
(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (i). Find x. 
Answer: (i) Probability \( = \frac{x}{12} \).
(ii) New total \( = 18 \). New white \( = x + 6 \). New probability \( = \frac{x+6}{18} \).
Given: \( \frac{x+6}{18} = 2 \left( \frac{x}{12} \right) \)

\( \implies \) \( \frac{x+6}{18} = \frac{x}{6} \)

\( \implies \) \( x+6 = 3x \)

\( \implies \) \( 2x = 6 \)

\( \implies \) \( x = 3 \).

Question. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is \( \frac{1}{4} \). The probability of selecting a blue ball at random from the same jar is \( \frac{1}{3} \). If the jar contains 10 orange balls, find the total number of balls in the jar. 
Answer: \( P(\text{Red}) + P(\text{Blue}) + P(\text{Orange}) = 1 \)
\( \frac{1}{4} + \frac{1}{3} + P(\text{Orange}) = 1 \)

\( \implies \) \( P(\text{Orange}) = 1 - \frac{7}{12} = \frac{5}{12} \)
Let total balls be \( N \).
\( \frac{10}{N} = \frac{5}{12} \)

\( \implies \) \( 5N = 120 \)

\( \implies \) \( N = 24 \).

Question. A bag contains 18 balls out of which x balls are red.
(i) If one ball is drawn at random from the bag, what is the probability that it is not red?
(ii) If 2 more red balls are put in the bag, the probability of drawing a red ball will be \( \frac{9}{8} \) times the probability of drawing a red ball in the first case. Find the value of x. 
Answer: (i) \( P(\text{not Red}) = \frac{18 - x}{18} \).
(ii) Case 1: \( P(\text{Red}) = \frac{x}{18} \).
Case 2: New total \( = 20 \). New red \( = x + 2 \). \( P(\text{Red}) = \frac{x+2}{20} \).
Given: \( \frac{x+2}{20} = \frac{9}{8} \left( \frac{x}{18} \right) \)

\( \implies \) \( \frac{x+2}{20} = \frac{x}{16} \)

\( \implies \) \( 16x + 32 = 20x \)

\( \implies \) \( 4x = 32 \)

\( \implies \) \( x = 8 \).

Question. A bag contains 24 balls out of which x are white. If one ball is drawn at random the probability of drawing a white ball is y. 12 more white balls are added to the bag. Now if a ball is drawn from the bag, the probability of drawing the white ball is \( \frac{5}{3}y \). Find the value of x.
Answer: \( y = \frac{x}{24} \).
New total \( = 36 \). New white \( = x + 12 \).
New probability \( = \frac{x+12}{36} = \frac{5}{3} \left( \frac{x}{24} \right) \)

\( \implies \) \( \frac{x+12}{36} = \frac{5x}{72} \)

\( \implies \) \( 2(x+12) = 5x \)

\( \implies \) \( 2x + 24 = 5x \)

\( \implies \) \( 3x = 24 \)

\( \implies \) \( x = 8 \).

Question. Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is
(i) a king (ii) of red colour
(iii) a face card (iv) a queen 
Answer: Total removed \( = 2 \) (red Q) \( + 2 \) (black J) \( = 4 \). Remaining \( = 52 - 4 = 48 \).
(i) Kings \( = 4 \). Probability \( = \frac{4}{48} = \frac{1}{12} \).
(ii) Original red \( = 26 \). Removed red Q \( = 2 \). Remaining red \( = 24 \). Probability \( = \frac{24}{48} = \frac{1}{2} \).
(iii) Original face cards \( = 12 \). Removed \( = 2 \) (Q) \( + 2 \) (J) \( = 4 \). Remaining face cards \( = 8 \). Probability \( = \frac{8}{48} = \frac{1}{6} \).
(iv) Original queens \( = 4 \). Removed red Q \( = 2 \). Remaining queens \( = 2 \). Probability \( = \frac{2}{48} = \frac{1}{24} \).

Question. Cards numbered 1 to 30 are put in a bag. A card is drawn at random from this bag. Find the probability that the number on the drawn card is
(i) not divisible by 3.
(ii) a prime number greater than 7.
(iii) not a perfect square number. 
Answer: Total outcomes \( = 30 \).
(i) Divisible by 3: {3, 6, 9, 12, 15, 18, 21, 24, 27, 30} i.e., 10. Not divisible \( = 20 \). Probability \( = \frac{20}{30} = \frac{2}{3} \).
(ii) Primes > 7: {11, 13, 17, 19, 23, 29} i.e., 6. Probability \( = \frac{6}{30} = \frac{1}{5} \).
(iii) Perfect squares: {1, 4, 9, 16, 25} i.e., 5. Not perfect square \( = 25 \). Probability \( = \frac{25}{30} = \frac{5}{6} \).

Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is
(i) a card of spade or an ace.
(ii) a black king.
(iii) neither a jack nor a king
(iv) either a king or a queen 
Answer: (i) \( \frac{16}{52} = \frac{4}{13} \). (ii) \( \frac{2}{52} = \frac{1}{26} \). (iii) \( \frac{44}{52} = \frac{11}{13} \). (iv) \( \frac{8}{52} = \frac{2}{13} \).

Question. A box contains cards bearing numbers from 6 to 70. If one card is drawn at random from the box, find the probability that it bears
(i) a one digit number. (ii) a number divisible by 5.
(iii) an odd number less than 30.
(iv) a composite number between 50 and 70.

Answer: Total numbers from 6 to 70 = \( 70 - 6 + 1 = 65 \)
(i) One digit numbers are 6, 7, 8, 9. Total = 4. Probability = \( \frac{4}{65} \)
(ii) Numbers divisible by 5 are 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70. Total = 13. Probability = \( \frac{13}{65} = \frac{1}{5} \)
(iii) Odd numbers less than 30 starting from 6 are 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29. Total = 12. Probability = \( \frac{12}{65} \)
(iv) Composite numbers between 50 and 70: Total numbers between 50 and 70 are 19 (51 to 69). Primes in this range are 53, 59, 61, 67. Count = 4. Composite numbers = \( 19 - 4 = 15 \). Probability = \( \frac{15}{65} = \frac{3}{13} \)

Question. In answering a question of MCQ test with 4 choices per question, one of them being correct, a student knows the answer, guesses or copies the answer. A student guesses the answer. What is the probability that his answer is correct?

Answer: There are 4 choices, and only one is correct.
If a student guesses, the probability of choosing the correct answer is \( \frac{1}{4} \).

Question. Honey goes to school either by a car driven by his driver or uses his bicycle. Probability that he will use the car is \( \frac{3}{7} \). What is the probability that he will use his bicycle for going to the school?

Answer: Probability (Car) + Probability (Bicycle) = 1
\( \implies \) \( \frac{3}{7} \) + Probability (Bicycle) = 1
\( \implies \) Probability (Bicycle) = \( 1 - \frac{3}{7} = \frac{4}{7} \)

Question. Samsung Electronics has launched two new mobile hand sets: Set-I. and set-II. Set-I is cheaper as compared to set-II. But set-II has an built-in device to recharge the battery with auto-cut power supply when it is fully charged. In a lot there are 250 pieces set-I and 100 set-II. If a mobile is picked at random:
(a) find the probability of getting set-I
(b) find the probability of getting set-II.

Answer: Total number of pieces = \( 250 + 100 = 350 \)
(a) P(getting set-I) = \( \frac{250}{350} = \frac{5}{7} \)
(b) P(getting set-II) = \( \frac{100}{350} = \frac{2}{7} \)

Question. A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest.

Answer: Total persons = 12
Number of extremely kind persons = \( 12 - (3 + 6) = 3 \)
(i) P(extremely patient) = \( \frac{3}{12} = \frac{1}{4} \)
(ii) P(extremely kind or honest) = \( \frac{3 + 6}{12} = \frac{9}{12} = \frac{3}{4} \)