![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\nQuestion 1.<\/p>\n
(A) 1
\n(B) \\(\\frac{3}{4}\\)
\n(C) \\(\\frac{1}{2}\\)
\n(D) 0
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nAn event that cannot occur has 0 probability, such an event is called impossible event.<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) \\(\\frac{1}{3}\\)
\n(B) 0.1
\n(C) 3%
\n(D) \\(\\frac{17}{16} \\)
\nAnswer:
\n(D) \\(\\frac{17}{16}\\)<\/p>\n
Explanation:
\nProbability of any event cannot be more than one or negative as \\(\\frac{17}{16}\\) > 1.<\/p>\n
Question 3.<\/p>\n
(A) 0.0001
\n(B) 0.001
\n(C) 0.01
\n(D) 0.1
\nAnswer:
\n(A) 0.0001<\/p>\n
Explanation:
\nThe probability of the event, which is very unlikely to happen, will be very close to zero. So it’s probability is 0.0001 which is minimum amongst the given values.<\/p>\n
Question 4.<\/p>\n
(A) p – 1
\n(B) p
\n(C) 1 – p
\n(D) \\(1-\\frac{1}{p}\\)
\nAnswer:
\n(C) 1 – p<\/p>\n
Explanation:
\nProbability of an event + Probability of its complementary event = 1
\n\u2234 p + Probability of complement = 1
\nProbability of complement = 1 – p<\/p>\n
<\/p>\n
Question 5.<\/p>\n
(A) less than 100
\n(B) less than 0
\n(C) greater than 1
\n(D) anything but a whole number
\nAnswer:
\n(B) less than 0<\/p>\n
Explanation:
\nProbability lies between 0 and 1 and when it is converted into percentage it will be between 0 and 100. So, cannot be negative.<\/p>\n
Question 6.<\/p>\n
(A) P(A) < 0 (B) P(A) > 1
\n(C) 0 \u2264 P(A) \u2264 1
\n(D) -1 \u2264 P(A) \u2264 1
\nAnswer:
\n(C) 0 \u2264 P(A) \u2264 1<\/p>\n
Explanation:
\nAs the probability of an event lies between 0 and 1.<\/p>\n
Question 7.<\/p>\n
(A) \\(\\frac{3}{26}\\)
\n(B) \\(\\frac{3}{13}\\)
\n(C) \\(\\frac{2}{13}\\)
\n(D) \\(\\frac{1}{2}\\)
\nAnswer:
\n(A) \\(\\frac{3}{26}\\)<\/p>\n
Explanation:
\nIn a deck of 52 cards, there are 26 red cards.
\nNumber of red face cards = 3 of hearts + 3 of diamonds = 6
\nSo, probability of having a red face card
\n= (A) \\(\\frac{6}{52}\\) = (A) \\(\\frac{3}{26}\\)<\/p>\n
Question 8.<\/p>\n
(A) \\(\\frac{1}{7}\\)
\n(B) \\(\\frac{2}{7}\\)
\n(C) \\(\\frac{3}{7}\\)
\n(D) \\(\\frac{5}{7}\\)
\nAnswer:
\n(A) \\(\\frac{1}{7}\\)
\nNumber of weeks = \\(\\frac{365}{7}\\) = 52\\(\\frac{1}{7}\\) = 52 weeks
\nNumber of days left = 1
\nFor example, it may be any of 7 days which from Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday; so,
\nT(E) = 7
\nF(E) = 1 (Sunday)
\np(F) = \\(\\frac{F(E)}{T(E)}\\) = \\(\\frac{1}{7}\\)<\/p>\n
Asertion and Reased Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Question 3.<\/p>\n Answer: Explanation: Case – Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) \\(\\frac{1}{26}\\) Explanation: Question 2.<\/p>\n (A) \\(\\frac{1}{26}\\) Question 3.<\/p>\n (A) \\(\\frac{1}{26}\\) Question 4.<\/p>\n (A) \\(\\frac{3}{13}\\) Explanation: Question 5.<\/p>\n (A) \\(\\frac{1}{26}\\) Explanation: II. Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n (A) \\(\\frac{1}{26}\\) Explanation: Question 2.<\/p>\n (A) 1 Explanation: Question 3.<\/p>\n (A) 1 Question 4.<\/p>\n (A) \\(\\frac{5}{9}\\) Question 5.<\/p>\n Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8 ?<\/p>\n (A) 1 Probability Class 10 MCQ Questions with Answers Question 1. If an event that cannot occur, then its probability is: (A) 1 (B) (C) (D) 0 Answer: (D) 0 Explanation: An event that cannot occur has 0 probability, such an event is called impossible event. Question 2. Which of the following cannot be the probability of …<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, the probability of each is \\(\\frac{1}{2}\\).
\nReason (R): When we toss a coin, there are two possible outcomes: head or tail. Therefore, the probability of each outcome is i.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion:
\nFrom 1 to 100 numbers, there are 50 even and 50 odd numbers.
\nTotal number of outcomes T(E) = 100 Number of outcomes favourable for event E (even numbers) = F (E) = 50
\nso, P (E) = \\(\\frac{50}{100}\\) = \\(\\frac{1}{2}\\)
\nSimilarly, the probability of getting odd numbers =\\(\\frac{1}{2}\\).
\nHence the probability of getting odd and even each = \\(\\frac{1}{2}\\).
\nHence, the given statement is true.
\n\u2234 Assertion is correct.
\nIn case of reason:
\nSince, there are two outcomes equal in all manners. So, probability of both head and tail is equal to \\(\\frac{1}{2}\\). each.
\nHence, the given statement is true.
\n\u2234 Reason is correct:
\nHence, both assertion and reason are correct but reason is not correct explanation for assertion.<\/p>\n<\/p>\nAssertion (A): If P(F) = 0.20, then the probability of ‘not E’ is 0.80.
\nReason (R): If two dice are thrown together, then the probability of getting a doublet is \\(\\frac{5}{6}\\).<\/h2>\n
\n(C) A is true but R is false
\nExplanation:
\nIn case of assertion:
\nP(E) = 0.20
\nP(not E) = 1 – P(E)
\n\u2234Assertion is correct.
\nIn case of reason:
\nTotal number of possible outcomes = 62 <\/sup> = 36
\nE : (doublets are (1,1), (2,2), (3, 3), (4,4), (5,5), (6,6)
\nNo. of favourable outcomes to E = 6
\nP(a doublet)
\n= \\(\\frac{Number of outcomes favourable to E}{Total number of outcomes}\\).
\n=\\(\\frac{6}{36}\\). = \\(\\frac{1}{6}\\).
\n\u2234Reason is incorrect:
\nHence, assertion is correct but reason is incorrect.<\/p>\nAssertion (A): The probability that a number selected at random from the number 1, 2, 3, , 15
\nis a multiple of 4, is \\(\\frac{1}{3}\\)..
\nReason (R): Two different coins are tossed simultaneously. The probability of getting at least one head is\\(\\frac{3}{4}\\).<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of reason:
\nn(S) = 15
\nn(A) = 3
\np(A) = \\(\\frac{n(A)}{n(S) }\\) = \\(\\frac{3}{15}\\) = \\(\\frac{1}{5}\\)
\nS = HH, HT, TH,TT
\nA = HH, HT, TH
\nn(S) = 4
\nn(A) = 3
\np(A) = \\(\\frac{n(A)}{n(S) }\\) = \\(\\frac{3}{4 }\\)
\n\u2234Reason is incorrect:
\nHence, assertion is correct but reason is incorrect.<\/p>\n
\nI. Read the following text and answer the following questions on the basis of the same: On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-15-Probability-1.png?resize=290%2C203&ssl=1\")
<\/p>\nFind the probability of getting a king of red colour.<\/h2>\n
\n(B) \\(\\frac{1}{13}\\)
\n(C) \\(\\frac{1}{52}\\)
\n(D) \\(\\frac{1}{4}\\)
\nAnswer:
\n(A) \\(\\frac{1}{26}\\)<\/p>\n
\nNo. of cards of a king of red colour = 52
\nProbability of getting a king of red colour
\n= \\(\\frac{No. of king of red colour}{Total number of cards }\\)
\n= \\(\\frac{2}{52}\\) = \\(\\frac{1}{26}\\)<\/p>\nFind the probability of getting a face card.<\/h2>\n
\n(B) \\(\\frac{1}{13}\\)
\n(C) \\(\\frac{2}{13}\\)
\n(D) \\(\\frac{3}{13}\\)
\nAnswer:
\n(D) \\(\\frac{3}{13}\\)<\/p>\n<\/p>\nFind the probability of getting a jack of hearts.<\/h2>\n
\n(B) \\(\\frac{1}{52}\\)
\n(C) \\(\\frac{3}{52}\\)
\n(D) \\(\\frac{3}{26}\\)
\nAnswer:
\n(B) \\(\\frac{1}{52}\\)<\/p>\nFind the probability of getting a red face card.<\/h2>\n
\n(B) \\(\\frac{1}{13}\\)
\n(C) \\(\\frac{1}{52}\\)
\n(D) \\(\\frac{1}{4}\\)
\nAnswer:
\n(A) \\(\\frac{3}{13}\\)<\/p>\n
\nNo. of face card = 13
\nTotal no of cards = 52
\nProbability of getting a face card
\n= \\(\\frac{ No. of face cards }{Total no. of cards}\\)
\n= \\(\\frac{12}{52}\\) = \\(\\frac{3}{13}\\)<\/p>\nFind the probability of getting a spade.<\/h2>\n
\n(B) \\(\\frac{1}{13}\\)
\n(C) \\(\\frac{1}{52}\\)
\n(D) \\(\\frac{1}{4}\\)
\nAnswer:
\n(D) \\(\\frac{1}{4}\\)<\/p>\n
\nNo. of face card = 13
\nTotal no of cards = 52
\nProbability of getting a face card
\n= \\(\\frac{ No. of face cards }{Total no. of cards}\\)
\n= \\(\\frac{13}{52}\\) = \\(\\frac{1}{4}\\)<\/p>\n
\nRahul and Ravi planned to play Business (board game) in which they were supposed to use two dice.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/02\/MCQ-Questions-for-Class-10-Maths-Chapter-15-Probability-2.png?resize=190%2C233&ssl=1\")
<\/p>\nRavi got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?<\/h2>\n
\n(B) \\(\\frac{5}{36}\\)
\n(C) \\(\\frac{1}{18}\\)
\n(D) 0
\nAnswer:
\n(B) \\(\\frac{5}{36}\\)<\/p>\n
\nThe outcomes when two dice are thrown together are:
\n= (1, 1), (1, 2), (1, 3), (1,4), (1, 5), (1,6)
\n(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
\n(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
\n(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
\n(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
\n(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
\nTotal outcomes = 36
\nNo. of outcomes when the sum is 8 = 5
\nProbability = \\(\\frac{5}{36}\\)<\/p>\nRahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13?<\/h2>\n
\n(B) \\(\\frac{5}{36}\\)
\n(C) \\(\\frac{1}{18}\\)
\n(D) 0
\nAnswer:
\n(D) 0<\/p>\n
\nNo. of outcomes when the sum is 13 = 0
\nTotal outcomes =36
\nProbability =\\(\\frac{0}{35}\\) = 0<\/p>\n<\/p>\nNow it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12?<\/h2>\n
\n(B) \\(\\frac{5}{36}\\)
\n(C) \\(\\frac{1}{18}\\)
\n(D) 0
\nAnswer:
\n(A) 1
\nExplanation:
\nNo. of outcomes when the sum is less than or equal to 12 = 36
\nTotal outcomes = 36
\nProbability = \\(\\frac{36}{36}\\) = 1<\/p>\n<\/p>\nRahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7?<\/h2>\n
\n(B) \\(\\frac{5}{36}\\)
\n(C) \\(\\frac{1}{6}\\)
\n(D)0
\nAnswer:
\n(C) \\(\\frac{1}{6}\\)<\/p>\n
\n(B) \\(\\frac{5}{36}\\)
\n(C) \\(\\frac{1}{18}\\)
\n(D) \\(\\frac{5}{18}\\)
\nAnswer:
\n(D) \\(\\frac{5}{18}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"