Probability Class 10 MCQ Questions with Answers
Question 1.
If an event that cannot occur, then its probability is:
(A) 1
(B) \(\frac{3}{4}\)
(C) \(\frac{1}{2}\)
(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 an event ?
(A) \(\frac{1}{3}\)
(B) 0.1
(C) 3%
(D) \(\frac{17}{16} \)
Answer:
(D) \(\frac{17}{16}\)
Explanation:
Probability of any event cannot be more than one or negative as \(\frac{17}{16}\) > 1.
Question 3.
An event is very unlikely to happen. Its probability is closest to:
(A) 0.0001
(B) 0.001
(C) 0.01
(D) 0.1
Answer:
(A) 0.0001
Explanation:
The 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.
Question 4.
If the probability of an event is p, then the probability of its complementary event will be:
(A) p – 1
(B) p
(C) 1 – p
(D) \(1-\frac{1}{p}\)
Answer:
(C) 1 – p
Explanation:
Probability of an event + Probability of its complementary event = 1
∴ p + Probability of complement = 1
Probability of complement = 1 – p
Question 5.
The probability expressed as a percentage of a particular occurrence can never be:
(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number
Answer:
(B) less than 0
Explanation:
Probability lies between 0 and 1 and when it is converted into percentage it will be between 0 and 100. So, cannot be negative.
Question 6.
If P(A) denotes the probability of an event A, then:
(A) P(A) < 0 (B) P(A) > 1
(C) 0 ≤ P(A) ≤ 1
(D) -1 ≤ P(A) ≤ 1
Answer:
(C) 0 ≤ P(A) ≤ 1
Explanation:
As the probability of an event lies between 0 and 1.
Question 7.
If a card is selected from a deck of 52 cards, then the probability of its being a red face card is :
(A) \(\frac{3}{26}\)
(B) \(\frac{3}{13}\)
(C) \(\frac{2}{13}\)
(D) \(\frac{1}{2}\)
Answer:
(A) \(\frac{3}{26}\)
Explanation:
In a deck of 52 cards, there are 26 red cards.
Number of red face cards = 3 of hearts + 3 of diamonds = 6
So, probability of having a red face card
= (A) \(\frac{6}{52}\) = (A) \(\frac{3}{26}\)
Question 8.
The probability that a non-leap year selected at random will contain 53 Sundays is:
(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{3}{7}\)
(D) \(\frac{5}{7}\)
Answer:
(A) \(\frac{1}{7}\)
Number of weeks = \(\frac{365}{7}\) = 52\(\frac{1}{7}\) = 52 weeks
Number of days left = 1
For example, it may be any of 7 days which from Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday; so,
T(E) = 7
F(E) = 1 (Sunday)
p(F) = \(\frac{F(E)}{T(E)}\) = \(\frac{1}{7}\)
Asertion and Reased Based MCQs
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True
Question 1.
Assertion (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 \(\frac{1}{2}\).
Reason (R): When we toss a coin, there are two possible outcomes: head or tail. Therefore, the probability of each outcome is i.
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
In case of assertion:
From 1 to 100 numbers, there are 50 even and 50 odd numbers.
Total number of outcomes T(E) = 100 Number of outcomes favourable for event E (even numbers) = F (E) = 50
so, P (E) = \(\frac{50}{100}\) = \(\frac{1}{2}\)
Similarly, the probability of getting odd numbers =\(\frac{1}{2}\).
Hence the probability of getting odd and even each = \(\frac{1}{2}\).
Hence, the given statement is true.
∴ Assertion is correct.
In case of reason:
Since, there are two outcomes equal in all manners. So, probability of both head and tail is equal to \(\frac{1}{2}\). each.
Hence, the given statement is true.
∴ Reason is correct:
Hence, both assertion and reason are correct but reason is not correct explanation for assertion.
Question 2.
Assertion (A): If P(F) = 0.20, then the probability of ‘not E’ is 0.80.
Reason (R): If two dice are thrown together, then the probability of getting a doublet is \(\frac{5}{6}\).
Answer:
(C) A is true but R is false
Explanation:
In case of assertion:
P(E) = 0.20
P(not E) = 1 – P(E)
∴Assertion is correct.
In case of reason:
Total number of possible outcomes = 62 = 36
E : (doublets are (1,1), (2,2), (3, 3), (4,4), (5,5), (6,6)
No. of favourable outcomes to E = 6
P(a doublet)
= \(\frac{Number of outcomes favourable to E}{Total number of outcomes}\).
=\(\frac{6}{36}\). = \(\frac{1}{6}\).
∴Reason is incorrect:
Hence, assertion is correct but reason is incorrect.
Question 3.
Assertion (A): The probability that a number selected at random from the number 1, 2, 3, , 15
is a multiple of 4, is \(\frac{1}{3}\)..
Reason (R): Two different coins are tossed simultaneously. The probability of getting at least one head is\(\frac{3}{4}\).
Answer:
(D) A is false and R is True
Explanation:
In case of reason:
n(S) = 15
n(A) = 3
p(A) = \(\frac{n(A)}{n(S) }\) = \(\frac{3}{15}\) = \(\frac{1}{5}\)
S = HH, HT, TH,TT
A = HH, HT, TH
n(S) = 4
n(A) = 3
p(A) = \(\frac{n(A)}{n(S) }\) = \(\frac{3}{4 }\)
∴Reason is incorrect:
Hence, assertion is correct but reason is incorrect.
Case – Based MCQs
Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. 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.
Question 1.
Find the probability of getting a king of red colour.
(A) \(\frac{1}{26}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Answer:
(A) \(\frac{1}{26}\)
Explanation:
No. of cards of a king of red colour = 52
Probability of getting a king of red colour
= \(\frac{No. of king of red colour}{Total number of cards }\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)
Question 2.
Find the probability of getting a face card.
(A) \(\frac{1}{26}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{2}{13}\)
(D) \(\frac{3}{13}\)
Answer:
(D) \(\frac{3}{13}\)
Question 3.
Find the probability of getting a jack of hearts.
(A) \(\frac{1}{26}\)
(B) \(\frac{1}{52}\)
(C) \(\frac{3}{52}\)
(D) \(\frac{3}{26}\)
Answer:
(B) \(\frac{1}{52}\)
Question 4.
Find the probability of getting a red face card.
(A) \(\frac{3}{13}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Answer:
(A) \(\frac{3}{13}\)
Explanation:
No. of face card = 13
Total no of cards = 52
Probability of getting a face card
= \(\frac{ No. of face cards }{Total no. of cards}\)
= \(\frac{12}{52}\) = \(\frac{3}{13}\)
Question 5.
Find the probability of getting a spade.
(A) \(\frac{1}{26}\)
(B) \(\frac{1}{13}\)
(C) \(\frac{1}{52}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)
Explanation:
No. of face card = 13
Total no of cards = 52
Probability of getting a face card
= \(\frac{ No. of face cards }{Total no. of cards}\)
= \(\frac{13}{52}\) = \(\frac{1}{4}\)
II. Read the following text and answer the following questions on the basis of the same:
Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice.
Question 1.
Ravi 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?
(A) \(\frac{1}{26}\)
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) 0
Answer:
(B) \(\frac{5}{36}\)
Explanation:
The outcomes when two dice are thrown together are:
= (1, 1), (1, 2), (1, 3), (1,4), (1, 5), (1,6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total outcomes = 36
No. of outcomes when the sum is 8 = 5
Probability = \(\frac{5}{36}\)
Question 2.
Rahul 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?
(A) 1
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) 0
Answer:
(D) 0
Explanation:
No. of outcomes when the sum is 13 = 0
Total outcomes =36
Probability =\(\frac{0}{35}\) = 0
Question 3.
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 less than or equal to 12?
(A) 1
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) 0
Answer:
(A) 1
Explanation:
No. of outcomes when the sum is less than or equal to 12 = 36
Total outcomes = 36
Probability = \(\frac{36}{36}\) = 1
Question 4.
Rahul 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?
(A) \(\frac{5}{9}\)
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{6}\)
(D)0
Answer:
(C) \(\frac{1}{6}\)
Question 5.
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 ?
(A) 1
(B) \(\frac{5}{36}\)
(C) \(\frac{1}{18}\)
(D) \(\frac{5}{18}\)
Answer:
(D) \(\frac{5}{18}\)