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# Class 9 Maths MCQs on Probabilty Chapter 15 with Solutions for CBSE Board Exam 2021-22

Class 9 Maths MCQs on probability chapter 15 is created here for helping the students of class 9  in preparation of their unit tests and CBSE exams. All MCQ questions are selected here as per the latest maths sample paper published by CBSE Board for 2021-22. MCQs on probability are selected as per the standard CBSE maths sample paper,so the study of these MCQs is mandatory for every student. All questions are explained here with the easiest method therefore every student can understand. You can also study other study inputs like science and maths NCERT Solutions.

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## Class 9 Maths MCQs on Probabilty Chapter 15 with Solutions for CBSE Board Exam 2021-22

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Q1. If a dice is thrown in the air then its probability of getting an even number is:

(a) 2/3       (b) 1/3        (c) 1/2       (d) 1/4

Ans. When the dice is thrown once,then total possible outcomes are 1,2,3,4,5,6 (i.e 6)

The even numbers are 2,4,6

P(even number) = the number of even numbers/total possible outcomes = 3/6 = 1/2

Q2.If a dice is thrown in the air then the probabilty of getting at least 4 is:

(a) 2/3        (b) 3/4         (c) 1/2        (d) 1/3

Ans. (a) 2/3

When a dice is thrown once then total possible outcomes are 1,2,3,4,5,6 i.e 6

The outcomes which are at least 4 are 1,2,3,4 i.e 4

P(at least 4) = the numbers showing the digits at least 4/total possible outcomes = 4/6 =2/3

Q3. If two coin are thrown in the air then the probability of getting two heads is :

(a) 2/3        (b) 1/2       (c) 1/3       (d) 1/4

Ans. Total outcomes when two coins are thrown simultaneously are (HH),(TT),(TH),(HT)

Two heads are (HH) i.e 1

P(2 heads ) = number of  two heads /total outcomes = 1/4

Q4.If three coins are thrown simultaneously then what is the probability of getting exactly three heads.

(a) 1/3        (b) 2/3         (c) 2/5         (d) 3/8

Ans. The outcomes when three coins are tossed simultaneously are (HHH),(TTT),(HHT),(TTH),(HTH),(THT),(THH),(HTT) [i.e 8]

The outcomes with exactly two heads are (HHT),(HTH),(THH) [i.e 3]

P( exaactly two heads) = the number of two heads/no. of total outcomes = 3/8

Q5. If the probabilty of raining a day is 0.25 then the probability of not raining is :

(a) 0.85        (b) 0.75       (c) 1/4       (d) 3/5

Ans. (b) 0.75

Let the event of raining is E

The probabilty of raining ,P(E) is 0.25

The probabilty of not raining is = 1 – 0.25 =0.75[P(E) + P(not E) = 1]

Q6.If a card is drawn from a dec of a 52 cards then the probabilty of getting a red card is:

(a) 1/2      (b) 1/4        (c) 3/4        (d) 4/5

Ans. (a) 1/2

The total number of cards are 52

The number of red cards are = 13 diamonds + 13 heart = 26

P(red card) = no. of red cards/total number of cards = 26/52 = 1/2

Q7.If a card is drawn from a well-suffled dec of 52 cards then the probability of getting an ace is:

(a) 1/13    (b) 2/13      (c) 1/2      (d) 1/4

Ans. (a) 1/13

The total number of cards are 52

The number of ace = 4

P(card is ace ) = no. of aces/total number of cards = 4/52 = 1/13

Q8. A fair dice is thrown once,then the probability of getting a composite number:

(a) 1/3       (b) 2/3        (c) 3/4      (d) 2/5

Ans. The total outcomes of the event are 1,2,3,4,5,6 (i.e 6)

The composite numbers are 2,4,6 (i.e 3)

P(getting a composite number) =no. of composite numbers/total outcomes = 3/6 = 1/2

Q9.The probabilty of occuring 53 sundays in a leap year is:

(a) 1/7        (b) 2/7      (c) 3/7       (d) 4/7

Ans. (a) 2/7

The total number of days in a leap year are 366

In 366 days there are  52 weeks and 2 days

The number of days which can make 53 sundays are =2

Total number of days in a week are = 7

P( no. of 53 sundays ) = no. of days which can make 53 sudays/total number of days in a week =2/7

Q10. The probabilty of choosing a girl out of 40 students of a class is 5/8,find the number of boys in the class.

(a) 20        (b) 15        (c) 30         (d) 26

Ans.(b) 15

Let the number of girls in the class are =x

Total student in a class are = 40

Probability of choosing a girl,P(G) = 5/8

P(G) = number of girls in the class /no. of total students in the class = x/40

x/40 = 5/8

x = 40 × (5/8) = 25

Number of boys = total students – number of girls =40 -25 = 15

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