Class 10 Maths MCQs -Chapter 15 Probability for Term 1 CBSE Board with Solutions - Future Study Point

# Class 10 Maths MCQs -Chapter 15 Probability for Term 1 CBSE Board with Solutions

Class 10 Maths MCQs -Chapter 15 Probability for Term 1 CBSE Board with Solutions are created here for helping the students of 10 class in their CBSE Term 1 exam 2021.Class 10 maths MCQs -Chapter 15 Probability are selected from the last year’s question papers as per the latest format of the maths question paper 2021 CBSE Term 1 exam.The solutions of the Maths MCQs of the chapter 15 is solved by an expert of maths in a easiest way so every students can understand these solutions in a proper ways.

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## Class 10 Maths MCQs -Chapter 15 Probability for Term 1 CBSE Board with Solutions

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Q1. If P(E) = 0.02, then what is the probability of    ‘not E ’

(a) 0.9         (b) 0.98         (c) 9.8         (d) 0.97

Ans.(b) 0.98

The sum of probability of an event and probability not happening of the same event is equal to 0

P(E) + P(not E) = 1

0.98 + P(not E) = 1

P(not E) = 1 – 0.98 = 0.02

Q2. A bag has 5 red balls, 4 blue balls, and 6 green balls , if marble is picked up at randomly then what is the probability of not getting a green marble .

(a) 3/5         (b) 4/5        (c) 2/5        (d) 4/15

Ans. Total number of marbles = 5 red balls + 4 blue balls + 6 green ball = 15 balls

P(green balls) = (number of green balls)/(number of total balls) = 6/15 = 2/5

P(not green balls) =1- P(green balls)= 1 -2/5 = 3/5

Q3. If two dice are thrown in the air,the probability of getting the sum as 5 will be:

(a) 3/5          (b) 5/36         (c) 1/9         (d) 7/36

Ans. When two dice are thrown ,the total number of outcomes are 6² = 36

The sum of the digits getting 5 = (1,4),(2,3),(3,2),(4,1) i.e 4

P(getting the sum 5) = (the number of outcomes ,the sum is 5)/total outcomes =4/36 =1/9

Q4. If two coins are thrown simultaneously, then what is the probability of getting at least 2 heads

(a) 2/3       (b) 1/4          (c) 5/9       (d)   1/9

Ans.(b) 1/4

If two coins are thrown simultaneously, then the total outcomes are (TT),(HH),(TH),(HT) i.e 4

The outcomes of getting atleast 2 heads = (HH) i.e 1

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

Q4.In a bag there are 8 red balls, 5 blue balls and certain number of green balls.If the probability of getting a green ball is 2/3,then find the total number of balls in the bag.

(a) 39          (b) 35         (c) 40          (d) 29

Ans.(a) 39

Let the number of green balls in the bag is x

Total number of balls in the bag = 8 red ball + 5 blue balls + x green balls = 13 + x

The probabilty of green balls = no of green balls/Total balls =2/3(given)

P(green balls) = 2/3 = x/(13+x)

26 + 2x = 3x

x = 26

Hence total balls in the bag = 26 green balls +8 red balls +5 blue balls =39 balls

Q5. If two dice are thrown in the air,then what is the probability of getting the sum of the digits greater than 8.

(a) 7/36       (b) 5/18      (c) 11/18       (d) 5/36

Ans.(b) 5/18

When two dice are thrown,then no. of total outcomes = 6² = 36

The outcomes in which sum is greater than 8 are (3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6) i.e 10

P( sum greater than 8) = the number of outcomes in which sum is greater than 8/total outcomes

= 10/36 = 5/18

Q6. If two dice are thrown in the air,then what is the probability of getting the sum of the digits at least 8.

(a) 2/7        (b) 3/14         (c) 5/12         (d) 5/24

Ans.(c) 5/12

When two dice are thrown,then no. of total outcomes = 6² = 36

The outcomes in which sum is at least  8 are (2,6),(3,5),(4,4),(5,3),(6,2),(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6) i.e 15

P( sum is at least 8) = the number of outcomes in which sum is at least 8/total outcomes

= 15/36 = 5/12

Q7.If two dice are thrown in the air,then what is the probability of getting the sum of the digits at most 8.

(a) 2/7        (b) 5/18     (c) 13/18     (d) 7/18

Ans.(c) 13/18

When two dice are thrown,then no. of total outcomes = 6² = 36

The outcomes in which sum is at least 8 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),(4,1),(4,2),(4,3),(4,4),(5,1),(5,2),(5,3),(6,1),(6,2)i.e 26

P( sum is at most 8) = the number of outcomes in which sum is at most 8/total outcomes

= 26/36 = 13/18

Q8. If a three coins are tossed siumltaneously,the probability of getting at least 2 heads is:

(a) 1/8         (b) 3/8        (c) 1/2         (d) 5/9

Ans.(c) 1/2

The total outcomes when three dice are thrown simultaneously are (2³ =8) (HHH),(TTT),(HHT),(THH),(HTH),(THT),(TTH),(HTT)

The outcomes in which there are at least 3 heads are (HHT),(THH),(HTH),(HHH)

P( getting at least 2 head) = the outcomes in with at least 2 heads/total outcomes = 4/8 = 1/2

Q9.In a family there are 3 children,the probabilty of having atleast two boys is

Ans.

Total possible out comes are (2³ =8) (BBB),(GGG),(BBG),(GBB),(BGB),(GBG),(GGB),(BGG)

The outcomes with at least two boys are (BBB),(BBG),(GBB),(BGB) i.e 4

P( at least two boys) = the outcomes with at least two boys/total outcomes =4/8 = 1/2

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