To know the opinion of the student about the subject statistic, a survey of $$200$$ students was conducted. The data is recorded in the following table Opinion Like Dislike No. of Students $$135$$ $$65$$ Find the probability that a student chosen at random $$(i)$$ likes statistics, $$(ii)$$ does not like it.

$$(i) \displaystyle \frac{27}{40} $$ $$(ii) \displaystyle \frac{13}{40} $$

$$(i) \displaystyle \frac{13}{40} $$ $$(ii) \displaystyle \frac{19}{40} $$

$$(i)\displaystyle \frac{17}{40} $$ $$(ii) \displaystyle \frac{29}{40} $$

MCQ Questions for Class 9 Maths Probability Quiz 2

In a survey of

$364$ children aged

$19-36$ months, it was found that

$91$ liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is :

0%
$0.25$
0%
$0.50$
0%
$0.75$
0%
$0.80$

Explanation

Favorable cases

$=364 - 91 =273$

Total cases

$= 364$

$\therefore$ Probability

$= \dfrac { 273 }{ 364 }=0.75$

A coin is tossed $150$ times and the outcomes are recorded. The frequency distribution of the outcomes $H$ (i.e., head) and $T$ (i.e., tail) is given below :

Outcome	$H$	$T$
Frequency	$85$	$65$

Find the value of $P(H)$ , i.e., probability of getting a head in a single trial.

0%
$P(H) =0.769$ (approx)
0%
$P(H) =0.663$ (approx)
0%
$P(H) =0.567$ (approx)
0%
None of these

Explanation

Total number of trials

$= 150$
Chances or trials which favour the outcome

$H=85$

$P(H) =$

$\displaystyle \frac{85}{150}$

$=$

$0.567$ (approx)

To know the opinion of the student about the subject statistic, a survey of

$200$ students was conducted.

The data is recorded in the following table

Opinion	Like	Dislike
No. of Students	$135$	$65$

Find the probability that a student chosen at random

$(i)$ likes statistics, $(ii)$ does not like it.

0%
$(i) \displaystyle \frac{13}{40}$

$(ii) \displaystyle \frac{19}{40}$
0%
$(i) \displaystyle \frac{27}{40}$

$(ii) \displaystyle \frac{13}{40}$
0%
$(i)\displaystyle \frac{17}{40}$

$(ii) \displaystyle \frac{29}{40}$
0%
None of these

Explanation

Total number of students

$= 200$
(i) Number of students who like the subject of statistics

$= 135$
The probability that a student likes that subject

$=$

$\displaystyle\frac{135}{200}$

$=$

$\displaystyle \frac{27}{40}$

(ii) Number of students who dislike the subject of statics

$=$ 65
The probability that a student dislikes the subject

$=$

$\displaystyle \frac{65}{200}$ =

$\displaystyle \frac{13}{40}$

On one page of a telephone directory there were

$200$ telephone numbers. The frequency distribution of their unit place digit is given in the following table:

Digit	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
Frequency	$22$	$26$	$22$	$22$	$20$	$10$	$14$	$28$	$16$	$20$

0%

$0.15$
0%

$0.17$
0%

$0.18$
0%
None of these

A couple of dice rolled. What is the empirical probability of getting the sum as

$5$ ?

0%
$\dfrac{3}{2}$
0%
$\dfrac{1}{9}$
0%
$\dfrac{5}{2}$
0%
$\dfrac{7}{2}$

Explanation

Refer below diagram for sample space of two dice rolled to get sum of

$5$

When a pair of dice rolled possibilities for getting

$5$ is

$(1, 4), (2, 3), (3, 2), (4, 1)$

Total number events occurred when we roll the dice

$= 36$

So, the empirical probability formula is as follows,

$P(E) =$

$\dfrac{4}{36}= \dfrac{1}{9}$

A survey was conducted by car manufacturing company in a metropolitan city on

$1000$ persons having monthly income from Rs.

$30,001$ to Rs.

$50,000$ . The data about the number of persons in various categories is as under:

Monthly income	Number of Cars
(in rupees)	$1$	$2$	More than $2$
$30,001-40,000$	$400$	$50$	$25$
$40,001-50,000$	$100$	$300$	$125$

Find the probability that a person selected at random in the income slab

$40,001-50,000$ have more than

$2$ cars.

0%
$0.125$
0%
$0.225$
0%
$0.325$
0%
None of thesee

Explanation

A survey was conducted by car manufacturing company in a metropolitan city on

$1000$ persons having monthly income from Rs.

$30,001$ to Rs.

$50,000$
In the income slab

$40,001-50,000$ having more than

$2$ cars in the company is

$125$
Then the probability that a person

$=\dfrac{125}{1000}=0.125$

You have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, if the probability of getting a non-blue sector is

$\dfrac{4}{m}$ . Then, the value of

$m$ is

0%
$5$
0%
$4$
0%
$3$
0%
$1$

Explanation

Out of 5 sectors, the pointer can stop at any of the sectors in 5 ways.

$\therefore$ Total number of outcomes =

$5$

There are 4 non-blue sectors in the spinning wheel, out of which one can be obtained in 4 ways.

So, the required probability =

$\dfrac{4}{5}$

Hence the value of

$m$ is

$5$

The record of a weather station shows that out of the past $250$ consecutive days, its weather forecasts were correct $175$ times. What is the probability that on a given day it was correct?

0%
$0.9$
0%
$0.8$
0%
$0.7$
0%
$0.6$

Explanation

Total no. of days for which record of weather is given

$=250$
No. of days for which forecasts were correct

$= 175$

$\therefore$ favorable days

$= 175$

$\therefore$ probability

$= \displaystyle \frac {175}{250} = 0.7$

A box has tokens numbered

$3$ to

$100$ . If a token is taken out at random the chance that the number is divisible by

$7$ is

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{7}$

Explanation

The tokens are numbered from

$3$ to

$100$

There are

$98$ numbers from

$3$ to

$100$

In this range the numbers divisible by

$7$ are

$7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91$ and

$98$

The number of numbers divisible by

$7$ is

$14$

$P$ (number divisible by

$7$ )

$=\displaystyle \frac{14}{98}=\frac{1}{7}$

Hence, the required probability is

$\dfrac{1}{7}$

A box contains

$32$ coloured marbles.

$8$ of them are red marbles and the rest are either blue or green marbles. A marble is drawn at random. Calculate the probability of drawing a marble which is not red in colour.

0%
$\dfrac {2}{3}$
0%
$\dfrac {5}{8}$
0%
$\dfrac {3}{4}$
0%
$\dfrac {7}{16}$

Explanation

The marbles not in red color are

$32-8=24$

Then probability of not drawing red marbles is,

$P(E)=\dfrac{24}{32}\\\\\ \ \ \ \ \ \ \ =\dfrac{3}{4}$

Find the chance that a non-leap year contains

$53$ Saturdays

0%
$\dfrac {2}{7}$
0%
$\dfrac {1}{7}$
0%
$\dfrac {2}{365}$
0%
$\dfrac {1}{365}$

Explanation

non-leap year has

$365$ days which is divided into

$7 \times 52=364 + 1$ days. There will be

$52$ fridays, one for each week except when the first day of the year is a Friday. That probability is

$P(E)=\dfrac{1}{7}$

A card is drawn from a well shuffled deck of

$52$ cards. Find the probability of getting a club card?

0%
$\dfrac {1}{2}$
0%
$\dfrac {3}{4}$
0%
$\dfrac {1}{4}$
0%
$1$

Explanation

a well shuffled deck of 52 cards there are 13 club card

Hence

$P(E)=\dfrac{13}{52}=\dfrac{1}{4}$

Observe the table given.

Club	English	Maths	Science
Number of students	$99$	$121$	$154$

A student is chosen randomly from the group as the co-ordinator of all the clubs. What is the probability that the student is from the science club?

0%
$\dfrac {7}{17}$
0%
$\dfrac {9}{34}$
0%
$\dfrac {10}{17}$
0%
$\dfrac {11}{34}$

Explanation

A student is chosen randomly from the group as the coordinator of all the clubs the probability that the student is from the science club

$P(E)=\dfrac{student_science}{total_students}=\dfrac{154}{154+99+121}=\dfrac{7}{17}$

From a normal pack of cards, a card is drawn at random. Find the probability of getting a jack or a king.

0%
$\dfrac {2}{52}$
0%
$\dfrac {1}{52}$
0%
$\dfrac {2}{13}$
0%
$\dfrac {1}{26}$

Explanation

From a normal pack of cards, a card is drawn at random

There are 4 kings and 4 jacks in a standard

$52$ card deck. So the probability of drawing a king or a jack would be

$P(E)=\dfrac{8}{52}=\dfrac{2}{13}$

A bag contains a number of marbles of which

$80$ are red,

$24$ are white and the rest are blue. If the probability of randomly selecting a blue marble from this bag is

$\dfrac {1}{5}$ , how many blue marbles are there in the bag?

0%
$25$
0%
$26$
0%
$27$
0%
$28$

Explanation

A bag contains a number of marbles of which 80 are red, 24 are white and the rest are blue. If the probability of randomly selecting a blue marble from this bag is

$\dfrac{1}{5}$

The Probabaility of red marble is

$P(E)=\dfrac{1}{5}=$

$\dfrac{blue}{red+blue+white}$ =

$\dfrac{blue}{80+24+blue}$

from Above we get

$4$ blue

$=104$

blue

$=26$

A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing a face card.

0%
$\dfrac { 8 }{ 13 }$
0%
$\dfrac { 3 }{ 13 }$
0%
$\dfrac { 4 }{ 13 }$
0%
$\dfrac { 1 }{ 4 }$

Explanation

There are

$12$ face cards in 52 well-shuffled deck of playing cards

therefore

$P(E)=\dfrac{card_{face}}{card_{total}}=$

$\dfrac{12}{52}=$

$\dfrac{3}{13}$

A card is drawn at random from a pack of well shuffled

$52$ cards. What is the probability of getting a queen of red suit?

0%
$\dfrac { 1 }{ 26 }$
0%
$\dfrac { 1 }{ 13 }$
0%
$\dfrac { 1 }{ 52 }$
0%
$\dfrac { 1 }{ 14 }$

Explanation

there are

$2$ queen of red suit.

A card is drawn at random from a pack of well shuffled 52 cards.

$P(E)=\dfrac{2}{52}=\dfrac{1}{26}$

In a single throw of two dice, find the probability that neither a doublet nor a total of 8 will appear.

0%
$\cfrac{11}{36}$
0%
$\cfrac{5}{18}$
0%
$\cfrac{13}{18}$
0%
$\cfrac{3}{16}$

Explanation

Doublet = $\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$

Total $8$ will be from the combination of $\{(2,6),(6,2),(3,5),(5,3),(4,4)\}$

Total no. of combinations $6+5-1=10$ ( $-1$ because $(4, 4)$ taken two times)

So, no. of possibilities that neither a doublet nor a total of 8 will be $=36-10=26$

Total no. possibilities $=6\times 6=36$

So, Probability $=\dfrac{26}{36}=\dfrac{13}{18}$

Bag contains

$100$ tickets bearing numbers

$1-100$ . A ticket is taken out randomly, find the probability getting of a even numbered ticket.

0%
$0.5$
0%
$0.4$
0%
$1.0$
0%
$0.05$

Explanation

Total number of tickets,

$n(S)=100$

Favorable outcomes for even number,

$n(E)=50$

The probability getting of an even numbered ticket,

$\dfrac {n(E)}{n(S)}=\dfrac {50}{100}=0.5$

In a non-leap year the probability of getting

$53$ Sundays or

$53$ Tuesdays or

$53$ Thursdays is.

0%
$\dfrac {1}{7}$
0%
$\dfrac {2}{7}$
0%
$\dfrac {3}{7}$
0%
$\dfrac {4}{7}$

Explanation

Non-leap year has

$365$ days

$1$ year

$=52$ weeks

$=52\times 7 days$

$=364$ days [It contains

$52$ Sundays, Mondays and so on].

$\therefore 1$ day will be left sample space for this

$1$ day

$=\{$ Mon, Tue, Wed, Thurs, Fri, Sat, Sun

$\}$

$P[ 53$ Sundays

$] =\dfrac { 1 }{ 7 }$

$P[ 53$ Tuesdays

$] =\dfrac { 1 }{ 7 }$

$P[53$ Thursdays

$] =\dfrac { 1 }{ 7 }$

$P(53$ Sundays or

$53$ Tuesdays or

$53$ Thursdays

$)$

$=P(53$ Sundays

$)+P(53$ Tuesdays

$)+P(53$ Thursdays

$)$

$=\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 7 } +\dfrac { 1 }{ 7 }$

$={ \dfrac { 3 }{ 7 } }$

CBSE Questions for Class 9 Maths Probability Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Probability Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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