A die is thrown $$400$$ times, the frequency of the outcomes of the events are given as under. outcome $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ Frequency $$70$$ $$65$$ $$60$$ $$75$$ $$63$$ $$67$$ Find the probability of occurrence of an odd number.

The probability of occurrence of odd number$$=\dfrac{200}{299}$$

MCQ Questions for Class 9 Maths Probability Quiz 1

A die having six faces is tossed

$80$ times and the data is as below:

Outcome	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$10$	$20$	$10$	$28$	$8$	$4$

Find $P (1)$ .

0%
$0.175$
0%
$0.135$
0%
$0.145$
0%
$0.125$

Explanation

Since, Total tosses

$=$

$80$
Getting no. of

$1$ 's in total tosses

$=$

$10$

$\therefore$ P

$(1) = \dfrac {10}{80} = 0.125$

There are

$40$ students in a class and their results is presented as below :

Result (Pass/Fail)	Pass	Fail
Number of Students	$30$	$10$

If a student chosen at random out of the class, find the probability that the student has passed the examination

0%
$0.12$
0%
$0.36$
0%
$0.65$
0%
$0.75$

Explanation

Total number of chances

$= 40$
Chances or trials which favour a student to pass

$= 30$
The probability of the required event, i.e., the student has passed the examination

$=$

$\displaystyle \frac{30}{40}$

$= 0.75$

$400$ students of class $X$ of a school appeared in a test of $100$ marks in the subject of social
studies and the data about the marks secured is as below :

Marks secured	Number of Students
$0-25$	$50$
$26-50$	$220$
$51-75$	$100$
Above $75$	$30$
Total number of students	$400$

If the result card of a student he picked up at random, what is the probability that the student has secured more than $50$ marks.

0%
$0.586$
0%
$0.75$
0%
$0.325$
0%
$0.1$

Explanation

Total number of students, i.e , the total frequency

$= 400$
The total number of students who secured more than

$50$ marks

$100 + 30 = 130$
Probability that the marks secured are more than

$50 =$

$\displaystyle \frac{130}{400}$ =

$\displaystyle \frac{1.3}{4}$ =

$0.325$

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%
$P (H) = 0.75$ (approx)
0%
$P (H) = 0.9$ (approx)
0%
$P (H) = 0.15$ (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)

A die is thrown

$200$ times and the outcomes

$1, 2, 3, 4, 5, 6$ have frequencies as below:

Outcome	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$40$	$38$	$43$	$29$	$28$	$22$

Find the probabilities of getting a number more than $1$ and less than $6$ in a toss (trial).

0%
$0.65$
0%
$0.55$
0%
$0.69$
0%
None of these

Explanation

Since, Total cases

$= 200$
Getting a no. more than

$1$ and less than

$6\, (2,3,4,5) = 38+43+29+28 = 138$

$\therefore$ probability

$=$

$\dfrac {138}{200} = 0.69$

There are $500$ packets in a large box and each packet contains $4$ electronic devices in it. On testing, at the time of packing, it was noted that there are some faulty pieces in the packets. The data is as below:

No. of faulty devices in a packet	Number of packets
$0$	$300$
$1$	$100$
$2$	$50$
$3$	$30$
$4$	$20$
Total number of packets	$500$

If one packet is drawn from the box, what is the probability that all the four devices in the packet are without any fault?

0%
$0.3$
0%
$0.75$
0%
$0.6$
0%
$0.95$

Explanation

When the packet has all the four devices without fault, it means the number of faulty devices in the packet is

$0$ .
Number of chances which are favourable to

$0$ are

$300$ as given in the table above. Thus, the probability of packet containing all the four devices without any fault

$=$

$\displaystyle \frac{300}{500}$

$=$

$\displaystyle \frac{3}{5}$

$=$

$0.6$

In a shooting game, John shoots the balls

$20$ times out of

$40$ trials. What is the empirical probability of the shooting event?

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

Explanation

The total number of trials

$= 40$
John shoots the balls only

$20$ times.
So, the empirical probability formula is as follows,

$P(E) =$

$\dfrac{20}{40}= \dfrac{1}{2}$

Find probability of bag chosen out random contains more than

$5\ kg$ .

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

Explanation

Total number of bags

$=10$

Number of bags contain flour more than

$5kg =7$

They are :

$5.05,5.08,5.03,5.06,5.08,5.04,5.07$

$P(A)=\dfrac {Number\ of\ favorable\ outcomes}{Total\ Number\ of\ possible\ outcomes}$

$\therefore P(A)=\dfrac{7}{10}=0.7$

There are $40$ students in a class and their results is presented as below :

Result (Pass/Fail)	Pass	Fail
Number of Students	$30$	$10$

If a student chosen at random out of the class, find the probability that the student has passed the examination.

0%
$0.75$
0%
$0.6$
0%
$0.45$
0%
$0.30$

Explanation

Total number of chances

$= 40$
Chances or trials which favour a student to pass

$= 30$
The probability of the required event, i.e., the student has passed the examination

$= \displaystyle \frac{30}{40} = 0.75$

What is the probability that there are

$5$ Mondays in the month of February 2016?

0%
$0$
0%
$\dfrac{1}{7}$
0%
$\dfrac{2}{7}$
0%
None of the above

Explanation

There are

$29$ days in February

$2016$ .

There will be

$4$ weeks and

$1$ day. This one day can be any one from mon, tue, wed, thurs, fri, sat, sun.

So, probability of having

$5$ Mondays

$=\dfrac { 1 }{ 7 }$

When a coin is tossed at random, then the probability of getting a head is ________.

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

Explanation

Tossing of a coin at random can give only two results, i.e. Heads or Tails.

$\Rightarrow$ Total number of outcomes possible

$= 2$

Probability

$= \dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}} = \dfrac{1}{2}$

Two men hit at a target with probabilities

$\dfrac{1}{2}$ and

$\dfrac{1}{3}$ respectively. What is the probability that exactly one of them hits the target?

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{2}{3}$

Explanation

${ E }_{ 1 }$ denotes the event when first man hits the target

${ E }_{ 2}$ denotes the event when second man hits the target

$P({ E }_{ 1 })=\cfrac { 1 }{ 2 } ,P({ E }_{ 2 })=\cfrac { 1 }{ 3 }$

Probability that exactly one of them hits the target,

$P=P({ E }_{ 1 })P({ \overset { \_ }{ E } }_{ 2 })+P({ E }_{ 2 })P({ \overset { \_ }{ E } }_{ 1 })$

$=\dfrac { 1 }{ 2 } \times \dfrac { 2 }{ 3 } +\dfrac { 1 }{ 3 } \times \dfrac { 1 }{ 2 } =\dfrac { 1 }{ 2 }$

In a single throw of a die, the probability of getting a multiple of

$3$ is ____________.

0%
$\displaystyle\frac{1}{2}$
0%
$\displaystyle\frac{1}{3}$
0%
$\displaystyle\frac{1}{6}$
0%
$\displaystyle\frac{3}{4}$

Explanation

Possible outcomes on rolling a dice are

$1,2,3,4,5,6$

Out of the possible outcomes, multiple of

$3$ are

$3$ and

$6$ .

Probability

$=$ Favourable Outcomes

$\div$ Total Outcomes

$= \dfrac{2}{6} = \dfrac{1}{3}$

A die is tossed

$80$ times and the number

$3$ is obtained

$14$ times. Now, a dice is tossed at random, then the probability of getting the number

$3$ is ________.

0%
$\dfrac {7}{40}$
0%
$\dfrac {3}{14}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {3}{40}$

Explanation

Total number of times dice was rolled

$= 80$

Number of times

$3$ is obtained

$= 14$

Probability

$= \dfrac{\text{Favourable Outcomes}}{\text{Total Outcomes}}$

$\Rightarrow$ Probability of obtaining

$3 = \dfrac{14}{80} =\dfrac{7}{40}$

A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table show the result of $1000$ cases :

Distance in $Km$	Frequency
Less than $4000$	$20$
$4000$ to $9000$	$210$
$9000$ to $14000$	$325$
More than $14000$	$445$

If you buy a tyre of this company what is the probability that it will need to be replaced after it has covered somewhere between $4000\, km$ and $14000\, km$ ?

0%
$0.65$
0%
$0.625$
0%
$0.125$
0%
None of these

Explanation

Probability that it will need to be replaced before it has covered

$4000$ km

$= \dfrac {20}{1000} = 0.02$

Probability that it will last more than

$9000 km = \dfrac { 325 + 445 } {1000} = \dfrac {770}{1000} = 0.77$

Probability that it will need to be replaced after it has covered somewhere between $4000$ km and $14000$ km = $\dfrac { 210 + 325 } {1000} = \dfrac {535}{1000} = 0.535$

No. of faulty devices in a packet	Number of packets
$0$	$300$
$1$	$100$
$2$	$50$
$3$	$30$
$4$	$20$
Total number of packets	$500$

If one packet is drawn from the box, what is the probability that all the four devices in the packet are without any fault?

0%
$0.5$
0%
$0.6$
0%
$0.8$
0%
$0.9$

Explanation

When the packet has all the four devices without fault, it means the number of faulty devices in the packet is

$0$ . Number of chances which are favourable to

$0$ are

$300$ as given in the table above.
Thus, the probability of packet containing all the four devices without any fault

$=$

$\displaystyle \frac{300}{500}$

$=$

$\displaystyle \frac{3}{5}$

$= 0.6$

Fifty seeds were selected at random from each of $5$ bags $A, B, C, D, E$ of seeds, and were kept under standardised conditions equally favourable to germination. After $20$ days, the number of seeds which had germinated in each collection were counted and recorded as follow :

Bag

$A$

$B$

$C$

$D$

$E$

Number of seeds

germinated

$40$

$48$

$42$

$39$

$41$

What is the probability of germination of

$(i)$ more than $40$ seeds in a bag?

$(ii)$ $49$ seeds in a bag?

$(iii)$ more than $35$ seeds in a bag?

0%
$(i)\, 0.690$
$(ii)\, 0.09$
$(iii)\, 1$
0%
$(i)\, 0.80$
$(ii)\, 0.006$
$(iii)\, 1$
0%
$(i)\, 0.70$
$(ii)\, 0.001$
$(iii)\, 1$
0%
$(i)\, 0.60$
$(ii)\, 0$
$(iii)\, 1$

A die is thrown

$400$ times, the frequency of the outcomes of the events are given as under.

outcome	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$70$	$65$	$60$	$75$	$63$	$67$

Find the probability of occurrence of an odd number.

0%
The probability of occurrence of odd number $=\dfrac{5}{7}$
0%
The probability of occurrence of odd number $=\dfrac{9}{2}$
0%
The probability of occurrence of odd number $=\dfrac{193}{400}$
0%
The probability of occurrence of odd number $=\dfrac{200}{299}$

Explanation

Sum of frequencies

$= 400$

Odd numbers are

$1, 3, 5$

Therefore, frequency of all odd numbers

$= 70 + 60 + 63 = 193$

$P(\text{event})=\dfrac{\text{Frequency of occurring of event}}{\text{The total number of trials}}$

Therefore , probability of occurrence of odd number

$=\dfrac{193}{400}$

The probability that a two digit number selected at random will be a multiple of '

$3$ ' and not a multiple of '

$5$ ' is

0%
$\displaystyle \frac{2}{15}$
0%
$\displaystyle \frac{4}{15}$
0%
$\displaystyle \frac{1}{15}$
0%
$\displaystyle \frac{4}{90}$

Explanation

$\Rightarrow$ Total number of two digit number

$=90$

$\Rightarrow$ Total number of two digit number which is divisible by

$3=30$

Out of this there are

$6$ numbers divisible by

$15$

$(15,30,45,60,75,90)$ which are also divisible by

$5.$

$\Rightarrow$ Total two digit number which are divisible by

$3$ but not

$5=30-6=24$

$\Rightarrow$ Required probability

$=\dfrac{24}{90}=\dfrac{4}{15}$

Two coins are tossed

$1000$ times and the outcomes are recorded as below:

No of heads	$2$	$1$	$0$
Frequency	$200$	$550$	$250$

Based on this information, the probability for at most one head is

$\dfrac{a}{b}$ , where

$(a,b)=1$ . Then

$\dfrac{a}{b}$ is

0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{4}{5}$
0%
$\dfrac{3}{4}$

Explanation

Total number of outcomes

$= 1000$
Number of favourable outcomes

$= 800$
Probability =

$\dfrac{800}{1000}$ =

$\dfrac{4}{5}$

CBSE Questions for Class 9 Maths Probability Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Probability Quiz 1 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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