M.C.Q

Question 11 Mark

A dice is rolled 600 times and the occurence of the outcomes 1, 2, 3, 4, 5 and 6 are given below:

Outcome	1	2	3	4	5	6
Frequency	200	30	120	100	50	100

The probability of geeting a prime number is:

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{49}{60}$
$\frac{39}{125}$

Answer

$\frac{1}{3}$

Solution:
Prime numbers in 1, 2, 3, 4, 5, 6 are: 2, 3, 5.
Number of times 2, 3, 5 occur = 30 + 120 + 50 = 200
Total number of cases = 200 + 30 + 120 + 100 + 50 + 100 = 600
Required probability $=\frac{\text{Cases when we obtained (2, 3, 5)}}{\text{Total no. of cases}}$
$=\frac{200}{600}=\frac{1}{3}$

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Question 21 Mark

A coin is tosses 1000 times, if the probability of getting a tail is $\frac{3}{8},$ how many times head is obtained?

525
375
625
725

Answer

625

Solution:
Probability of getting a tail $=\frac{3}{8}$
⇒ Probability of geeting a head $=1-\frac{3}{8}=\frac{5}{8}$
Also,
Probability of getting a head $=\frac{\text{No.of heads obtained}}{\text{Total no. of trials}}$
$\Rightarrow\ \frac{5}{8}=\frac{\text{No. of heads obtained}}{1000}$
⇒ No. of heads obtainted $=\frac{5}{8}\times1000=625$

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Question 31 Mark

Which of the following cannot be the probability of an event?

$\frac{1}{3}$
$\frac{3}{5}$
$\frac{5}{3}$
$1$

Answer

$\frac{5}{3}$

Solution:
The probability of an event always lies between 0 and 1.
Since $\frac{5}{3}>1,$ it cannot be the probability of an event.

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Question 41 Mark

Two coins are tassed simultaneously. The probability of geeting atmost one head is:

$\frac{1}{4}$
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{5}{4}$

Answer

$\frac{3}{4}$

Solution:
If two coins are tossed simultaneously, then possible cases are HH, TH, HT, TT.
Total number of cases = 4
Number of favorable cases (atmost one head) = (HT, TH, TT) = 3
Now,
Probability of getting atmost one head $=\frac{3}{4}$

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Question 51 Mark

In a football match, Ronaldo makes 4 goals from 10 penalty kicks. The probability of converting a penalty kick into a goal by Ronaldo, is:

$\frac{1}{4}$
$\frac{1}{6}$
$\frac{1}{3}$
$\frac{2}{5}$

Answer

$\frac{2}{5}$

Solution:
Probability that Ronaldo makes a goal
$=\frac{\text{Number of goal made in all kicks}}{\text{Total number of kicks}}$
$=\frac{4}{10}=\frac{2}{5}$

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Question 61 Mark

A bag contains 50 coins and each coin is marked from 51 to 100. One coin is picked at random. The probability that the number on the coin is not a prime number, is:

$\frac{1}{5}$
$\frac{3}{5}$
$\frac{2}{5}$
$\frac{4}{5}$

Answer

$\frac{4}{5}$

Solution:
Prime numbers from 51 to 100:
53, 59, 61, 67, 71, 73, 79, 83, 89, 97
⇒ Number of prime numbers = 10
⇒ Numver of non-prime numbers = 50 - 10 = 40
Total numbers = 50
Thus, probability of getting no-prime number $=\frac{40}{50}=\frac{4}{5}$

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Question 71 Mark

The percentage of attendance of different classes in a year in a school is given below:

Class	X	IX	VIII	VII	VI	V
Attendance	30	62	85	92	76	55

What is the probability that the class attendance is more than 75%?

$\frac{1}{6}$
$\frac{1}{3}$
$\frac{5}{6}$
$\frac{1}{2}$

Answer

$\frac{1}{2}$

Solution:
Total number of classes = 6
Number of classes having attendance > 75% = VIII, VII, VI = 3
⇒ Required probability $=\frac{3}{6}=\frac{1}{2}$

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Question 81 Mark

The probability of an event of a trial is:

1
0
Less than 1
More than 0

Answer

Less than 1

Solution:
Probability of an event $=\frac{\text{No. of favorable cases}}{\text{Total no. of cases}}$
Since number of favoravle cases can not be greater than total number of cases,
Probability < 1.

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Question 91 Mark

The probability of an impossible event is:

1
0
Less than 0
Greater than 1

Answer

0

Solution:
The probability of an impossible event is always zero since the chances of occurring of that event is zero.

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Question 101 Mark

The probability of a certain event is:

0
1
Greater than 1
Less than 0

Answer

1

Solution:
The chance of occuring of an certain event is always 100%.
Probability $=\frac{\text{No. of favorable cases}}{\text{Total no. of cases}}$
Thus, for a certain event,
Number of favorable cases = Total number of cases
⇒ Probability = 1

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