M.C.Q

Question 11 Mark

80 bulbs are selected at random from a lot and their lifetime in hours is recorded as under.

Lifetime (in hours)	300	500	700	900	1100
Frequency	10	12	23	25	10

One bulb is selected at random from the lot. What is the probability that the selected bulb has a life more than 500 hours?

$\frac{27}{40}$
$\frac{29}{40}$
$\frac{5}{16}$
$\frac{11}{40}$

Answer

$\frac{29}{40}$

Solution:
Total number of bulbs = 80
Probability that selected bulb has a life more than 500 hours $=\frac{23+25+10}{80}$
$=\frac{58}{80}$
$=\frac{29}{40}$

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Question 21 Mark

Two coins are tossed simultaneously 600 times to get:
2 heads : 234 times, 1 head : 206 times, 0 head : 160 times.
If two coins are tossed at random, what is the probability of getting at least one head?

$\frac{103}{300}$
$\frac{39}{100}$
$\frac{11}{15}$
$\frac{4}{15}$

Answer

$\frac{11}{15}$

Solution:
Total number of outcomes = 600
Probability of getting at least 1 head = Probability of getting 1 head + Probability of getting 2 heads
$=\frac{206}{600}+\frac{234}{600}$
$=\frac{440}{600}$
$=\frac{11}{15}$

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Question 31 Mark

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:

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

Answer

$\frac{3}{4}$

Solution:
Total number of children = 364
Number of children who like to eat potato chips = 91
⇒ Number of children who do not like to eat potato chips = 364 - 91 = 273
$\therefore$ Required probability $=\frac{273}{364}=\frac{3}{4}$

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Question 41 Mark

In a medical examination of students of a class, the following blood groups are recorded:

Blood group	A	B	AB	O
Number of students	11	15	8	6

From this class, a student is chosen at random. What is the probability that the chosen student has blood group AB?

$\frac{13}{20}$
$\frac{3}{8}$
$\frac{1}{5}$
$\frac{11}{40}$

Answer

$\frac{1}{5}$

Solution:
Total number of students = 11 + 15 + 8 + 6 = 40
Number of students having blood group AB = 8
$\therefore$ Required probaility $=\frac{8}{40}=\frac{1}{5}$

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Question 51 Mark

It is given that the probability of winning a game is 0.7. What is the probability of losing the game?

0.8
0.3
0.35
0.15

Answer

0.3

Solution:
Let E be the event of winning a game.
Then, P(E) = 0.7
We know that, P(E) + P(E') = 1
Where E' is the event of losing a game.
⇒ 0.7 + P(E') = 1
⇒ P(E') = 0.3
So, the probability of losing the game is 0.3.

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Question 61 Mark

The table given below shows the month of birth of 36 students of a class.

Month of birth	Jan.	Feb.	March	April	May	June	July	Aug.	Sept.	Oct.	Nov.	Dec.
No. of students	4	3	5	0	1	6	1	3	4	3	4	3

A student is chosen at random from the class. What is the probability that the chosen student was born in October?

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{1}{4}$
$\frac{1}{12}$

Answer

$\frac{1}{12}$

Solution:
Let E be the event of choosing a student born in october.
P(E) $=\frac{\text{Number of students born in october}}{\text{Toatl number of student}}$
$=\frac{3}{36}$
$=\frac{1}{12}$

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Question 71 Mark

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. What is the probability that in a given delivery, the ball does not hit the boundary?

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

Answer

$\frac{4}{5}$

Solution:
Let E be the event where the ball does not hit a boundary.
The batsman hits a boundary 6 times.
So, he does not hit a boundary 30 - 6 = 24 times
P(E) $=\frac{\text{Number of times the ball does not hit a boundary}}{\text{Total number of balls played}}$
$=\frac{24}{30}$
$=\frac{4}{5}$

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Question 81 Mark

80 bulbs are selected at random from a lot and their lifetime is hours is recorded as under.

Lifetime (in hours)	300	500	700	900	1100
Frequency	10	12	23	25	10

One bulb is selected at random from the lot. What is the probability that its life is 1150?

$\frac{1}{80}$
$\frac{7}{16}$
$1$
$0$

Answer

0

Solution:
Total number of bulbs = 80
Number of bulbs having life of 1150 hours = 0
$\therefore$ Required probability = 0

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Question 91 Mark

In a sample survey of 645 people, it was found that 516 people have a high school certificate. If a person is chosen at random, what is the probability that he / she has a high school certificate?

$\frac{1}{2}$
$\frac{3}{5}$
$\frac{7}{10}$
$\frac{4}{5}$

Answer

$\frac{4}{5}$

Solution:
Total number of people = 645
Number of people having high school certificate = 516
$\therefore$ Required probability $=\frac{516}{645}=\frac{4}{5}$

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Question 101 Mark

To know the opinion of the students about the subject Sanskrit, a survey of 200 students was conducted. The data is recorded as under.

Opinion	like	dislike
Number of students	135	65

What is the probability that a student chosen at random does not like it?

$\frac{13}{27}$
$\frac{27}{40}$
$\frac{13}{40}$
$\frac{27}{13}$

Answer

$\frac{13}{40}$

Solution:
Total number of students = 200
Number of students who does not like Sanskrit = 65
$\therefore$ Required probability $=\frac{65}{200}=\frac{13}{40}$

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Question 111 Mark

Two coins are tossed 1000 times and the outcomes are recorded as given below:

Number of heads	2	1	0
Frequency	200	550	250

Now, if two coins are tossed at random, what is the probability of getting at most one head?

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

Answer

$\frac{4}{5}$

Solution:
Total number of outcomes = 1000
Probability of getting at most 1 head = Probability of getting at most 0 head + Probability of getting at 1 head
$=\frac{250}{1000}+\frac{550}{1000}$
$=\frac{800}{1000}$
$=\frac{4}{5}$

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Question 121 Mark

A coin is tossed 60 times and the tail appears 35 times. In a random throw of a coin, what is the probability of getting a head?

$\frac{7}{12}$
$\frac{12}{7}$
$\frac{5}{12}$
$\frac{1}{25}$

Answer

$\frac{5}{12}$

Solution:
Total number of times the coin is tossed = 60
Number of times tail appears = 35
So, the number of times head appears = 60 - 35 = 25
Required probability $=\frac{\text{Number of times head appears}}{\text{Total number of rimes the coin is tossed}}$
$=\frac{25}{60}$
$=\frac{5}{12}$

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Question 131 Mark

A bag contains 5 red, 8 black and 7 white balls. One ball is chosen at random. What is the probability that the chosen ball is black?

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

Answer

$\frac{2}{5}$

Solution:
Total number of balls in the bag = 5 + 8 + 7 = 20
Number of black balls = 8
Required probability $=\frac{\text{Number of black balls}}{\text{Total number of balls}}$
$=\frac{8}{20}$
$=\frac{2}{5}$

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Question 141 Mark

In 50 throws of a dice, the outcomes were noted as shown below:

Outcome	1	2	3	4	5	6
Number of times	8	9	6	7	12	8

A dice is thrown at random. What is the probability of getting an even number?

$\frac{12}{25}$
$\frac{3}{50}$
$\frac{1}{8}$
$\frac{1}{2}$

Answer

$\frac{12}{25}$

Solution:
Let E be the event of getting an evan number.
So, E = {2, 4, 6}
2 appears 9 times, 4 appears 7 times and 6 appears 8 times.
So, the total number of times an even number comes up is 9 + 7 + 8 = 24
P(E) $=\frac{\text{Number of times an even number comes up}}{\text{Total number of times the die is thrown}}$
$=\frac{24}{50}$
$=\frac{12}{25}$

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Question 151 Mark

A bag contains 16 cards bearing numbers 1, 2, 3, ..., 16 respectively. One card is chosen at random. What is the probability that the chosen card bears a number divisible by 3?

$\frac{3}{16}$
$\frac{5}{16}$
$\frac{11}{16}$
$\frac{13}{16}$

Answer

$\frac{5}{16}$

Solution:
Total number of cards in the bag = 16
Number divisible by 3 are {3, 6, 9, 12, 15}
Number of card that bears numbers divisible by 3 = 5
Required probability $=\frac{5}{16}$

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Question 161 Mark

The outcomes of 65 throws of a dice were noted as shown below:

Outcome	1	2	3	4	5	6
Number of times	8	10	12	16	9	10

A dice is thrown at random. What is the probability of getting a prime number?

$\frac{3}{35}$
$\frac{3}{5}$
$\frac{31}{65}$
$\frac{36}{65}$

Answer

$\frac{31}{65}$

Solution:
Let E be the event of getting a prime number.
So, E = {2, 3, 5}
2 appears 10 times, 3appears 12 times and 5 appears 9 times.
So, the total number of times a prime number comes up is 10 + 12 + 9 = 31.
P(E) $=\frac{\text{Number of times a prime number comes up}}{\text{Total number of times the die is thrown}}$
$=\frac{31}{65}$

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