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Probability Test - 13

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Question 1
1 / -0

An unbiased die is thrown once. The probability of getting a prime number is

A
1/4

B
1/2

C
1/5

D
1/3

Solution

Number of possible outcomes = {2, 3, 5} = 3
Number of Total outcomes = 6
∴ Probability of getting a prime number = 3/6 = 1/2
Hence, the correct option is (B).
Question 2
1 / -0

Two dice are thrown simultaneously. The probability of getting a doublet is

A
1/2

B
2/3

C
1/36

D
1/6

Solution

Hint:- Probabilty of tuberable event-
$\Rightarrow \frac{P o s s i b l e e l e m e n t}{T o t a l n o . o f e l e m e n t}$
Solution:-
Doublet means getting same number on both dice simultaneously
Doublets $= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$
Number of possible outcomes $= 6$
Total number of ways to throw a dice $= 36$
Probability of getting a doublet $= 6 / 36 = 1 / 6$
Hence, the correct option is (D).
Question 3
1 / -0

When a die is thrown, the probability of getting an odd number less than 3 is

A
1/2

B
1/6

C
1/4

D
1/3

Solution

Hint:- On a die phase total six number are present three are odd and three are even.
Solution:- odd outcomes are 1,3,5 to getting a odd number are less then 3
so only possible odd number is 1
Hence Required probability = $\frac{number of way's}{total number}$
$\Rightarrow \frac{1}{6}$
Note:- These type of question always read carefully means less then or grater then.
Hence, the correct option is (B).
Question 4
1 / -0

If ‘p’ is the probability of an event, then p satisfies

A
0 < p ⩽ 1

B
0 < p < 1

C
0 ≤ p ≤ 1

D
0 ⩽ p < 1

Solution

Hint:- The probability of an impossible event =0 and the probability of ansured event $= 1$
Solution:- Probability of any event is lie between impossible probability and possible probability.
$i . e, 0 \leq P \leq 1$
Hence, the correct option is (C).
Question 5
1 / -0

The probability of getting 3 head or 3 tails in tossing a coin 3 times is

A
1/2

B
3/8

C
1/4

D
1/8

Solution

Hint:- Probability to getting head or tail on a coin is half.
Total outcomes = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT) = 8
Number of possible outcomes = (HHH) or (TTT) = 2
Required Probability = 2/8 = 1/4
Hence, the correct option is (C).
Question 6
1 / -0

A box contains 3 blue balls, 2 white balls and 4 red balls. If a ball is drawn at random from the box, the probability of getting a white ball is

A
2/9

B
3/9

C
4/9

D
1

Solution

Hint:- Probability is numerical descriptions of how likely an event is to occur.
Solution:-
Blue Ball=3
Red Ball =4
White Ball=2
Total number of ball =9
Probability to getting white ball-
$\Rightarrow \frac{n o . o f w h i t e b a l l}{T o t a l n u m b e r o f b a l l}$
$\Rightarrow \frac{2}{9}$
Hence, the correct option is (A).
Question 7
1 / -0

A die is thrown once. The probability of getting an even number and a multiple of 3 is

A
1/2

B
1/5

C
1/6

D
1/3

Solution

Hint:- In a die total six phase and 1-6 numbr are denoted on die.
Solution:-
On a die the even numben are 2,4,6 and multiple by it means only 6 is multiple of 3
So number of possible outcome $= 6 = 1$
Number of total outcomes $= 6$
So the Required Probability $= \frac{1}{6}$
Note:- on a die phase 3 odd and 3 even number are Present.
Hence, the correct option is (C).
Question 8
1 / -0

The probability that a student will score centum in Mathematics is 3/10. The probability that the student will not score centum is

A
6/10

B
1/3

C
1/2

D
7/10

Solution

Hint:- Probability is a branch of mathematics concerning numerical descriptions of how likely an event is to occur or how likely it is that a preposition is true.
Solution:-
Given $P (E) = \frac{3}{10}$
Probability that the student will not score centum is
P (not E) = 1-P(E)
$\Rightarrow 1 - \frac{3}{10} \Rightarrow \frac{7}{10}$
Note:- Also directly find if the probability of any event is $\frac{3}{10}$ and to total probability is 1.
Hence, the correct option is (D).
Question 9
1 / -0

If S is the sample space of a random experiment, then p(S) =

A
0

B
1/4

C
1/8

D
1

Solution

Hint:- $P (\frac{A}{B}) \Rightarrow P (A \cap B)$
Solution:-
$P (\frac{S}{A}) = 1$
$\frac{P (S \cap A)}{P (A)}$
'S' is a sample space. So the intersection of 'S' with 'A'

So $\frac{P (A)}{P (A)} = 1$
Note:- 'S' is a sample space.
Hence, the correct option is (D).
Question 10
1 / -0

A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random, the probability that it is not red is

A
1/2

B
5/12

C
3/12

D
3/4

Solution

Hint:- Probability of tuberable event-
$\Rightarrow \frac{P o s s i b l e e l e m e n t}{T o t a l n o . o f e l e m e n t}$
Solution:-
Numbers of balls which are not red $= 12 - 3 = 9$
Number of possible outcomes $= 9$
Number of Total outcomes $= 12$
Required Probability $= 9 / 12 = 3 / 4$
Hence, the correct option is (D).
Question 11
1 / -0

An unbiased die is thrown once. The probability of getting a prime or composite number is

A
1

B
5/6

C
1/2

D
1/6

Solution

Hint:- Prime numben on die are 2,3,5 amd composite is 4,6
Solution:- Prime numbers on a die are 2,3,5
composite numbers on a die are 4,6
Prime and Composite numbers on a die $= 2, 3, 4, 5, 6$
Number of possible outcomes $= 5$
Number of Total outcomes $= 6$
Required Probability $= 5 / 6$
Hence, the correct option is (B).
Question 12
1 / -0

One card is drawn from a well shuffled pack of 52 cards. The probability of getting an ace is

A
2/13

B
1/52

C
1/13

D
4/13

Solution

Hint:- use P(A)= $\frac{f a v o r a b l e o u t c o m e}{t o t a l n u m b e r o f f a v o r a b l e}$
Solution:- total number of favorable outcomes
$n (s) = 52$
probability to getting an ace card $= 4$
So the number of favorable outcomes $n (A) = 4$
$P (A) = \frac{n (A)}{n (S)} \Rightarrow \frac{4}{52} \Rightarrow \frac{1}{13}$
Note:- four all are present in card box.
Hence, the correct option is (C).
Question 13
1 / -0

If ‘φ’ is an impossible event, then p(φ) =

A
1

B
0

C
1/2

D
1/4

Solution

Hint:- An event which has no chance of occuring is called an impossible event.
Solution:- If φis an impossible event, then its probability is zero.
Hence, the correct option is (B).
Question 14
1 / -0

There are 8 defective items in a sample of 24 items. One item is drawn at random. The probability that it is a non defective item is

A
1

B
1/3

C
2/3

D
1/2

Solution

Hint:- Probability is numerical descriptions of how likely an event is to occur.
Solution:-
Total number of item =24
Defective item =8
Non defective= 24-8= 16
Probability to get non defective = $\frac{16}{24} = \frac{2}{3}$
Note:- First find the non defective item.
Hence, the correct option is (C).
Question 15
1 / -0

A number is selected at random from the numbers 7, 3, 9, 7, 9, 5, 7, 9, 9, 5. The probability that the selected number is their average is

A
7/10

B
5/10

C
3/10

D
1/10

Solution

Hint: - Add up all numbers then divide by how many numbers there are and you will get the average of the numbers.
Average of given numbers = $\frac{7 + 3 + 9 + 7 + 9 + 5 + 7 + 9 + 9 + 5}{10} = \frac{70}{10} = 7$
(There are 3 times 7 in the given numbers)
Therefore Number of outcomes $= 3$
Number of total outcomes $= 10$
$∴$ Required Probability $= \frac{3}{10}$
Note:- Find the average of given numbers carefully.
Hence, the correct option is (C).

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Probability Test - 13

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Probability Test - 13

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SHARING IS CARING

Question 1

1/4

1/2

1/5

1/3

Solution

Question 2

1/2

2/3

1/36

1/6

Solution

Question 3

1/2

1/6

1/4

1/3

Solution

Question 4

0 < p ⩽ 1

0 < p < 1

0 ≤ p ≤ 1

0 ⩽ p < 1

Solution

Question 5

1/2

3/8

1/4

1/8

Solution

Question 6

2/9

3/9

4/9

1

Solution

Question 7

1/2

1/5

1/6

1/3

Solution

Question 8

6/10

1/3

1/2

7/10

Solution

Question 9

0

1/4

1/8

1

Solution

Question 10

1/2

5/12

3/12

3/4

Solution

Question 11

1

5/6

1/2

1/6

Solution

Question 12

2/13

1/52

1/13

4/13

Solution