Probability(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 9 of 78 pages

Question 81. If there are 10 apples, 10 oranges and 10 mangoes in a fruit basket, then the probability of not picking a mango is

\frac{1}{3}

True
False
0.94
0.98

Discuss Question

Answer: Option B. -> False
:
B
Number of mangoes = 10
Total number of fruits = 10 + 10 + 10 = 30
Probability of picking a mango =

\frac{Number of mangoes}{Total number of fruits}

Probability of picking a mango =

\frac{10}{30} = \frac{1}{3}

Probability of not picking a mango =

1 - \frac{1}{3} = \frac{2}{3}

Question 82. If 20 students in a class failed an exam, the probability of a student selected at random getting passed in an exam is 0.6. The number of students in the class is = ___ .

Discuss Question

:
Let the number of students in a class be

x

.
The probability of students passing an exam = 0.6
Number of students who failed an exam = 20
Number of students who passed an exam

= x - 20

Probability of a student passing an exam

= \frac{Number of students who passed an exams}{Number of students in a class}

0.6 = \frac{x - 20}{x}

0.6 x = x - 20

0.4 x = 20

x = 50

The number of students in the class is 50.

Question 83. An action which results in one or several outcomes is called as a _______.

outcome
trial
sample space
none of the above

Discuss Question

Answer: Option B. -> trial
:
B
.

Question 84. A die is rolled twice. Find the probability that 5 will come up both the times.

136
1836
636
436

Discuss Question

Answer: Option A. -> 136
:
A
Number of all possible outcomes = (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3), (2,4),(2,5),(2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4),(6,5) and (6,6).
The number of all possible outcomes = 36.
There is only one case (5,5) when 5 comes up both the times.

Probability that 5 will come up both the times = \frac{Number of times 5 comes up both the times}{Number of all possible outcomes}

∴

P (5 will come up both the times) =

\frac{1}{36}

Question 85. A coin was flipped 100 times. What is the maximum probability of occurrence of tail?

0.5
1
0
0.75

Discuss Question

Answer: Option B. -> 1
:
B
A coin is tossed 100 times and consider the case when you only get tail 100 times and no head. So, in all the trials you will have atail.
Probability of getting a tail =

\frac{Number of times tails appear}{Number of times a coin is tossed}

Probability of getting a tail =

\frac{100}{100} = 1

Question 86. A coin was tossed 1500 times and out of these, 100 times, the coin stuck somewhere so the result was not counted. The frequency of occurrence of a head was 700. Find the probability of occurrence of a tail.

0.46
0.5
0.3
1

Discuss Question

Answer: Option B. -> 0.5
:
B
A coin was tossed 1500 times but 100 times result was not counted.
So, the total number of times result counted = 1400
Number of times head comes = 700
Number of times tail comes = 1400 - 700 = 700
Probability of occurrence of tail

= \frac{Number of times tail comes}{Total number of times result counted}

Probability of occurrence of tail

= \frac{700}{1400}

Probability of occurrence of tail = 0.5

Question 87. When a coin is tossed, what is the probability of occurrence of "head" as the outcome?

Discuss Question

Answer: Option D. -> 12
:
D
When you flip a coin, there are two outcomes either head or tail. So the probability of occurrence of the head is the same as the probability of occurrence of the tail that is

\frac{1}{2}

Question 88. What will be the probability that two friends have the same birthday date in a normal year ?

173
1365
7365
12365

Discuss Question

Answer: Option B. -> 1365
:
B
Number of days in a year = 365
Therefore, total possible outcomes = 365
If they have birthday on same day, then number of favourable outcomes =1
Therefore, required probability

= \frac{1}{365}

Question 89. A goldsmith can choose between 3 metals A, B or C to make an artifact. What is the probability that he chooses C?

Discuss Question

Answer: Option A. -> 13
:
A
Since the goldsmith can choose amongst A, B and C, he has 3 possibilities of which the favorable outcome is choosing C. Therefore, probability that he chooses C is

\frac{1}{3}

Question 90. A man is known to speak truth is 75% cases. If he throws an unbiased die and tells his friend that it is a six, then the probability that it is actually a six, is

Discuss Question

Answer: Option D. -> 38
:
D
Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have