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Test: Probability And Expected Value By Mathematical Expectation- 4

40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Probability And Expected Value By Mathematical Expectation- 4

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Attempt Test: Probability And Expected Value By Mathematical Expectation- 4 | 40 questions in 40 minutes | Mock test for CA Foundation preparation | Free important questions MCQ to study Quantitative Aptitude for CA CPT for CA Foundation Exam | Download free PDF with solutions

QUESTION: 1

If probability of drawing a spade from a well-shuffled pack of playing cards is ¼ then the probability that of the card drawn from a well-shuffled pack of playing cards is ‘not a spade’ is

A.
1
B.
½
C.
¼
D.
¾

Solution:

QUESTION: 2

Probability of the sample space is

A.
0
B.
½
C.
1
D.
none

Solution:

QUESTION: 3

Sum of all probabilities is equal to

A.
0
B.
½
C.
¾
D.
1

Solution:

QUESTION: 4

Let a sample space be S = {X₁, X₂, X₃} which of the fallowing defines probability space on S ?

A.
P(X₁)= ¼ , P(X₂)= 1/₃ , P(X₃)= 1/₃
B.
P(X₁₎= 0, P(X₂)= 1/₃, P(X₃)= 2/₃
C.
P(X₁)= 2/₃ , P(X₂)= 1/₃ , P(X₃)= 2/₃
D.
none

Solution:

QUESTION: 5

Let P be a probability function on S = {X₁ , X₂ , X₃} if P(X₁)= ¼ and P(X₃) = 1/₃ then P (X₂) is equal to

A.
5/12
B.
7/12
C.
3/4
D.
none

Solution:

QUESTION: 6

The chance of getting a sum of 10 in a single throw with two dice is

A.
10/36
B.
1/12
C.
5/36
D.
none

Solution:

Option (b) 1 /2 is correct.

Explanation:-

{ There are a total of 36 combinations on throw of two dice.

The sum of 10 could be obtained as :

(6+4, 4+6, 5+5) that is in 3 ways.

Therefore, the required probability here is :- 3 / 36 = 1/ 12
=> 1/ 12

QUESTION: 7

The chance of getting a sum of 6 in a single throw with two dice is

A.
3/36
B.
4/36
C.
6/36
D.
5/36

Solution:

QUESTION: 8

P (B/A) defines the probability that event B occurs on the assumption that A has happened

A.
Yes
B.
no
C.
both
D.
none

Solution:

QUESTION: 9

The complete group of all possible outcomes of a random experiment given an ________ set of events.

A.
mutually exclusive
B.
exhaustive
C.
both
D.
none

Solution:

QUESTION: 10

When the event is ‘certain’ the probability of it is

A.
0
B.
1/2
C.
1
D.
none

Solution:

QUESTION: 11

The classical definition of probability is based on the feasibility at subdividing the possible outcomes of the experiments into

A.
mutually exclusive and exhaustive
B.
mutually exclusive and equally likely
C.
exhaustive and equally likely
D.
mutually exclusive,exhaustive and equally likely cases

Solution:

QUESTION: 12

Two unbiased coins are tossed. The probability of obtaining ‘both heads’ is

A.
¼
B.
2/4
C.
¾
D.
none

Solution:

QUESTION: 13

Two unbiased coins are tossed. The probability of obtaining one head and one tail is

A.
¼
B.
2/4
C.
¾
D.
none

Solution:

QUESTION: 14

Two unbiased coins are tossed. The probability of obtaining both tail is

A.
2/4
B.
3/4
C.
¼
D.
none

Solution:

QUESTION: 15

Two unbiased coins are tossed. The probability of obtaining at least one head is

A.
¼
B.
2/4
C.
¾
D.
none

Solution:

QUESTION: 16

When 3 unbiased coins are tossed. The probability of obtaining 3 heads is

A.
2/4
B.
¼
C.
¾
D.
1/8

Solution:

Correct Answer :- d

Explanation :

Total no. Of outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT

No. Outcomes of all heads: 1

Prob. Of all heads = no. Of outcomes of all heads/ total no. Of outcomes

i.e (1/8)

QUESTION: 17

When unbiased coins are tossed. The probability of getting both heads or both tails is

A.
½
B.
¾
C.
¼
D.
none

Solution:

QUESTION: 18

Two dice with face marked 1, 2, 3, 4, 5, 6 are thrown simultaneously and the points on the dice are multiplied together. The probability that product is 12 is

A.
4/36
B.
5/36
C.
12/36
D.
none

Solution:

QUESTION: 19

A bag contain 6 white and 5 black balls. One ball is drawn. The probability that it is white is

A.
5/11
B.
1
C.
6/11
D.
1/11

Solution:

QUESTION: 20

Probability of occurrence of at least one of the events A and B is denoted by

A.
P(AB)
B.
P(A+B)
C.
P(A/B)
D.
none

Solution:

QUESTION: 21

Probability of occurrence of A as well as B is denoted by

A.
P(AB)
B.
P(A+B)
C.
P(A/B)
D.
none

Solution:

QUESTION: 22

Which of the following relation is true ?

A.
P(A)– P(A^C)= 1
B.
P(A)+ P(A^C)= 1
C.
P(A) P(A^C)= 1
D.
none

Solution:

QUESTION: 23

If events A and B are mutually exclusive, the probability that either A or B occurs is given by

A.
P(A+B)= P(A)– P(B)
B.
P(A+B)(A)+ P(B)– P(AB)
C.
P (A+B)= P(A)– P(B)+ P(AB)
D.
P(A+B)= P(A)+ P(B)

Solution:

QUESTION: 24

The probability of occurrence of at least one of the 2 events A and B (which may not be mutually exclusive) is given by

A.
P(A+B)= P(A)– P(B)
B.
P(A+B)= P(A)+ P(B)– P(AB)
C.
P(A+B)= P(A)– P(B)+ P(AB)
D.
P(A+B)= P(A+B)= P(A)+P (B)

Solution:

QUESTION: 25

If events A and B are independent, the probability of occurrence of A as well as B is given by

A.
P(AB)= P(A/B)
B.
P(AB)= P(A)
C.
P(AB)= P(A)P(B)
D.
None

Solution:

QUESTION: 26

For the condition P(AB)= P(A)P(B) two events A and B are said to be

A.
dependent
B.
independent
C.
equally like
D.
none

Solution:

QUESTION: 27

The conditional probability of an event B on the assumption that another event A has actually occurred is given by

A.
P(B/_A)= P(AB)/P(A)
B.
P(A/_B)= P(AB)/ P(B)
C.
P(B/_A)= P(AB)
D.
P(A/_B)= P(AB)/ P(A)P(B)

Solution:

QUESTION: 28

In a throw of coin what is the probability of getting tails.

A.
1
B.
2
C.
1/2
D.
0

Solution:

Total cases = [H,T] - 2
Favourable cases = [T] -1
So probability of getting tails = 1/2

QUESTION: 29

Demand of products per day for three days are 21, 19, 22 units and their respective probabilities are 0.29, 0.40, 0.35. profit per unit is $0.50 then expected profits for three days are

A.
21, 19, 22
B.
21.5, 19.5, 22.5
C.
0.29, 0.40, 0.35
D.
3.045, 3.8, 3.85

Solution:

QUESTION: 30

If P (A)= 1, P(B)= 1, the events A & B are

A.
not equally likely
B.
mutually exclusive
C.
equally likely
D.
none

Solution:

QUESTION: 31

If events A and B are independent then

A.
A^Cand B^Care dependent
B.
A^C and B are dependent
C.
A and B^Care dependent
D.
A^Cand B^Care also independent

Solution:

QUESTION: 32

A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is a card of spade or an ace.

A.
4/13
B.
10/13
C.
26/13
D.
8/13

Solution:

There are 52 cards in total.

Hence, total Outcomes =52

Probability of the card drawn is a card of spade or an Ace:

Total cards of Spade =13

Total Aces =4

Number of Aces of Spade =1

Therefore,

Probability of the card drawn is a card of spade or an Ace:

13/52 + 4/52 - 1/52 = 16/52 = 4/13

QUESTION: 33

When a die is tossed, the sample space is

A.
S =(1,2,3,4,5)
B.
S =(1,2,3,4)
C.
S =(1,2,3,4,5,6)
D.
none

Solution:

QUESTION: 34

Find the expectation of a random variable X?

A.
0.5
B.
3.5
C.
2.5
D.
1.5

Solution:

QUESTION: 35

If events A and B are independent and P(A)= 2/3 , P(B)= 3/5 then P(A+B)is equal to

A.
13/15
B.
6/15
C.
1/15
D.
none

Solution:

QUESTION: 36

The expected no. of head in 100 tosses of an unbiased coin is

A.
100
B.
50
C.
25
D.
none

Solution:

QUESTION: 37

A and B are two events such that P(A)= 1/₃, P(B) = ¼, P(A+B)= 1/₂, than P(B/_A) is equal to

A.
¼
B.
1/3
C.
1/2
D.
none

Solution:

QUESTION: 38

Probability mass function is always

A.
0
B.
greater than 0
C.
greater than equal to 0
D.
less than 0

Solution:

QUESTION: 39

What is the probability of getting an even number when a dice is thrown?

A.
1/6
B.
1/3
C.
1/2
D.
1/4

Solution:

The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6)
So, n (S) = 6 E is the event of getting an even number.
So, n (E) = 3
Probability of getting an even number P (E) = Total number of favorable outcomes/Total number of outcomes
n(E) / n(S) = 3/6 = ½

QUESTION: 40

The sum of probability mass function is equal to