CA Foundation Exam  >  CA Foundation Tests  >  Test: Probability And Expected Value By Mathematical Expectation- 4 - CA Foundation MCQ

Test: Probability And Expected Value By Mathematical Expectation- 4 - CA Foundation MCQ


Test Description

30 Questions MCQ Test - Test: Probability And Expected Value By Mathematical Expectation- 4

Test: Probability And Expected Value By Mathematical Expectation- 4 for CA Foundation 2024 is part of CA Foundation preparation. The Test: Probability And Expected Value By Mathematical Expectation- 4 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Probability And Expected Value By Mathematical Expectation- 4 MCQs are made for CA Foundation 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Probability And Expected Value By Mathematical Expectation- 4 below.
Solutions of Test: Probability And Expected Value By Mathematical Expectation- 4 questions in English are available as part of our course for CA Foundation & Test: Probability And Expected Value By Mathematical Expectation- 4 solutions in Hindi for CA Foundation course. Download more important topics, notes, lectures and mock test series for CA Foundation Exam by signing up for free. 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 for CA Foundation Exam | Download free PDF with solutions
Test: Probability And Expected Value By Mathematical Expectation- 4 - 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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 1

 

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 2

Probability of the sample space is

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 3

Sum of all probabilities is equal to

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 3
  • In probability theory, the sum of all possible outcomes of a random event equals 1.
  • This represents the certainty that one of the outcomes will occur.
  • For example, when flipping a coin, the outcomes are heads or tails, and their probabilities add up to 1.
  • Options like 0, 1/2, or 3/4 do not represent the total probability of all outcomes.
  • Therefore, the correct answer is 1, indicating total certainty.
Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 4

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - 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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 6

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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 6

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 7

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 8

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 9

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 10

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 11

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 12

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 13

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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 13

To find the probability of obtaining one head and one tail when two unbiased coins are tossed, follow these steps:

  1. Determine the total number of possible outcomes:

    When two unbiased coins are tossed, each coin can land as either heads (H) or tails (T). The possible outcomes are:

    • HH
    • HT
    • TH
    • TT

    Thus, there are a total of 4 possible outcomes.

  2. Determine the favorable outcomes:

    The favorable outcomes for getting one head and one tail are:

    • HT
    • TH

    There are 2 favorable outcomes.

  3. Calculate the probability:

    The probability is the number of favorable outcomes divided by the total number of possible outcomes:

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 14

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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 14
  • When tossing two unbiased coins, each coin has two outcomes: heads (H) or tails (T).
  • The possible outcomes for two coins are: HH, HT, TH, TT.
  • Among these outcomes, only one is both tails (TT).
  • Thus, the probability of getting both tails is the number of favorable outcomes (1) divided by the total outcomes (4): 1/4.
  • The correct answer is 1/4.
Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 15

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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 15
  • When two unbiased coins are tossed, the possible outcomes are: HH, HT, TH, TT.
  • Out of these outcomes, HH, HT, and TH have at least one head.
  • Only TT has no heads.
  • So, there are 3 favorable outcomes (at least one head) out of 4 total outcomes.
  • The probability is 3/4, which corresponds to option 3.
Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 16

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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 16

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)

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 17

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - 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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 19

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 20

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 21

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 22

Which of the following relation is true ?

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 23

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - 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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 25

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 26

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 27

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 28

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

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 28

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

Test: Probability And Expected Value By Mathematical Expectation- 4 - 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

Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 30

If P (A)= 1/2, P(B)= 1/2, the events A & B are

Detailed Solution for Test: Probability And Expected Value By Mathematical Expectation- 4 - Question 30

Given:


  • P(A) = 1/2
  • P(B) = 1/2

The events A and B are:


  • Equally likely

Explanation:


  • Probability of both events A and B is 1/2.
  • This means each event has an equal chance of occurring.
  • Hence, events A and B are equally likely.

Correct answer: A

View more questions
Information about Test: Probability And Expected Value By Mathematical Expectation- 4 Page
In this test you can find the Exam questions for Test: Probability And Expected Value By Mathematical Expectation- 4 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Probability And Expected Value By Mathematical Expectation- 4, EduRev gives you an ample number of Online tests for practice

Top Courses for CA Foundation

Download as PDF

Top Courses for CA Foundation