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Test: Probability - Commerce MCQ


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25 Questions MCQ Test - Test: Probability

Test: Probability for Commerce 2024 is part of Commerce preparation. The Test: Probability questions and answers have been prepared according to the Commerce exam syllabus.The Test: Probability MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Probability below.
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Test: Probability - Question 1

The conditional probability of an event E, given the occurrence of the event F is given by

Detailed Solution for Test: Probability - Question 1

The conditional probability of an event E, given the occurrence of the event F is given by :

Test: Probability - Question 2

A coin is tossed three times, if E : head on third toss , F : heads on first two tosses. Find P(E|F)

Detailed Solution for Test: Probability - Question 2

S ={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}
E = {HHH,HTH,THH,TTH}
F = {HHH,HHT}
E∩F = {HHH}

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Test: Probability - Question 3

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Detailed Solution for Test: Probability - Question 3

Required probability : P(BB) = P(B) X P(B/B) ……………{B means black card}

Test: Probability - Question 4

A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Detailed Solution for Test: Probability - Question 4

Let 
E1 and E2 are events that a person has disease and the person has not disease.

Let A  = event that the test result is positive






Test: Probability - Question 5

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is

Detailed Solution for Test: Probability - Question 5

The possible values of X are 0 , 1 and 2 .
As we know that pack of cards contains 4 aces.

Therefore , the probability distribution of X is : 

Therefore required value of
E(X) is = 0+ 384/2652 + 24/2652 = 408/2652 = 2/13.

Test: Probability - Question 6

The conditional probability of an event E, given the occurrence of the event F

Detailed Solution for Test: Probability - Question 6

As the probability of any event always lies between 0 and 1. Therefore , 0 ≤ P (E|F) ≤ 1.

Test: Probability - Question 7

A coin is tossed three times, if E : at least two heads , F : at most two heads. Find P(E|F)

Detailed Solution for Test: Probability - Question 7


Test: Probability - Question 8

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Detailed Solution for Test: Probability - Question 8

Total oranges = 15., Good Oranges = 12. Let G stands for a good orange. Therefore , Required Probability = P(GGG) = P(G).P(G/G).P(G/GG) 

Test: Probability - Question 9

There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

Detailed Solution for Test: Probability - Question 9

Let 
E1 , E2 and E3 and are events of selection of a two headed coin , biased coin and unbiased coin respectively. 

Let A = event of getting head. 




 

Test: Probability - Question 10

Which of the following conditions do Bernoulli trials satisfy?

Detailed Solution for Test: Probability - Question 10

Bernoulli trials satisfies the finite number of independent trials .

Test: Probability - Question 11

The conditional probability of an event E’s complement E’, given the occurrence of the event F

Detailed Solution for Test: Probability - Question 11

As the total probability of an event is always 1 . therefore , P (E′|F) = 1 – P (E|F).

Test: Probability - Question 12

A coin is tossed three times, E : at most two tails , F : at least one tail. Find P(E|F)

Detailed Solution for Test: Probability - Question 12


Test: Probability - Question 13

Given that the events A and B are such that P(A) = 1/2 , P (A ∪ B) = 3/5 and P(B) = p. Find p if they are mutually exclusive

Detailed Solution for Test: Probability - Question 13

Test: Probability - Question 14

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Detailed Solution for Test: Probability - Question 14

Let 
E1 , E2 and E3 and are events of selection of a scooter driver , car driver and truck driver respectively. 



Test: Probability - Question 15

Which of the following conditions do Bernoulli trials satisfy?

Detailed Solution for Test: Probability - Question 15

Bernoulli trials are true only when Each trial has exactly two outcomes: success or failure and the probability of success or failure remains the same in each trial.

Test: Probability - Question 16

If E, F and G are events then P ((E ∪ F)|G) =

Detailed Solution for Test: Probability - Question 16

If E, F and G are events then P ((E ∪ F)|G) represents the conditional probability of the given event . therefore P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G) .

Test: Probability - Question 17

A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

Detailed Solution for Test: Probability - Question 17


Test: Probability - Question 18

Given that the events A and B are such that P(A) = 1/2 P (A ∪ B) = 3/5 and P(B) = p. Find p if they independent.

Detailed Solution for Test: Probability - Question 18

Since A and B are independent events. Therefore


Test: Probability - Question 19

A random variable is a real valued function whose domain is the.

Detailed Solution for Test: Probability - Question 19

A random variable is a real valued function whose domain is the sample space of a random experiment .

Test: Probability - Question 20

A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of 5 successes?

Detailed Solution for Test: Probability - Question 20

p = probability of success = 3/6 = ½ . q = probability of failure = 1 – p = 1 – ½ = ½ . let x be the number of successes , then x has the binomial distribution with : n = 6 , p = ½ , q = ½ . 

Test: Probability - Question 21

If E and F are events then P (E ∩ F) =

Detailed Solution for Test: Probability - Question 21

If E and F are events then P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0.

Test: Probability - Question 22

A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Detailed Solution for Test: Probability - Question 22

n(S)=36.
Let A = event of getting sum 8. 
= {(2,6),(3,5),(4,4),(5,3),(6,2),}
And B = event of getting a number less than 4 on red die.
={(1,1),(2,1),(3,1),(4,1),(5,1),(5,1), (1,2),(2,2),(3,2),(4,2),(5,2),(6,2), (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)}.

Test: Probability - Question 23

Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4. Find P(A ∩ B)

Detailed Solution for Test: Probability - Question 23

Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4 

Test: Probability - Question 24

Let X be a random variable assuming values x1, x2,....,xn with probabilities p1, p2, ...,pn, respectively such that . Mean of X denoted by μ is defined as

Detailed Solution for Test: Probability - Question 24

Let X be a random variable assuming values x1, x2,....,xn with probabilities p1, p2, ...,pn, respectively such that  Mean of X denoted by μ is defined as: 

Test: Probability - Question 25

A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of at least 5 successes?

Detailed Solution for Test: Probability - Question 25

p = probability of success = 3/6 = ½ .
q = probability of failure = 1 – p = 1 – ½ = ½ 
. let x be the number of successes , then x has the binomial distribution with :
n = 6 , p = ½ , q = ½ .


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