Probability MCQ (Multiple Choice Questions)

1. What is the probability of an impossible event?
a) 1
b) 0
c) Insufficient data
d) Not defined
View Answer

Answer: b
Explanation: If the probability of an event is 0, then it is called as an impossible event.

2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a) 34
b) 16
c) 13
d) 12
View Answer

Answer: a
Explanation: Total outcomes = (HH, HT, TH, TT)
Favorable outcomes = (TT, HT, TH)
At most one head refers to a maximum one head,
Therefore, probability = 34.
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3. Who provided the definition for probability?
a) Archimedes
b) Einstein
c) Euclid
d) Simon Laplace
View Answer

Answer: d
Explanation: The definition of probability was given by Pierre Simon Laplace in the year 1795. Probability can be defined as the ratio of the number of favorable outcomes to the total number of possible outcomes.

4. A bag contains 5 green and 3 blue balls. Two balls are picked at random. What is the probability that both are of the same colour?
a) \(\frac{C_1^5 * C_1^3}{C_2^8}\)
b) 0.5
c) \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\)
d) \(\frac{C_2^5 * C_2^3}{C_2^8}\)
View Answer

Answer: c
Explanation:Total no.of balls = 5G+3B = 8
No.of ways in which 2 balls can be picked = 8C2
Probability of picking both balls as green = 5C2 /8C2
Probability of picking both balls as blue = 3C2 /8C2
∴ required probability \(\frac{C_2^5}{C_2^8}+\frac{C_2^3}{C_2^8}\).

5. Company A produces 10% defective products, Company B produces 20% defective products and C produces 5% defective products. If choosing a company is an equally likely event, then find the probability that the product chosen is defective.
a) 0.11
b) 0.21
c) 0.22
d) 0.12
View Answer

Answer: d
Explanation: Let A be the event of selecting a defective item. Let Ei be the event of selecting a company. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3)
\( = (\frac{1}{3})(\frac{10}{100}) + (\frac{1}{3})(\frac{20}{100}) + (\frac{1}{3})(\frac{5}{100}) \)
\( = \frac{0.35}{3} = 0.12. \)

6. A problem in mathematics is given to three students A, B and C. If the probability of A solving the problem is 12 and B not solving it is 14. The whole probability of the problem being solved is 6364 then what is the probability of solving it?
a) 12
b) 78
c) 164
d) 18
View Answer

Answer: b
Explanation:
Let A be the event of A solving the problem
Let B be the event of B solving the problem
Let C be the event of C solving the problem
Given P(a) = 12, P(~B) = 14 and P(A ∪ B ∪ C) = 63/64

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We know P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)

= 1 – P(ABC)

= 1 – P(A) P(B) P(C)

Let P(C) = p
ie 6364 = 1 – (12)(14)(p)

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= 1 – p8
⇒ P =1/8 = P(C)
⇒P(C) = 1 – P = 1 – 18 = 78.

7. Which of the following is a table with all possible values of a random variable and its corresponding probabilities?
a) Probability Density Function
b) Probability Mass Function
c) Probability Distribution
d) Cumulative distribution function
View Answer

Answer: c
Explanation: The given statement is the definition of a probability distribution.

8. The probability that at least one of the events Q and R occur is 0.6. If Q and R have probability of occurring together as 0.2, then P(Q) + P(R) is?
a) 1.2
b) 0.8
c) 0.4
d) Indeterminate
View Answer

Answer: a
Explanation: Given : P(Q ∪ R) = 0.6, P(Q ∩ R) = 0.2
P(Q ∪ R) + P(Q ∩ R) = P(Q) + P(R)
2 – (P(Q ∪ R) + P(Q ∩ R)) = 2 – (P(Q) + P(R))
= (1 – P(Q)) + (1 – P(R))
2 – (0.6 + 0.2) = P(Q) + P(R)
P(Q) + P(R) = 2 – 0.8
= 1.2

9. Let there be two newly launched phones A and B. The probability that phone A has good battery life is 0.7 and the probability that phone B has a good battery life is 0.8. Then find the probability that a phone has good battery life.
a) 0.45
b) 0.85
c) 0.75
d) 0.65
View Answer

Answer: c
Explanation: Let A be the event thtat a phome has a good battery life and Ei be the event that a phone is selected. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( = (\frac{1}{2})(0.7) + (\frac{1}{2})(0.8) \)
= 0.75.

10. Suppose box A contains 4 green and 5 black coins and box B contains 6 green and 3 black coins. A coin is chosen at random from the box A and placed in box B. Finally, a coin is chosen at random from among those now in box B. What is the probability a blue coin was transferred from box A to box B given that the coin chosen from box B is green?
a) 1429
b) 1529
c) 710
d) 12
View Answer

Answer: b
Explanation: Let E represent the event of moving a black coin from box A to box B. We want to find the probability of a black coin which was moved from box A to box B given that the coin chosen from B was green. The probability of choosing a green coin from box A is P(R) = 79 and the probability of choosing a black coin from box A is P(B) = 59. If a green coin was moved from box A to box B, then box B has 7 green coins and 3 black coins. Thus the probability of choosing a green coin from box B is 710. Similarly, if a black coin was moved from box A to box B, then the probability of choosing a green coin from box B is 610.
Hence, the probability that a black coin was transferred from box A to box B given that the coin chosen from box B is green is given by
\(P(E|R)=\frac{P(R|E)*P(E)}{P(R)}\)
=\(\frac{(\frac{6}{10})*(\frac{5}{9})}{(\frac{7}{10})*(\frac{4}{9})+(\frac{6}{10})*(\frac{5}{9})}\)
= 1529.
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11. Which of the following mentioned standard Probability density functions is applicable to discrete Random Variables?
a) Rayleigh Distribution
b) Exponential Distribution
c) Poisson Distribution
d) Gaussian Distribution
View Answer

Answer: c
Explanation: Poisson Distribution is applicable to discrete Random Variables.

12. In a discrete probability distribution, the sum of all probabilities is always?
a) 1
b) 0
c) Undefined
d) Infinite
View Answer

Answer: a
Explanation: It is based on the basic axiom of probability distribution.

13. If the probability of hitting the target is 0.3, find mean and variance.
a) 0.3, 0.16
b) 0.3, 0.21
c) 0.6, 0.16
d) 0.6, 0.24
View Answer

Answer: b
Explanation: p = 0.3
q = 1-p
= 1-0.3 = 0.67
Therefore, mean = p = 0.3 and
Variance = pq = (0.3) (0.7) = 0.21.

14. If ‘p’, ‘q’ and ‘n’ are probability of success, failure and number of trials respectively in a Binomial Distribution, what is its Standard Deviation?
a) (np)2
b) \(\sqrt{np}\)
c) \(\sqrt{npq}\)
d) \(\sqrt{pq}\)
View Answer

Answer: c
Explanation: The variance (V) for a Binomial Distribution is given by V = npq
Standard Deviation = \(\sqrt{variance} = \sqrt{npq}\).

15. Find the Expectation of a Hypergeometric Distribution such that the probability that a 4-trial hypergeometric experiment results in exactly 2 successes, when the population consists of 16 items.
a) 1/3
b) 1/8
c) 1/4
d) 1/2
View Answer

Answer: d
Explanation: In Hypergeometric Distribution the Mean or Expectation E(X) is given as
E(X) = n*k /N
Here n = 4, k = 2, N = 16.
Hence E (X) = 1/2.
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16. What is the probability of a sure event?
a) 1
b) 0
c) 1/4
d) 1/2
View Answer

Answer: a
Explanation: Sure event has all outcomes in favour i.e. number of outcomes in favour equal to the total number of possible outcomes.
And probability is ratio of number of outcomes in favour to the total number of possible outcomes.
So, probability of sure event will be 1.

17. If P(C) = 5/13, P(D) = 7/13 and P(C∩D) = 3/13, evaluate P(C|D).
a) 2/7
b) 3/5
c) 3/7
d) 1/7
View Answer

Answer: c
Explanation: We know that P(C|D) = P(C∩D) / P(D). (By formula for conditional probability)
Which is equivalent to (3/13) / (7/13), hence the value of P(C|D) = 3/7.

18. In a bucket there are 5 blue, 15 yellow, and 25 orange balls. If the ball is picked up randomly, find the probability that it is neither yellow nor blue?
In a bucket there are 5 purple, 15 grey and 25 green balls. If the ball is picked up randomly, find the probability that it is neither grey nor purple?
a) \(\frac{51}{43}\)
b) \(\frac{2}{7}\)
c) \(\frac{5}{9}\)
d) \(\frac{12}{13}\)
View Answer

Answer: c
Explanation: If the ball is neither yellow nor blue then it must be orange. There are 45 balls in total of which 25 are orange and so the probability of picking a purple ball is \(\frac{25}{45} = \frac{5}{9}\).

19. The annual salaries of workers in a manufacturing factory are normally distributed with a mean of Rs. 48,000 and a standard deviation of Rs. 1500. Find the probability of workers who earn between Rs. 35,000 and Rs. 52,000.
a) 20%
b) 42.1%
c) 64%
d) 76.2%
View Answer

Answer: b
Explanation: For x = 45000, z = -2 and for x = 52000, z = 0.375. Now, area between z = -2 and z = 0.375 is equal to 0.421 or 42.1% earn between Rs. 45,000 and Rs. 52,000.

20. What is the probability of getting a number greater than 6 on dice?
a) 1
b) 1/3
c) 1/2
d) 0
View Answer

Answer: d
Explanation: Since number greater than 6 cannot appear on dice so probability of getting a number greater than 6 is zero.

21. How many outcomes are possible when drawing a card from deck of cards?
a) 1
b) 13
c) 52
d) 26
View Answer

Answer: c
Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K.

22. A number is selected from the first 20 natural numbers. Find the probability that it would be divisible by 3 or 7?
a) \(\frac{7}{20}\)
b) \(\frac{12}{37}\)
c) \(\frac{24}{67}\)
d) \(\frac{19}{46}\)
View Answer

Answer: a
Explanation: Let X be the event that the number selected would be divisible by 3 and Y be the event that the selected number would be divisible by 7. Then A u B denotes the event that the number would be divisible by 3 or 7. Now, X = {3, 9, 12, 15, 18} and Y = {7, 14} whereas S = {1, 2, 3, …,20}. Since A n B = Null set, so that the two events A and B are mutually exclusive and as such we have,
P(A u B) = P(A) + P(B) ⇒ P(A u B) = \(\frac{5}{20} + \frac{2}{20}\)
Therefore, P(A u B) = \(\frac{7}{20}\).

23. A programmer has a 95% chance of finding a bug every time she compiles his code, and it takes her three hours to rewrite the code every time she discovers a bug. Find the probability that she will finish her program by the end of her workday. (Assume that a workday is 9 hours)
a) 44%
b) 76%
c) 28%
d) 37%
View Answer

Answer: c
Explanation: In this instance, a success is a bug-free compilation, and a failure is the discovery of a bug. The programmer needs to have 0, 1 or 2 failures, so her probability of finishing the program is: P(X=0) + P(X=1) + P(X=2) = (0.95)0(0.1) + (0.95)1(0.1) + (0.95)2(0.1) = 0.28% = 28%.

24. In a Poisson Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by?
a) m = p
b) m = np(1-p)
c) m = (np)2
d) m = np
View Answer

Answer: d
Explanation: For a discrete probability function, the mean value or the expected value is given by
Mean(μ)=\(\sum_{x=0}^n x p(x)\)
For Poisson Distribution P(x)=\(\frac{e^{-m}m^x}{x!}\) substitute in above equation and solve to get µ = m = np.

25. If E and F are two events such that P(a) = 0.2, P(b) = 0.6 and P(E/F) = 0.2 then the value of P(E /~F) is ___________
a) 0.8
b) 0.2
c) 13
d) 0.5
View Answer

Answer: b
Explanation: For independent events,
P(E/~F) = P(a) = 0.2.

26. A jar contains ‘y’ black colored balls and ‘x’ yellow colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is black and second one is yellow?
a) \(\frac{y^2-y}{x^2+y^2+2xy-(x+y)}\)
b) \(\frac{xy-y}{x^2+y^2+2xy-(x-y)}\)
c) \(\frac{xy-y}{x^2+y^2+2xy-(x+y)}\)
d) \(\frac{xy}{x^2+y^2+2xy-(x+y)}\)
View Answer

Answer: d
Explanation: Number of black balls = y
Number of Yellow balls = x
Total number of balls = x + y
Probability of Black ball first = y / x + y
No. of balls remaining after removing one = x + y – 1
Probability of pulling secondball as Yellow=\(\frac{x}{x+y-1}\)
Required porbability=\(\frac{y}{(x+y)}\frac{x}{(x+y-1)}=\frac{xy}{x^2+y^2+2xy-(x+y)}\)

27. The probability that person A completes all the tasks assigned is 50% and that of person B is 20%. Find the probability that all the tasks are completed.
a) 0.35
b) 0.45
c) 0.15
d) 0.25
View Answer

Answer: a
Explanation: Let A be the event that all tasks are completed and Ei is the event that a person is selected.
Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\( =(\frac{1}{2})(\frac{50}{100}) + (\frac{1}{2})(\frac{20}{100}) \)
= 0.35

28. Find the number of ways of arranging the letters of the words DANGER, so that no vowel occupies odd place.
a) 144
b) 96
c) 36
d) 48
View Answer

Answer: a
Explanation: The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

29. If the probability that a bomb dropped from a place will strike the target is 70% and if 10 bombs are dropped, find mean and variance?
a) 4, 1.6
b) 0.4, 0.16
c) 7, 2.1
d) 0.7, 0.21
View Answer

Answer: c
Explanation: Here, p = 70% = 0.7 and q = 1-p = 30% = 0.4 and n = 10
Therefore, mean = np = 7
Variance = npq = (10)(0.7)(0.3)
= 2.1.

30. Consider Jack draws 3 cards from a pack of 52 cards. What is the probability of getting no kings?
a) 0.8726
b) 0.7862
c) 0.8762
d) 0.7826
View Answer

Answer: d
Explanation: The Random Experiment follows hypergeometric distribution with,
N = 52 since there are 52 cards in a deck.
k = 4 since there are 4 kings in a deck.
n = 3 since we randomly select 3 cards from the deck.
x = 0 since we want no kings.
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
h(0; 52, 4, 3) = [4C0] [48C3] / [52C3]
h(0; 52, 4, 3) = 0.7826.

31. Which of the following cannot be the value of probability?
a) 1/2
b) 0
c) -1
d) 1
View Answer

Answer: c
Explanation: Since probability is ratio of number of outcomes in favour to the total number of possible outcomes. So, it cannot be negative.

32. A coin is tossed up 4 times. The probability that tails turn up in 3 cases is?
a) \(\frac{1}{4} \)
b) \(\frac{1}{6} \)
c) \(\frac{1}{3} \)
d) \(\frac{1}{2} \)
View Answer

Answer: d
Explanation: p=0.5 (Probability of tail)
q=1-0.5=0.5
n=4 and x is binomial variate.
P (X=x) = nCx px qn-x.
P (X=3) = 4C3 (0.5)3 = 12.

33. Suppose 5 men out of 100 men and 10 women out of 250 women are colour blind, then find the total probability of colour blind people. (Assume that both men and women are in equal numbers.)
a) 0.5
b) 0.05
c) 0.045
d) 0.45
View Answer

Answer: c
Explanation: Let A be the event of selecting a colour blind person and Ei be the event of selecting a person. Then,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2)
\(= (\frac{1}{2})(\frac{5}{100}) + (\frac{1}{2})(\frac{10}{250}) \)
= 0.045.

34. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 14
b) 34
c) 16
d) 12
View Answer

Answer: a
Explanation:S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
All 3 same outcomes mean either all head or all tail
Total outcomes = 8
Favourable outcomes = {HHH, TTT} = 2
∴ required probability = 28 = 14.

35. A problem is given to 5 students A, B, C, D, E. If the probability of solving the problem individually is 1/2, 1/3, 2/3, 1/5, 1/6 respectively, then find the probability that the problem is solved.
a) 0.57
b) 0.27
c) 0.47
d) 0.37
View Answer

Answer: d
Explanation: Let M be the event that the problem is solved. Let Ei be the event that a student is chosen. Then,
P(M) = P(E1) P(M|E1) + P(E2) P(M|E2) + P(E3) P(M|E3) + P(E4) P(M|E4) + P(E5) P(M|E5)
\( = (\frac{1}{5})(\frac{1}{2}) + (\frac{1}{5})(\frac{1}{3}) + (\frac{1}{5})(\frac{2}{3}) + (\frac{1}{5})(\frac{1}{5}) + (\frac{1}{5})(\frac{1}{6}) \)
= 0.37.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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