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Q1. A die is tossed twice. The probability of getting 1, 2, 3 or 4 on the first toss and 4, 5, or 6 on the second toss is:

• 1/2
• 2/5
• 1/3
• 1/9

Solution

Here, the events are independent.  

Q2. A die is rolled. If the outcome is an odd number then the probability that it is the number greater than 1 is

• 2/3
• 1/3
• 3/8
• 5/6

Solution

Sample space S={1,2,3,4,5,6} A=getting odd number={1,3,5} B=getting a number greater than 1={2,3,4,5,6} Required probability is P(B/A)=

Q3. If A and B are independent events, such that P(A)= 0.6, P(B) = 0.3, then P(A and not B) = ...

• 0.18
• 0.12
• 0.42
• 0.72

Solution

Since, A and B are independent events P(A and not B) =

Q4. If S is the sample space associated with a random experiment. Then a real valued function which assign to each element of S, a unique real number is called a …... 

• Real numbers
• Non-negative integers
• Random variable
• Random experiment

Solution

 If S is the sample space associated with a random experiment. Then a real valued function which assign to each element of S, a unique real number is called a random variable.  

Q5. In a class 25% of student appears for medical entrance, 35% for engineering and 15% for both. One student is selected at random. The probability that he appear for engineering, if it is known that he appears for medical is:   

• 3/5
• 3/7
• 5/7
• 1/4

Solution

 Here, P(A) = 25/100, P(B) = 35/100, = 15/100 Required probability = P(B/A)

Q6. If a random variable X assumes all possible values in a given interval, then it is called ………

• Discrete random variable
• Continuous random variable
• Random experiment
• Random variable

Solution

If a random variable X assumes all possible values in a given interval, then it is called continuous random variable

Q7. In a class 45% students read English, 30%read French and 20% read both English and French.One student is selected at random,the probability that he reads English if it is known that he reads French

• 1/3
• 2/3
• 5/6
• None of the Above

Solution

P(E)= then P(E/F)==

Q8. The mean number of tails in three tosses of a fair coin is:

• 3/2
• 3
• 6
• 1/6

Solution

Here, n = 3 and p(tail) = p(head) = ½ So, n = 3 and p = ½, then mean number of tails = np = 3(1/2) = 3/2

Q9. In a meeting, 60% of the members favour and 40% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X). 

• 0.6 and 0.24
• 0.6 and 0
• 0.4 and 0.24
• 0.6 and 0.44

Solution

Here, the distribution of random variable X is: P(X = 0) = 40/100 = 0.4 and P(X = 1) = 60/100 = 0.6 X 0 1 P(X) 0.4 0.6

Q10. If a random variable X assumes only a finite or countable infinite set of values, then it is called ………

• Discrete random variable
• Continuous random variable
• Random experiment
• Random variable

Solution

If a random variable X assumes only a finite or countable infinite set of values, then it is called discrete random variable.

Q11. The conditional probability P(E_i |A) is called a ……… probability of the hypothesis E_i.

• Hypothesis
• Posteriori
• Causes
• Beyes'

Solution

 The following terminology is generally used when Bayes' theorem is applied. The events E₁, E₂, ..., E_n are called hypotheses. The probability P(E_i) is called the priori probability of the hypothesis E_i . The conditional probability P(E_i |A) is called a posteriori probability of the hypothesis E_i.

Q12. If A and B are two events of sample space S associated with a random experiment, then occurrence of A provided B has already occurred is called conditional probability. It is denoted by

• P(A)P(B)
• P(A)/P(B)
• P(A/B)
• P(B/A)

Solution

If A and B are two events of sample space S associated with a random experiment, then occurrence of A provided B has already occurred is called conditional probability. It is denoted by P(A/B).

Q13. A class has 10 students whose ages are 15, 14, 16, 17, 19, 20, 16, 18, 20, and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. The standard deviation of X is:

• 2.11
• 4.41
• 4.45
• 17.5

Solution

Age 14 15 16 17 18 19 20 No. of student/s (X) 1 1 2 1 1 1 3 P(X) 0.1 0.1 0.2 0.1 0.1 0.1 0.3  

Q14. A random variable X has the following probability distribution: X 0 1 2 3 4 P(X) 0 k 2k 2k 3k Then k =?

• 1/8
• 0
• 8
• 4

Solution

S:, therefore 8k = 1   And k = 1/8

Q15. What is the probability of picking a spade from a normal pack of cards and rolling an odd number on a die?

• ¼  
• ½   
• 0
• 1/8

Solution

These are independent events, so the probabilities ¼ and ½ of the two respectively get multiplied.

Q16. In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

• 10^-1
• (1/2 ) ⁵
• (9/10) ⁵
• 9/10

Solution

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

Q17. A random variable X has the probability distribution: X 0 1 2 3 4 5 6 7 P(X) 0 0.1 0.2 0.2 0.3 0.01 0.02 0.17 Then the value of P(X > 4) is:

• 0.23
• 0.2
• 0.13
• 0.45

Solution

P(X > 4) = P(5) + P(6) + P(7) = 0.01 + 0.02 + 0.17 = 0.2

Q18. A bag contains 25 tickets numbered from 1 to 25. Two tickets are drawn one after another without replacement. The probability that both tickets will show even numbers is:

• 13/25
• 11/25
• 3/8
• 11/24

Solution

There are 12 even numbers from 1 to 25. Let A be the event of getting an even number on 1^st draw and B be the event of getting an even number in 2^nd draw.

Q19. Three coins are tossed. If at least two coins show head, the probability of getting one tail is:

• 1
• 1/3
• 2/3
• 3/4

Solution

Let S be the sample space, A be the event of getting one tail and B be the event of getting at least two heads, then  n(S) = 8, n(A) = 3, n(B) = 4, = 3 Required probability = P(A/B)

Q20. A die is tossed twice. The probability of getting 1, 2, 3 or 4 on the first toss and 4, 5, or 6 on the second toss is:

• 1/2
• 2/5
• 1/3
• 1/9

Solution

Let A be the event of getting 1, 2, 3 or 4 on the first dice and B be the event of getting 4, 5, or 6 on the second dice. Here, the events are independent.

Q21. Random variable is a real valued function whose domain is the sample space of a ……… and range is the set of ……...       

• Random variable, non-negative integers
• Real numbers, real numbers
• Random variable, real numbers
• Random experiment, real numbers

Solution

Random variable is a real valued function whose domain is the sample space of a random experiment and range is the set of real numbers.

Q22. Bayes theorem is applicable when the events are:

• pairwise disjoint
• exhaustive
• have nonzero probabilities
• pairwise disjoint, exhaustive and have nonzero probabilities

Solution

This is a prerequisite condition for Baye’s theorem.

Q23. The odds against a certain event are 5 to 2 and the odds in favour of another event independent of the former are 6 to 5. Find the probability that ‘at least one of the events’ happens:

• 2/7
• 6/ 11
• 12/77
• 52/77

Solution

P (E₁) = 2/7 P (E₂) = 6/11  P (at least one ) = 1 – P(none)                            

Q24. A and B take turn in throwing two dice;the first to throw 9 being awarded.The ratio of probabilities of their winning if A has the first throw is

• 7/8
• 9/8
• 8/7
• 9/10

Solution

When two dice are thrown then the sum will be 9 if we get (3,6),(4,5),(5,4),(6,3) so the probability of getting sum as 9 is 4/36=1/9 probability of not getting 9 is 8/9 If A starts the he can win  either in 1st chance or 3rd or 5th .....so on so the probability of winning of A is=   B can win either in 2nd chance or 4th or 6th so on the probability of winning of B is the required ratio is P(A)/P(B)=9/8

Q25.  The variance of probability distribution of number of doublets in three throws of a pair of dice is

• 4293/11664
• 4294/11664
• 4295/11664

Solution

Let p = throwing a doublet in a single throw of a pair of dice.q = not = throwing a doublet in a single throw of a pair of dice.p = 6 / 36  = 1/ 6  and q = 1 – p = 1 – 1 / 6 = 5 /6 .P( X = 0) = ³c₀ p⁰q³ = 1.q³ = (5/6)³ = 125 / 216.P( X = 1) = ³c₁ p¹q² = 3p.q² =3(1/6) (5/6)² = 75 / 216.P( X = 2) = ³c₂ p²q¹ = 3p².q¹ =3(1/6)² (5/6)¹ = 15 / 216P( X = 3) = ³c₃ p³q⁰ = 1.p³ = (1/6)³ = 1 / 216. The probability distribution table is given by E(X)=X*P(X)=0*125/216+1*75/216+2*15/216+3*1/216=118/216 E(X²)=X²*P(X)=0*125/216+1*75/216+4*15/216+9*1/216=144/216 VARIANCE Var(X)=E(X²)-[E(X)]²=144/216-(118/216)²=17180/46656=4295/11664

Q26. The number of adults living in homes on a randomly selected city block is described by the following probability distribution. Number of adults, x 1 2 3 4 or more Probability, P(x) 0.25 0.50 0.15 ??? What is the probability that 4 or more adults reside at a randomly selected home?

• .25
• .50
• .15
• .10

Solution

The sum of all the probabilities is equal to 1. Therefore, the probability that four or more adults reside in a home is equal to 1 - (0.25 + 0.50 + 0.15) or 0.10.

Q27. If A and B are two events such that P(A) = 0.3 and P(B) = 0.9 and P(B|A) = 0.6,then P(A|B) = ……

• 0.36
• 0.5
• 0.2
• 0.1

Solution

Q28. An urn contains 3 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. The possible values of X are:

• 0, 1 & 2
• 0, 1, 2 & 3
• 1, 2 & 3
• 1 & 2

Solution

Let X be the random variable denoting the number of black balls. Then X can take values 0, 1 & 2.

Q29. Trials of a random experiment are called Bernoulli trials, if they satisfy the condition/s:

• There should be a finite number of independent trials
• Each trial has exactly two outcomes: success or failure.
• The probability of success remains the same in each trial.
• All of the above

Solution

Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes: success or failure. (iv) The probability of success remains the same in each trial.

Q30. Let E and F be events of a sample space S of an experiment, then P(E'/F) = ...

• P(E')/P(F)
• P(E').P(F)
• 1 - P(E/F)
• P(E/F)'