Question bank

Part-A

Q1) Define random experiment. (5)

A1)

An experiment in which all the possible outcomes are known in advance but we cannot predict as to which of them will occur when we perform the experiment,

For example- Experiment of tossing a coin is random experiment as the possible outcomes head and tail are known in advance but which one will turn up is not known.

Similarly throwing a dice is also a random experiment as we do not know what will is the outcome but we know all the possible outcomes.

Drawing a card from a well shuffled pack of 52 playing cards ‘are the examples of random experiment.

Q2) What are sample space and sample points of a random experiment? (5)

A2) ASet of all possible outcomes of a random experiment is known as sample space and is usually denoted by S, and the total number of elements in the sample space is known as size of the sample space and is denoted by n(S).

(i) If we toss a coin then the sample space is

S = {H, T}, where H and T denote head and tail respectively and n(S) = 2.

(ii) If a die is thrown, then the sample space is

S = {1, 2, 3, 4, 5, 6} and n(S) = 6.

Sample Point

Each outcome of an experiment is visualised as a sample point in the sample space. Foe example-

(i) If a coin is tossed then getting head or tail is a sample point.

(ii) If a die is thrown twice, then getting (1, 1) or (1, 2) or (1, 3) or…or (6, 6) is

a sample point.

Q3) Explain the different types of events. (7)

A3)

Equally likely events: Two events are said to be ‘equally likely’, if one of them cannot be expected in preference to the other. For instance, if we draw a card from well-shuffled pack, we may get any card. Then the 52 different cases are equally likely.

Independent events: Two events may be independent, when the actual happening of one does not influence in any way the probability of the happening of the other.

Mutually Exclusive events: Two events are known as mutually exclusive, when the occurrence of one of them excludes the occurrence of the other. For example, on tossing of a coin, either we get head or tail, but not both.

Compound Event: When two or more events occur in composition with each other, the simultaneous occurrence is called a compound event. When a die is thrown, getting a 5 or 6 is a compound event.

Favourable Events: The events, which ensure the required happening, are said to be favourable events. For example, in throwing a die, to have the even numbers, 2, 4 and 6 are favourable cases.

Q4) Explain independent events. (5)

A4)

Events are said to be independent if happening or non-happening of any one event is not affected by the happening or non-happening of other events. For example, if a coin is tossed certain number of times, then happening of head in any trial is not affected by any other trial i.e. all the trials are independent.

Two events A and B are independent if and only if P(B|A) = P(B) i.e. there is no relevance of giving any information. Here, if A has already happened, even then it does not alter the probability of B.

If A and B are independent events then-

(Two disjoint events are not independent.)

Independence implies that

Q5) What is Bernoulli’s distribution. Explain it (8)

A5)

Let a product is tested which may be defective or non-defective, let p be the probability of non-defective and q = 1 - p be the probability of defective product.

And let X be a random variable which takes 1 if success occurs and 0 if failure occurs.

Therefore-

P[X = 1] = p

P[X = 0] = q = 1- p

This experiment is known as a Bernoulli trial and the random variable X is a Bernoulli variable.

Conditions for Bernoulli tests

1. A finite number of tests.

2. Each trial must have exactly two results: success or failure.

3. The tests must be independent.

4. The probability of success or failure must be the same in each test.

Bernoulli distribution-

A discrete random variable said to be follow a Bernoulli distribution with parameter p if its p.m.f is given by-

Bernoulli distribution in tabular form can be given as-

1. Mean of the Bernoulli distribution is p.

2. Variance = p(1-p)

Example: If X be a random variable following a Bernoulli distribution with parameter p = 0.6, then find its mean and variance.

Sol.

Mean = p = 0.6

Variance = p(1-p) = 0.6 (1 – 0.6) = 0.6 ×0.4 = 0.24

Q6) Write a short note on Poisson distribution. (8)

A6)

Poisson distribution was derived in 1837 by a French mathematician Simeon D Poisson (1731-1840).

Examples of Poisson distribution-

Poisson distribution is a limiting case of binomial distribution under certain conditions listed below-

1. n, the number of trials are infinitely large.

2. p, the probability of success for each trial is very small.

3. Np is finite quantity say

A random variable X is said to be follow Poisson distribution if it has the following probability mass function-

Moments of Poisson distribution-

1. First moment about origin- which Is known as mean.

2. Second moment about origin-

3. Third moment about origin-

4. Fourth moment about origin-

Note-

1. Poisson distribution is always positively skewed distribution.

2. Mean and variance of Poisson dist. Are always equal

For Poisson distribution-

Part-B

Q7)  Find the probability of getting 53 sundays in a randomly selected non-leap year. (5)

A7)

We know that there are 365 days in a non-leap year. Or 52 complete weeks and 1 over day

This over day may be one of the days

Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

So, the number of exhaustive cases = 7

Let A be the event of getting 53 Sundays

There will be 53 Sundays in a non leap year if and only if the over day is Sunday.

So that

Number of favourable cases for event A= 1

P(A) = Number of favourable cases / total cases = 1/7

Q8) Three horses A, B and C are in a race. A is twice as likely to win as B and B is twice as likely to win as C. What are the respective probabilities of their winning? (5)

A8)

Let x be the probability that A wins the race.

Probability that B wins the race = twice the probability of A’s winning

= 2(x) = 2 x

And probability of C’s winning = twice the probability of B’s winning

= 2(2x) = 4x

Now, as the sum of the probability of happening an event and that of its

Complementary event(s) is 1. Here, the complementary of A is the happening of B

Or C]

 x + 2x + 4x = 1, and hence x = 1/7

Therefore, the respective chances of winning A, B and C are 1/7, 2/7, 4/7

Q9) 25 lottery tickets are marked with first 25 numerals. A ticket is drawn at random. Find the probability that it is a multiple of 5 or 7. (5)

A9)

Let A be the event that the drawn ticket bears a number multiple of 5 and B be the event that it bears a number multiple of 7.

Therefore,

A = {5, 10, 15, 20, 25}

B = {7, 14, 21}

Here, as A

Therefore, A and B are mutually exclusive, and hence

Q10) A bag contains 4 red, 5 black and 2 green balls. One ball is drawn from the bag. Find the probability that-

(i) It is a red ball

(ii) It is not black

(iii) It is green or black              (5)

A10)

There are total 11 exhaustive cases.

And the favourable cases are 4, then the probability of getting a red ball is-

Similarly-

Probability of getting a ball which is not black is-

Probability of getting a green or black ball is-

Q11) A box contains 4 white and 2 black balls and a second box contains three balls of each colour. Now a bag is selected at random and a ball is drawn randomly from the chosen box. Then what will be the probability that the ball is white?  (5)

A11)

Here we have two mutually exclusive cases-

1. The first bag is chosen

2. The second bag is chosen

The chance of choosing the first bag is 1/2. And if this bag is chosen then the probability of drawing a white ball is 4/6.

So that the probability of drawing a white ball from the first bag is-

And the probability of drawing a white ball from the second bag is-

Here the events are mutually exclusive, then the required probability is-

Q12) A can hit a target 3 times in 5 shots, B 2 times in 5 shots, and C 3 times in 4 shots. All of them fire one shot each simultaneously at the target.

What is the probability that-

1. Two shots hit

2. At least two shots hit    (7)

A12)

1. Now probability that 2 shots hit the target-

2.

Probability of at least two shots hitting the target

Q13) A factory has two machines A and B making 60% and 40% respectively of the total production. Machine A produces 3% defective items, and B produces 5% defective items. Find the probability that a given defective part came from   (8)

A13)

We consider the following events:

A: Selected item comes from A.

B: Selected item comes from B.

D: Selected item is defective.

We are looking for . We know:

Now,

So we need

Since, D is the union of the mutually exclusive events and (the entire sample space is the union of the mutually exclusive events A and B)

Q14) Three urns contain 6 red, 4 black; 4 red, 6 black; 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red find the probability that it is drawn from the first urn. (8)

A14)

:The ball is drawn from urnI.

: The ball is drawn from urnII.

: The ball is drawn from urnIII.

R:The ball is red.

We have to find

Since the three urns are equally likely to be selected

Also,

From (i), we have

Q15) A die is rolled. If the outcome is a number greater than 3, what is the probability that it is a prime number?  (5)

A15)

The sample space of the experiment is

S = {1, 2, 3, 4, 5, 6}

Let A be event that the outcome is a number greater than 3 and B be the event that it is a prime number.

A = {4, 5, 6}, B = {2, 3, 5} and hence

P(A) =  3/6,  P(B) = 3/6, 1/6.

The required probability = P(B|A)

Q16) Find the probability distribution of the number of heads when three fair coins are tossed simultaneously.    (5)

A16)

Let X be the number of heads in the toss of three fair coins.

As the random variable, “the number of heads” in a toss of three coins may be

0 or 1 or 2 or 3 associated with the sample space

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT},

X can take the values 0, 1, 2, 3, with

P[X = 0] = P[TTT ] = 1/8

P[X = 1] = P[HTT, THT, TTH] = 3/8

P[X = 2] = P[HHT, HTH, THH] = 3/8

P[X = 3] = P [HHH] = 1/8

Probability distribution of X, i.e. the number of heads when three coins are tossed simultaneously is

Q17) A die is thrown 8 times then find the probability that 3 will show-

1. Exactly 2 times

2. At least 7 times

3. At least once                    (7)

A17)

As we know that-

Then-

1. Probability of getting 3 exactly 2 times will be-

2. Probability of getting 3 at least 7 or 8 times will be-

3. Probability of getting 3 at least once or (1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 times)-

Q18) The probability that an evening college student will graduate is 0.4. Determine the probability that out of 5 students (a) none, (b) one, and (c) atleast one will graduate.   (7)

A18)

Here

n = 5, p = 0.4 or 4/10, q = 0.6 = (6/10)

a)  The probability of zero success-

(b) The probability of one success-

(c) The probability of atleast one success

= 1– probability of no success

= 1– 0.078

= 0.922

Q19) If cars arriving at workshop follow the Poisson distribution. If the average number of cars arrivals during a specified period of an hour is 2.

Find the probabilities that during the given hour-

1. No car arrive

2. At least two cars arrive.    (5)

A19)

Here the average of car arrivals is - 2

So that mean = 2

Let X be the number of cars arriving during the given  hour,

By using Poisson distribution, we get-

So that the required probability-

1. P [no car will arrive] = P [X = x] =

2. P [At least two cars will arrive] = P [X≥2] = P [X =2] + P [X = 3] + ……….

= 1 - P [[X =1] + P [X =0]]

Q20) Find the expectations of the number of an unbiased die when thrown.  (5)

A20)

X can take the values-

1, 2, 3, 4, 5, 6

With

P[X = 1] = P[X = 2] = P[X = 3] = P[X = 4] = P[X = 5] = P[X = 6] = 1/6

The distribution table will be-

Hence the expectation of number on the die thrown is-

So that-