Sample Solution

IGNOU MST-003 Solved Assignment | 1st January 2025 to 31st December, 2025 | Probability Theory | PGDAST Sample Solution

MST-003 Solved Assignment 2025

State whether the following statements are True or False and also give the reason in support of your answer.
(a) Sample space of a (i) random experiment tossing two coins simultaneously and (ii) One coin two times is the same.
(b) Standard deviation of a random variable $X$ may take any real value, i.e. its value lies in the interval $(- \infty, \infty)$ .
(c) If events $E_{1}, E_{2}, E_{3}, E_{4}, \dots, E_{n}$ are mutually exclusive and exhaustive then $P (E_{1} \cup E_{2} \cup E_{3} \cup \dots \cup E_{n})$ will be greater than $1 / 2$ but less than 1.
(d) If S is sample space of a random experiment and E is an event defined on this sample space then $P (S ∣ E) = 1$ .
(e) If $X$ is a random variable having range set ${0, 1, 2, 3}$ then the set ${x \in S : X (x) = 0}$ is an event having at least one outcome of the random experiment.
There are 4 black, 3 blue and 8 red balls in an urn. Three balls are selected one by one without replacement. What is the probability that:
(i) First ball drawn is black, second one is red and third one is blue
(ii) All the three balls are of the same colour.
A random 5-card poker hand is dealt from a standard deck of cards. Find the probability (in terms of binomial coefficients) of getting a flush (all 5 cards being of the same suit: do not count a royal flush, which is a flush with an ace, king, queen, jack and 10).
Show that

f (x) = {(\frac{1}{2})}^{x + 1}, x = 0, 1, 2, 3, 4, 5, \dots

is a valid PMF for a discrete random variable. Also find out its CDF.

A group of 100 people are comparing their birthdays (as usual, assume their birthdays are independent and not on February 29, etc.). Find the expected number of pairs of people with the same birthday, and the expected number of days in the year on which at least two of these people were born.
Random variable $X$ follows Beta distribution with parameters $a = 3, b = 2$ and has pdf

f (x) = {\begin{cases} 12 x^{2} (1 - x), 0 < x < 1 \\ 0, otherwise \end{cases}

Find (i) CDF of X (ii)

P [0 < X < 1 / 2]

(iii) mean and variance of X without using direct formula for mean and variance.

Consider the joint PDF for the type of customer service $X (0 =$ telephonic hotline, $1 =$ Email) and of satisfaction score $Y (1 =$ unsatisfied, $2 =$ satisfied, $3 =$ very satisfied $)$ :

	Y
X	1	2	3
0	0	$1 / 2$	$1 / 4$
1	$1 / 6$	$1 / 12$	0

(a) Determine and interpret the marginal distributions of both X and Y .
(b) Calculate the

75 %

quantile for the marginal distribution of Y.
(c) Determine and interpret the conditional distribution of satisfaction level for

X = 1

.
(d) Are the two variables independent?
(e) Calculate and interpret the covariance of X and Y .
8. State Monty Hall problem and solve it.

Solution

Question:-1

(a) Sample space of a (i) random experiment tossing two coins simultaneously and (ii) one coin two times is the same.

True.
Justification: For (i) tossing two coins simultaneously, the sample space is all possible outcomes:

{H H, H T, T H, T T}

, where

H

is heads and

T

is tails. For (ii) tossing one coin two times, the outcomes are also

{H H, H T, T H, T T}

, as each toss is independent and can be heads or tails. The sample spaces are identical in both cases, as the order of tosses matters in both experiments, and the outcomes are indistinguishable in terms of the final pairs.

(b) Standard deviation of a random variable $X$ may take any real value, i.e., its value lies in the interval $(- \infty, \infty)$ .

False.
Counterexample: The standard deviation

σ

of a random variable

X

is defined as

σ = \sqrt{Var (X)}

, where

Var (X) = E [(X - E [X])^{2}]

. Since variance is an expected value of a non-negative quantity

(X - E [X])^{2}

Var (X) \geq 0

, and thus

σ \geq 0

. Therefore,

σ

cannot be negative, so it does not lie in

(- \infty, \infty)

but rather in

[0, \infty)

(c) If events $E_{1}, E_{2}, E_{3}, E_{4}, \dots, E_{n}$ are mutually exclusive and exhaustive, then $P (E_{1} \cup E_{2} \cup E_{3} \cup \dots \cup E_{n})$ will be greater than $1 / 2$ but less than 1.

False.
Justification: If events

E_{1}, E_{2}, \dots, E_{n}

are mutually exclusive and exhaustive, their union covers the entire sample space

S

, so

P (E_{1} \cup E_{2} \cup \dots \cup E_{n}) = P (S) = 1

. Since the probability equals 1, it is not strictly less than 1, contradicting the statement’s requirement that the probability is greater than

1 / 2

but less than 1.

(d) If $S$ is the sample space of a random experiment and $E$ is an event defined on this sample space, then $P (S ∣ E) = 1$ .

True.
Justification: The conditional probability is defined as

P (S ∣ E) = \frac{P (S \cap E)}{P (E)}

. Since

E \subseteq S

, we have

S \cap E = E

. Thus,

P (S \cap E) = P (E)

, and so

P (S ∣ E) = \frac{P (E)}{P (E)} = 1

, provided

P (E) > 0

. If

P (E) = 0

, the conditional probability is undefined, but the statement holds for any event

E

with positive probability, as is typical in such contexts.

(e) If $X$ is a random variable having range set ${0, 1, 2, 3}$ , then the set ${x \in S : X (x) = 0}$ is an event having at least one outcome of the random experiment.

False.
Counterexample: The set

{x \in S : X (x) = 0}

is the event where the random variable

X

takes the value 0. While this set is always an event (as it is a subset of the sample space defined by the preimage of 0 under

X

), it does not necessarily have at least one outcome. For example, consider a sample space

S = {s_{1}, s_{2}}

with

X (s_{1}) = 1

X (s_{2}) = 2

. The range of

X

{1, 2} \subseteq {0, 1, 2, 3}

, but

{x \in S : X (x) = 0} = \emptyset

, which is an event with no outcomes. Thus, the statement is false because the event may be empty.

Question:-2

There are 4 black, 3 blue and 8 red balls in an urn. Three balls are selected one by one without replacement. What is the probability that:

(i) First ball drawn is black, second one is red and third one is blue
(ii) All the three balls are of the same colour.

Answer:

To solve the problem, we need to calculate two probabilities for an urn containing 4 black, 3 blue, and 8 red balls, with three balls drawn one by one without replacement. The total number of balls is

4 + 3 + 8 = 15

(i) Probability that the first ball is black, the second is red, and the third is blue

We calculate the probability of drawing a black ball first, followed by a red ball, and then a blue ball, without replacement.

First ball (black): There are 4 black balls out of 15 total balls.

$P (black first) = \frac{4}{15}$
Second ball (red): After drawing a black ball, 14 balls remain, including 8 red balls.

$P (red second ∣ black first) = \frac{8}{14} = \frac{4}{7}$
Third ball (blue): After drawing a black and a red ball, 13 balls remain, including 3 blue balls.

$P (blue third ∣ black first, red second) = \frac{3}{13}$

The joint probability is the product of these conditional probabilities:

P (black, red, blue) = \frac{4}{15} \times \frac{8}{14} \times \frac{3}{13}

Simplify:

= \frac{4 \times 8 \times 3}{15 \times 14 \times 13} = \frac{96}{2730}

= \frac{96 \div 6}{2730 \div 6} = \frac{16}{455}

Thus, the probability for part (i) is:

\frac{16}{455}

(ii) Probability that all three balls are of the same color

We need to find the probability that all three balls drawn are either all black, all blue, or all red. Since the colors are mutually exclusive, we calculate the probability for each case and sum them.

Case 1: All three balls are black
- Number of ways to choose 3 black balls from 4: $(\binom{4}{3}) = 4$
- Number of ways to choose 3 balls from 15: $(\binom{15}{3}) = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$
- Probability: $P (all black) = \frac{(\binom{4}{3})}{(\binom{15}{3})} = \frac{4}{455}$
Case 2: All three balls are blue
- Number of ways to choose 3 blue balls from 3: $(\binom{3}{3}) = 1$
- Probability: $P (all blue) = \frac{(\binom{3}{3})}{(\binom{15}{3})} = \frac{1}{455}$
Case 3: All three balls are red
- Number of ways to choose 3 red balls from 8: $(\binom{8}{3}) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
- Probability: $P (all red) = \frac{(\binom{8}{3})}{(\binom{15}{3})} = \frac{56}{455} = \frac{8}{65}$
Total probability:

$P (all same color) = P (all black) + P (all blue) + P (all red)$

$= \frac{4}{455} + \frac{1}{455} + \frac{56}{455} = \frac{4 + 1 + 56}{455} = \frac{61}{455}$

The fraction

\frac{61}{455}

is in its simplest form (61 is prime, and 455 = 5 × 7 × 13, with no common factors).

Thus, the probability for part (ii) is:

\frac{61}{455}

Final Answer

(i) \frac{16}{455}, (ii) \frac{61}{455}

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