Maharashtra Board Class 12 Maths Part 2 Chapter 8 Miscellaneous Solutions

Exercise 8(I)

Question 1. F(x) is c.d.f. of discrete r.v. X whose p.m.f. is given by \(P(x) = k \binom{4}{x}\), for x = 0, 1, 2, 3, 4 & P(x) = 0 otherwise then F(5) =
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{4}\)
(d) 1
Answer: (d) 1
In simple words: The cumulative distribution function F(x) at a point includes the probability of all values up to and including that point. Since the random variable X is defined only up to x=4, F(5) would include all possible probabilities, making the total probability 1.

🎯 Exam Tip: Remember that the cumulative distribution function (c.d.f.) F(x) always approaches 1 as x approaches infinity (or the maximum value for discrete variables). For values beyond the defined range, F(x) will be 1.

Question 2. F(x) is c.d.f. of discrete r.v. X whose distribution is

then F(-3) =
(a) 0
(b) 1
(c) 0.2
(d) 0.15
Answer: (a) 0
In simple words: The cumulative distribution function F(x) at any point x is the probability that the random variable takes a value less than or equal to x. Since the smallest value X can take is -2, the probability of X being less than or equal to -3 is 0.

🎯 Exam Tip: For discrete random variables, F(x) for any value less than the minimum possible outcome of X is always 0. This is a fundamental property of c.d.f.s.

Question 3. X: number obtained on uppermost face when a fair die is thrown then E(X) =
(a) 3.0
(b) 3.5
(c) 4.0
(d) 4.5
Answer: (b) 3.5
In simple words: The expected value (mean) of the numbers obtained when a fair die is thrown is the average of all possible outcomes (1, 2, 3, 4, 5, 6), which is (1+2+3+4+5+6)/6 = 21/6 = 3.5.

🎯 Exam Tip: For a fair die, the expected value is simply the average of all faces. This concept applies to any discrete uniform distribution.

Question 4. If p.m.f. of r.v. X is given below.

then Var(X) =
(a) \(p^2\)
(b) \(q^2\)
(c) pq
(d) 2pq
Answer: (d) 2pq
In simple words: This probability mass function represents a binomial distribution with n=2 trials, where P(x) = \(\binom{2}{x} p^x q^{2-x}\). For a binomial distribution, the variance is given by npq. With n=2, the variance is 2pq.

🎯 Exam Tip: Recognize the pattern of the PMF. If it matches a known distribution (like binomial here), use its standard formulas for mean and variance to save calculation time.

Question 5. The expected value of the sum of two numbers obtained when two fair dice are rolled is
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (c) 7
In simple words: The expected value of the sum of two random variables is the sum of their individual expected values. For a single fair die, E(X) = 3.5. So, for two dice, E(X1 + X2) = E(X1) + E(X2) = 3.5 + 3.5 = 7.

🎯 Exam Tip: This question tests the linearity of expectation, E(X+Y) = E(X) + E(Y). This property simplifies calculations for sums of random variables significantly.

Question 6. Given p.d.f. of a continuous r.v. X as \(f(x) = \frac{x^2}{3}\) for -1 < x < 2 = 0 otherwise then F(1) =
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{3}{9}\)
(d) \(\frac{4}{9}\)
Answer: (b) \(\frac{2}{9}\)
In simple words: F(1) is the cumulative probability up to x=1, calculated by integrating the probability density function from the lower bound (-1) to 1. Integrating \(\frac{x^2}{3}\) from -1 to 1 gives \([\frac{x^3}{9}]_{-1}^{1} = \frac{1^3}{9} - \frac{(-1)^3}{9} = \frac{1}{9} - (-\frac{1}{9}) = \frac{2}{9}\).

🎯 Exam Tip: For a continuous random variable, the c.d.f. F(x) is the integral of the p.d.f. f(x) from negative infinity (or the lower bound of the domain) to x. Pay close attention to integration limits.

Question 7. X is r.v. with p.d.f. \(f(x) = \frac{k}{\sqrt{x}}\), 0 < x < 4 = 0 otherwise then E(X) =
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{2}{3}\)
(d) 1
Answer: (b) \(\frac{4}{3}\)
In simple words: First, find the constant k by integrating f(x) over its domain [0, 4] and setting the result to 1. Once k is found, calculate E(X) by integrating x*f(x) over the same domain. After finding k = \(\frac{1}{4}\), E(X) becomes \(\int_0^4 x \cdot \frac{1}{4\sqrt{x}} dx = \int_0^4 \frac{\sqrt{x}}{4} dx = \frac{1}{4} [\frac{2}{3}x^{3/2}]_0^4 = \frac{1}{4} \cdot \frac{2}{3} (4^{3/2}) = \frac{1}{6} \cdot 8 = \frac{4}{3}\).

🎯 Exam Tip: Always normalize the p.d.f. by finding the constant 'k' first. Then, use the definition of expected value E(X) = \(\int x f(x) dx\) over the valid range of X.

Question 8. If X follows \(B(20, \frac{1}{10})\) then E(X) =
(a) 2
(b) 5
(c) 4
(d) 3
Answer: (a) 2
In simple words: For a binomial distribution \(B(n, p)\), the expected value E(X) is given by the formula n times p. Here, n = 20 and p = \(\frac{1}{10}\), so E(X) = \(20 \times \frac{1}{10} = 2\).

🎯 Exam Tip: Recognize the parameters of the binomial distribution (n = number of trials, p = probability of success) and directly apply the formula E(X) = np for binomial random variables.

Question 9. If E(X) = m and Var(X) = m then X follows -
(a) Binomial distribution
(b) Poisson distribution
(c) Normal distribution
(d) none of the above
Answer: (b) Poisson distribution
In simple words: A key characteristic of the Poisson distribution is that its mean (E(X)) is equal to its variance (Var(X)), both denoted by the parameter 'm'.

🎯 Exam Tip: This is a definition-based question. Knowing the properties of common probability distributions, particularly the relationship between mean and variance, is crucial for quick identification.

Question 10. If E(X) > Var(X) then X follows -
(a) Binomial distribution
(b) Poisson distribution
(c) Normal distribution
(d) none of the above
Answer: (a) Binomial distribution
In simple words: For a binomial distribution, E(X) = np and Var(X) = npq. Since q (probability of failure) is always less than 1 (unless p=1), Var(X) = npq will always be less than E(X) = np (provided n and p are positive).

🎯 Exam Tip: Understand the relationships between mean and variance for different distributions: Binomial (E(X) > Var(X)), Poisson (E(X) = Var(X)), Negative Binomial (E(X) < Var(X)).

Exercise 8(II)

Question 1. The values of discrete r.v. are generally obtained by __________
Answer: counting
In simple words: Discrete random variables represent quantities that can be counted, such as the number of heads in coin flips or the number of defective items.

🎯 Exam Tip: Distinguish between discrete and continuous random variables: discrete values are obtained by counting, while continuous values are obtained by measurement.

Question 2. The values of continuous r.v. are generally obtained by __________
Answer: measurement
In simple words: Continuous random variables represent quantities that can be measured, such as height, weight, or temperature, and can take any value within a given range.

🎯 Exam Tip: This is a fundamental concept in probability. Remember that measurements result in continuous data, while counts result in discrete data.

Question 3. If X is discrete random variable takes the values \(X_1, X_2, X_3, ...... X_n\) then \(\sum_{i=1}^{n} P(x_i) = \) __________
Answer: 1
In simple words: The sum of all probabilities for all possible outcomes of a discrete random variable must always equal 1, representing the certainty that one of the outcomes will occur.

🎯 Exam Tip: This is the axiom of total probability and a defining characteristic of any probability distribution function.

Question 4. If f(x) is distribution function of discrete r.v. X with p.m.f. \(p(x) = \frac{x-1}{3}\) for x = 1, 2, 3, and p(x) = 0 otherwise then F(4) = __________
Answer: 1
In simple words: F(4) represents the cumulative probability up to and including x=4. Since the maximum value X can take is 3, F(4) includes all possible probabilities, which sums up to 1.

🎯 Exam Tip: Similar to Question 1(I), for any value beyond the maximum possible outcome of a discrete random variable, the c.d.f. F(x) will be 1.

Question 5. If f(x) is distribution function of discrete r.v. X with p.m.f. \(p(x) = k \binom{4}{x}\) for x = 0, 1, 2, 3, 4, and p(x) = 0 otherwise then F(-1) = __________
Answer: 0
In simple words: F(-1) represents the cumulative probability up to and including x=-1. Since the minimum value X can take is 0, the probability of X being less than or equal to -1 is 0.

🎯 Exam Tip: Similar to Question 2(I), for any value less than the minimum possible outcome of a discrete random variable, the c.d.f. F(x) will be 0.

Question 6. E(X) is considered to be __________ of the probability distribution of X.
Answer: centre of gravity
In simple words: The expected value E(X) represents the average or central tendency of a random variable, analogous to the center of mass in physics, hence referred to as the "centre of gravity".

🎯 Exam Tip: The expected value is the theoretical average of a random variable and gives a single representative value for the distribution.

Question 7. If X is continuous r.v. and F(\(x_i\)) = \(P(X \le x_i) = \int_{-\infty}^{x_i} f(x)dx\) then f(x) is called __________
Answer: Cumulative Distribution Function
In simple words: The function F(\(x_i\)) which represents the probability that a random variable X takes a value less than or equal to \(x_i\) is by definition the Cumulative Distribution Function (CDF).

🎯 Exam Tip: Understand the difference between the probability density function f(x) (which describes the likelihood of a continuous random variable taking on a given value) and the cumulative distribution function F(x) (which gives the probability that the random variable is less than or equal to a value).

Question 8. In Binomial distribution probability of success __________ from trial to trial.
Answer: remains constant/independent
In simple words: A core condition for a binomial distribution is that the probability of success (p) must remain the same for each independent trial.

🎯 Exam Tip: Remember the four key properties of a binomial experiment: fixed number of trials, independent trials, only two outcomes (success/failure), and constant probability of success.

Question 9. In Binomial distribution, if n is very large and probability success of p is very small such that np = m (constant) then __________ distribution is applied.
Answer: Poisson
In simple words: When the number of trials (n) is very large and the probability of success (p) is very small, while their product (np = m) is constant, the binomial distribution can be approximated by a Poisson distribution.

🎯 Exam Tip: This is a crucial approximation in probability theory. The Poisson distribution is often used to model rare events occurring over a fixed interval of time or space.

Exercise 8(III)

Question 1. If \(P(X = x) = k \binom{4}{x}\) for x = 0, 1, 2, 3, 4, then \(F(5) = \frac{1}{4}\) when f(x) is c.d.f.
Answer: False
In simple words: The maximum value for X is 4. F(5) is the cumulative probability up to x=5, which means it includes all possible probabilities. Therefore, F(5) must be 1, not \(\frac{1}{4}\).

🎯 Exam Tip: The cumulative distribution function F(x) for any value beyond the maximum possible outcome of a discrete random variable is always 1. This is a fundamental property of c.d.f.s.

Question 2.

If F(x) is c.d.f. of discrete r.v. X then F(-3) = 0.
Answer: True
In simple words: The smallest value X can take is -2. Therefore, the cumulative probability F(-3), which sums all probabilities for values less than or equal to -3, must be 0 because no such values exist in the distribution.

🎯 Exam Tip: The c.d.f. F(x) is 0 for any value x less than the minimum possible value of the random variable. This is a crucial boundary condition for c.d.f.s.

Question 3. X is the number obtained on the uppermost face when a die is thrown the E(X) = 3.5.
Answer: True
In simple words: For a fair six-sided die, the possible outcomes are 1, 2, 3, 4, 5, 6, each with a probability of \(\frac{1}{6}\). The expected value E(X) = \(\sum x P(x) = (1+2+3+4+5+6) \times \frac{1}{6} = \frac{21}{6} = 3.5\).

🎯 Exam Tip: For discrete uniform distributions, the expected value is simply the average of the minimum and maximum possible outcomes: (min + max) / 2.

Question 4. If p.m.f. of discrete r.v.X is

then E(X) = 2p.
Answer: True
In simple words: This probability mass function corresponds to a binomial distribution \(B(n=2, p)\). For a binomial distribution, the expected value E(X) is given by np. Here, n=2, so E(X) = 2p.

🎯 Exam Tip: Recognizing standard distribution patterns is key. If the PMF looks like \(\binom{n}{x}p^x q^{n-x}\), it's a binomial distribution, and you can apply its standard formulas for mean and variance.

Question 5. The p.m.f. of a r.v. X is \(p(x) = \frac{2x}{n(n+1)}\), x = 1, 2,......n = 0 otherwise, Then \(E(X) = \frac{2n+1}{3}\)
Answer: True
In simple words: The expected value E(X) is calculated as \(\sum x p(x)\). Substituting the given p(x) and simplifying the sum \(\sum_{x=1}^{n} x \cdot \frac{2x}{n(n+1)}\) leads to the formula \(\frac{2n+1}{3}\).

🎯 Exam Tip: For discrete distributions, E(X) = \(\sum x_i p(x_i)\). Be prepared to use summation formulas like \(\sum x^2 = \frac{n(n+1)(2n+1)}{6}\) for such problems.

Question 6. If \(f(x) = kx (1 – x)\) for 0 < x < 1 = 0 otherwise then k = 12
Answer: False
In simple words: To find k, integrate f(x) from 0 to 1 and set it equal to 1. \(\int_0^1 k(x-x^2) dx = k[\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = k(\frac{1}{2} - \frac{1}{3}) = k(\frac{1}{6})\). Setting this to 1, we get \(k=6\). So, the statement that k=12 is false.

🎯 Exam Tip: For any valid probability density function, the total probability over its entire domain must be 1. This is the normalization condition used to find unknown constants like 'k'.

Question 7. If \(X \sim B(n, p)\) and n = 6 and P(X = 4) = P(X = 2) then \(p = \frac{1}{2}\).
Answer: True
In simple words: For a binomial distribution, \(P(X=x) = \binom{n}{x} p^x q^{n-x}\). If \(P(X=4) = P(X=2)\) with n=6, then \(\binom{6}{4}p^4 q^2 = \binom{6}{2}p^2 q^4\). Since \(\binom{6}{4} = \binom{6}{2}\), this simplifies to \(p^4 q^2 = p^2 q^4\). Assuming \(p, q \ne 0\), we can divide by \(p^2 q^2\) to get \(p^2 = q^2\), which means \(p=q\). Since \(p+q=1\), it implies \(p = \frac{1}{2}\).

🎯 Exam Tip: When probabilities for different numbers of successes are equal in a binomial distribution, it often implies a symmetric distribution, which occurs when the probability of success (p) is 0.5.

Question 8. If r.v. X assumes values 1, 2, 3,........., n with equal probabilities then \(E(X) = \frac{n+1}{2}\)
Answer: True
In simple words: If X takes values 1, 2, ..., n with equal probability (which is \(\frac{1}{n}\) for each value), it's a discrete uniform distribution. The expected value for such a distribution is the average of the first and last values, which is \(\frac{1+n}{2}\).

🎯 Exam Tip: For a discrete uniform distribution on integers 1 to n, the expected value is simply the midpoint of the range. This is a common property to remember.

Question 9. If r.v. X assumes the values 1, 2, 3,........., 9 with equal probabilities, E(X) = 5.
Answer: True
In simple words: Here, n=9. Using the formula from Question 8, E(X) = \(\frac{n+1}{2} = \frac{9+1}{2} = \frac{10}{2} = 5\).

🎯 Exam Tip: This is a direct application of the expected value formula for a discrete uniform distribution. Identify 'n' correctly and apply the formula.

Exercise 8(IV)

Part - I

Question 1. Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
(i) An economist is interested in knowing the number of unemployed graduates in the town with a population of 1 lakh. Solution: X = No. of unemployed graduates in a town. The population of the town is 1 lakh \(\therefore\) X takes finite values \(\therefore\) X is a Discrete Random Variable \(\therefore\) Range of = {0, 1, 2, 4, .... 1,00,000}
(ii) Amount of syrup prescribed by a physician. Solution: X: Amount of syrup prescribed. \(\therefore\) X Takes infinite values \(\therefore\) X is a Continuous Random Variable.
(iii) A person on a high protein diet is interested in the weight gained in a week. Solution: X: Gain in weight in a week. X takes infinite values \(\therefore\) X is a Continuous Random Variable.
(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them. Solution: X: No. of female rats selected X takes finite values. \(\therefore\) X is a Discrete Random Variable. Range of X = {0, 1, 2, 3, 4, 5}
(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint. Solution: X: Speed of car in km/hr X takes infinite values \(\therefore\) X is a Continuous Random Variable.
In simple words: Discrete random variables are countable (like number of graduates or rats), while continuous random variables are measurable (like amount of syrup, weight, or speed), taking any value within a range.

🎯 Exam Tip: To differentiate, ask if the variable can be counted (discrete) or if it can take any value within an interval (continuous). The range for discrete variables is a set of distinct values; for continuous, it's an interval.

Question 2. The probability distribution of a discrete r.v. X is as follows.

(i) Determine the value of k.
(ii) Find P(X \(\le\) 4), P(2 < X < 4), P(X \(\ge\) 3). Solution: (i) Assuming that the given distribution is a p.m.f. of X \(\therefore\) Each P(X = x) \(\ge\) 0 for x = 1, 2, 3, 4, 5, 6 \(k \ge 0\) \(\Sigma P(X = x) = 1\)
\(k + 2k + 3k + 4k + 5k + 6k = 1\)
\(21k = 1\)
\(\therefore k = \frac{1}{21}\)
(ii) \(P(X \le 4) = 1 – P(X > 4)\)
\( = 1 – [P(X = 5) + P(X = 6)]\)
\( = 1 - [\frac{5}{21}+\frac{6}{21}]\)
\( = 1 - \frac{11}{21}\)
\( = \frac{10}{21}\)
\(P(2 < X < 6) = p(3) + p(4) + p(5)\)
\( = 3k + 4k + 5k\)
\( = \frac{3}{21}+\frac{4}{21}+\frac{5}{21}\)
\( = \frac{12}{21}\)
\( = \frac{4}{7}\)
(iii) \(P(X \ge 3) = p(3) + p(4) + p(5) + p(6)\)
\( = 3k + 4k + 5k + 6k\)
\( = 3 \times \frac{1}{21} + 4 \times \frac{1}{21} + 5 \times \frac{1}{21} + 6 \times \frac{1}{21}\)
\( = \frac{3}{21} + \frac{4}{21} + \frac{5}{21} + \frac{6}{21}\)
\( = \frac{18}{21}\)
\( = \frac{6}{7}\)
In simple words: For any valid probability distribution, the sum of all probabilities must be 1. Use this to find 'k'. Then, calculate probabilities for ranges by summing the individual probabilities within those ranges or by using the complement rule.

🎯 Exam Tip: Always verify that all probabilities are non-negative and sum to 1. When calculating probabilities for ranges, remember to correctly identify the included outcomes and use the complement rule (P(A) = 1 - P(A')) when it simplifies calculations.

Question 3. Following is the probability distribution of an r.v. X.

Find the probability that
(i) X is positive.
(ii) X is non-negative.
(iii) X is odd.
(iv) X is even. Solution: (i) P(X is positive) \(P(X = 0) = p(1) + p(2) + p(3)\) \( = 0.25 + 0.15 + 0.10\) \( = 0.50\)
(ii) P(X is non-negative) \(P(X \ge 0) = p(0) + p(1) + p(2) + p(3)\) \( = 0.20 + 0.25 + 0.15 + 0.10\) \( = 0.70\)
(iii) P(X is odd) \(P(X = -3, -1, 1, 3)\) \( = p(-3) + p(-1) + p(1) + p(3)\) \( = 0.05 + 0.15 + 0.25 + 0.10\) \( = 0.55\)
(iv) P(X is even) \( = 1 – P(X \text{ is odd})\) \( = 1 - 0.55\) \( = 0.45\)
In simple words: Calculate probabilities for categories (positive, non-negative, odd, even) by summing the individual probabilities of the outcomes that fall into each category. For even numbers, use the complement rule: 1 - P(odd).

🎯 Exam Tip: Carefully identify which values of X satisfy the given condition (e.g., positive means X > 0, non-negative means X \(\ge\) 0). The complement rule \(P(A) = 1 - P(A')\) is useful for categories like "even" if "odd" is already calculated.

Question 4. The p.m.f of a r.v. X is given by \[P(X = x) = \begin{cases} \binom{5}{x} \frac{1}{2^5}, & x = 0, 1, 2, 3, 4, 5. \\ 0 & \text{otherwise} \end{cases}\] Show that P(X \(\le\) 2) = P(X \(\ge\) 3). Solution: For x = 0, 1, 2, 3, 4, 5 \(P(X = x) = \frac{^5C_x}{2^5}\) \(\therefore P(X = 0) = \frac{^5C_0}{2^5} = \frac{1}{32}\) \(\therefore P(X=1) = \frac{^5C_1}{2^5} = \frac{5}{32}\) \(\therefore P(X = 2) = \frac{^5C_2}{2^5} = \frac{10}{32}\) \(\therefore P(X = 3) = \frac{^5C_3}{2^5} = \frac{10}{32}\) \(\therefore P(X = 4) = \frac{^5C_4}{2^5} = \frac{5}{32}\) \(\therefore P(X = 5) = \frac{^5C_5}{2^5} = \frac{1}{32}\) \(\therefore\) The given p.m.f. of X is

\(P(X \le 2) = p(0) + p(1) + p(2)\)
\( = \frac{1}{32} + \frac{5}{32} + \frac{10}{32} = \frac{16}{32} = \frac{1}{2}\)
\(P(X \ge 3) = p(3) + p(4) + p(5)\)
\( = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}\)
\(P(X \le 2) = P(X \ge 3)\)
In simple words: First, calculate the individual probabilities for each value of X using the given PMF. Then, sum the probabilities for X \(\le\) 2 (i.e., X=0, 1, 2) and separately sum them for X \(\ge\) 3 (i.e., X=3, 4, 5) to show they are equal.

🎯 Exam Tip: This problem demonstrates the symmetry of a binomial distribution when p=0.5. Calculate each term precisely and remember that \(\binom{n}{k} = \binom{n}{n-k}\), which is useful for symmetry checks.

Question 5. In the following probability distribution of an r.v. X

Find a and obtain the c.d.f. of X. Solution: Given distribution is p.m.f. of r.v. X \(\Sigma P(X = x) = 1\) \(\therefore p(1) + p(2) + p(3) + p(4) + p(5) = 1\)
\(\frac{1}{20} + \frac{3}{20} + a + 2a + \frac{1}{20} = 1\)
\(\therefore \frac{1}{20} + \frac{3}{20} + \frac{1}{20} + a + 2a = 1\)
\(\frac{5}{20} + 3a = 1\)
\(\therefore 3a = 1 - \frac{5}{20}\)
\(3a = \frac{15}{20}\)
\(\therefore a = \frac{15}{20 \times 3} = \frac{1}{4}\)
\(\implies 2a = \frac{2}{4} = \frac{10}{20} = \frac{1}{2}\)
\(\therefore\) The given p.m.f. is

Let F(x) be the c.d.f. of X
\(F(1) = p(1) = \frac{1}{20}\)
\(F(2) = p(1) + p(2) = \frac{1}{20} + \frac{3}{20} = \frac{4}{20} = \frac{1}{5}\)
\(F(3) = p(1) + p(2) + p(3)\)
\( = \frac{1}{20} + \frac{3}{20} + \frac{5}{20} = \frac{9}{20}\)
\(F(4) = p(1) + p(2) + p(3) + p(4)\)
\( = \frac{1}{20} + \frac{3}{20} + \frac{5}{20} + \frac{10}{20} = \frac{19}{20}\)
\(F(5) = p(1) + ... + p(5)\)
\(\therefore\) The c.d.f. is given as

In simple words: First, use the property that the sum of all probabilities in a PMF equals 1 to solve for the unknown constant 'a'. Then, construct the cumulative distribution function (CDF) by adding up the probabilities for each value of X sequentially.

🎯 Exam Tip: When given a PMF with unknown constants, the first step is always to use the normalization condition (\(\sum P(x) = 1\)) to find the constant. For CDF, remember F(x) = \(\sum_{t \le x} P(t)\).

Question 6. A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p.m.f. of X. Solution: A fair coin is tossed 4 times \(\therefore\) Sample space contains 16 outcomes Let X = Number of heads obtained \(\therefore\) X takes the values x = 0, 1, 2, 3, 4. \(\therefore\) The number of heads obtained in a toss is an even \(\binom{4}{x}\) for x = 0,1,2,3,4 \(\therefore ^4C_0 = 1, ^4C_1 = 4, ^4C_2 = 6, ^4C_3 = 4, ^4C_4 = 1\) \(\therefore\) The p.m.f. of X is given as

In simple words: Tossing a fair coin multiple times is a binomial experiment. With n=4 trials and p=\(\frac{1}{2}\) (probability of heads), X follows a binomial distribution. The PMF formula is \(P(X=x) = \binom{n}{x} p^x (1-p)^{n-x}\).

🎯 Exam Tip: For coin toss problems, always identify n (number of tosses) and p (probability of heads/tails). Since a fair coin implies p=0.5, the binomial distribution will be symmetric.

Question 7. Find the probability of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appearing in at least one toss. Solution: S: A die is tossed two times S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} n(S) = 36
(i) X: No. is greater than 4 Range of X = {0, 1, 2}

(ii) X: Six appears on aleast one die. Range of X = {0, 1, 2}

In simple words: For two dice tosses, list all 36 possible outcomes. Then, for each definition of success, count the number of outcomes that satisfy 0, 1, or 2 successes to determine their probabilities and construct the PMF.

🎯 Exam Tip: When dealing with multiple dice, it's helpful to list the sample space or use a grid to count outcomes. Define 'success' clearly for each part of the question to avoid errors in counting.

Question 8. A random variable X has the following probability distribution.

Determine (i) k, (ii) P(X < 3), (iii) P(X > 6), (iv) P(0 < X < 3). Solution: (i) It is a p.m.f. of r.v. X \(\Sigma p(x) = 1\) \(p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1\)
\(k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1\)
\(9k + 10k^2 = 1\)
\(10k^2 + 9k - 1 = 0\)
\(10k^2 + 10k - k - 1 = 0\)
\(\therefore 10k(k + 1) – 1(k + 1) = 0\)
\(\therefore (10k – 1) (k + 1) = 0\)
\(\therefore 10k – 1 = 0\) Or \(k + 1 = 0\)
\(\therefore k = \frac{1}{10}\) or \(k = -1\) k = -1 is not accepted, \(p(x) \ge 0, \forall x \in R\) \(\therefore k = \frac{1}{10}\)
(ii) \(P(X < 3) = p(1) + p(2)\)
\( = k + 2k\)
\( = 3k\)
\( = 3 \times \frac{1}{10}\)
\( = \frac{3}{10}\)
In simple words: First, use the fundamental property that all probabilities sum to 1 to form a quadratic equation in 'k'. Solve for 'k' and choose the positive value since probabilities cannot be negative. Then, calculate P(X < 3) by summing the probabilities for X=1 and X=2.

🎯 Exam Tip: Always remember that probabilities must be non-negative. If solving for an unknown constant yields multiple values, only select the one that ensures all probabilities are \(\ge 0\).

Question 9. The following is the c.d.f. of a r.v. X.

Find the probability distribution of X and P(-1 ≤ X ≤ 2).
Solution:

P(-1 ≤ X ≤ 2) = p(-1) + p(0) + p(1) + p(2)
= 0.2 + 0.15 + 0.10 + 0.10
= 0.55
In simple words: First, derive the p.m.f. (probability mass function) from the given c.d.f. (cumulative distribution function) by subtracting successive F(x) values. Then, sum the p.m.f. values for X from -1 to 2 to find P(-1 ≤ X ≤ 2).

🎯 Exam Tip: Remember that for a discrete random variable, p(x) = F(x) - F(x-1). Always ensure the sum of p(x) equals 1 and F(x) is non-decreasing.

Question 10. Find the expected value and variance of the r.v. X if its probability distribution is as follows.
(i)

Solution:
E(X) = \( \Sigma x \cdot p(x) \)
= \( 1 \left( \frac{1}{5} \right) + 2 \left( \frac{2}{5} \right) + 3 \left( \frac{2}{5} \right) \)
= \( \frac{1}{5} + \frac{4}{5} + \frac{6}{5} \)
= \( \frac{11}{5} \)
= 2.2
E(\(X^2\)) = \( \Sigma x^2 \cdot p(x) \)
= \( 1^2 \left( \frac{1}{5} \right) + 2^2 \left( \frac{2}{5} \right) + 3^2 \left( \frac{2}{5} \right) \)
= \( \frac{1}{5} + \frac{8}{5} + \frac{18}{5} \)
= \( \frac{27}{5} \)
= 5.4
Var (X) = E(\(X^2\)) – [E(X)]\(^2\)
= (5.4) - (2.2)\(^2\)
= 5.4 - 4.84
= 0.56
S.D. of X = \( \sigma_x \) = \( \sqrt{Var(X)} \) = \( \sqrt{0.56} \)
= 0.7483
In simple words: To find the expected value, multiply each possible value of X by its probability and sum them up. For variance, calculate the expected value of X squared, then subtract the square of the expected value of X. Standard deviation is simply the square root of the variance.

🎯 Exam Tip: Be careful with calculations, especially squaring and fractions. Ensure you use the correct formulas for E(X) and Var(X).

(ii)

Solution:
E(X) = \( \Sigma x \cdot p(x) \)
= \( (-1) \left( \frac{1}{5} \right) + 0 \left( \frac{2}{5} \right) + 1 \left( \frac{2}{5} \right) \)
= \( -\frac{1}{5} + 0 + \frac{2}{5} \)
= \( \frac{1}{5} \)
= 0.2
E(\(X^2\)) = \( \Sigma x^2 \cdot p(x) \)
= \( (-1)^2 \left( \frac{1}{5} \right) + 0^2 \left( \frac{2}{5} \right) + 1^2 \left( \frac{2}{5} \right) \)
= \( \frac{1}{5} + 0 + \frac{2}{5} \)
= \( \frac{3}{5} \)
= 0.6
Var (X) = E(\(X^2\)) – [E(X)]\(^2\)
= 0.6 - (0.2)\(^2\)
= 0.6 - 0.04
= 0.56
S.D. of X = \( \sigma_x \) = \( \sqrt{Var(X)} \) = \( \sqrt{0.56} \)
= 0.7483
In simple words: Follow the same process as part (i): compute E(X) by summing \(x \cdot p(x)\), and E(\(X^2\)) by summing \(x^2 \cdot p(x)\). Then, use the formula Var(X) = E(\(X^2\)) - [E(X)]\(^2\) to find the variance, and its square root for standard deviation.

🎯 Exam Tip: Pay attention to negative values of X, as squaring them makes them positive for E(\(X^2\)). Double-check arithmetic.

(iii)

Solution:
E(X) = \( \Sigma x \cdot p(x) \)
= \( 1 \left( \frac{1}{n} \right) + 2 \left( \frac{1}{n} \right) + 3 \left( \frac{1}{n} \right) + \ldots + n \left( \frac{1}{n} \right) \)
= \( \frac{1}{n} (1+2+3+\ldots+n) \)
= \( \frac{1}{n} \cdot \frac{n(n+1)}{2} \)
\[ \therefore \Sigma r = \frac{n(n+1)}{2} \]
= \( \frac{n+1}{2} \)
E(\(X^2\)) = \( \Sigma x^2 \cdot p(x) \)
= \( 1^2 \left( \frac{1}{n} \right) + 2^2 \left( \frac{1}{n} \right) + 3^2 \left( \frac{1}{n} \right) + \ldots + n^2 \left( \frac{1}{n} \right) \)
= \( \frac{1}{n} (1^2+2^2+3^2+\ldots+n^2) \)
= \( \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} \)
\[ \therefore \Sigma r^2 = \frac{n(n+1)(2n+1)}{6} \]
= \( \frac{(n+1)(2n+1)}{6} \)
Var(X) = E(\(X^2\)) – [E(X)]\(^2\)
= \( \frac{(n+1)(2n+1)}{6} - \left( \frac{n+1}{2} \right)^2 \)
= \( \frac{n+1}{2} \left[ \frac{2n+1}{3} - \frac{n+1}{2} \right] \)
= \( \frac{n+1}{2} \left[ \frac{2(2n+1) - 3(n+1)}{6} \right] \)
= \( \frac{n+1}{2} \left[ \frac{4n+2 - 3n - 3}{6} \right] \)
= \( \frac{n+1}{2} \left[ \frac{n-1}{6} \right] \)
= \( \frac{n^2-1}{12} \)
S.D. of X = \( \sigma_x \) = \( \sqrt{Var(X)} \) = \( \sqrt{\frac{n^2-1}{12}} \)
In simple words: For a uniform discrete distribution, calculate E(X) by summing the values of X multiplied by their constant probability (1/n). Then, find E(\(X^2\)) using the sum of squares formula. Finally, apply the variance formula using these results.

🎯 Exam Tip: Remember the summation formulas for the first 'n' natural numbers and the sum of their squares to efficiently solve such problems.

(iv)

Solution:
E(X) = \( \Sigma x \cdot p(x) \)
= \( 0 \left( \frac{1}{32} \right) + 1 \left( \frac{5}{32} \right) + 2 \left( \frac{10}{32} \right) + 3 \left( \frac{10}{32} \right) + 4 \left( \frac{5}{32} \right) + 5 \left( \frac{1}{32} \right) \)
= \( 0 + \frac{5}{32} + \frac{20}{32} + \frac{30}{32} + \frac{20}{32} + \frac{5}{32} \)
= \( \frac{80}{32} \)
= \( \frac{5}{2} \)
= 2.5
E(\(X^2\)) = \( \Sigma x^2 \cdot p(x) \)
= \( 0^2 \left( \frac{1}{32} \right) + 1^2 \left( \frac{5}{32} \right) + 2^2 \left( \frac{10}{32} \right) + 3^2 \left( \frac{10}{32} \right) + 4^2 \left( \frac{5}{32} \right) + 5^2 \left( \frac{1}{32} \right) \)
= \( 0 + \frac{5}{32} + \frac{40}{32} + \frac{90}{32} + \frac{80}{32} + \frac{25}{32} \)
= \( \frac{240}{32} \)
= \( \frac{15}{2} \)
= 7.5
Var(X) = E(\(X^2\)) – [E(X)]\(^2\)
= 7.5 - (2.5)\(^2\)
= 7.5 - 6.25
= 1.25
S.D. of X = \( \sigma_x \) = \( \sqrt{Var(X)} \)
= \( \sqrt{1.25} \)
= 1.118
In simple words: This distribution represents a binomial probability, where X values are the number of successes. Calculate E(X) by multiplying each X by its P(X=x) and summing. Then, calculate E(\(X^2\)) similarly. Finally, apply the variance formula: E(\(X^2\)) minus the square of E(X), and take the square root for the standard deviation.

🎯 Exam Tip: Recognize binomial probability mass functions; for B(n,p), E(X) = np and Var(X) = npq. For this specific case (n=5, p=0.5), E(X)=2.5 and Var(X)=1.25, which can serve as a quick check for your calculations.

Question 11. A player tosses two coins. He wins Rs. 10 if 2 heads appear, Rs. 5 if 1 head appears, and Rs. 2 if no head appears. Find the expected value and variance of the winning amount.
Solution:
S: Two fair coin are tossed
S = {HH, HT, TT, TH}
n(S) = 4
\(\therefore\) Range of X = {0, 1, 2} (where X is the number of heads)
\(\therefore\) Let Y = amount received corresponds to values of X

Expected winning amount
E(Y) = \( \Sigma py \) = \( \frac{22}{4} \) = Rs. 5.5
V(Y) = \( \Sigma py^2 \) – (\( \Sigma py \))\(^2\)
= \( \frac{154}{4} \) - (5.5)\(^2\)
= 38.5 - 30.25
= Rs. 8.25
In simple words: First, define the possible outcomes and their probabilities for tossing two coins. Then, map these outcomes to the winning amounts (Y) and their corresponding probabilities. Calculate the expected value E(Y) by summing \(Y \cdot P(Y)\). For variance, calculate E(\(Y^2\)) and use the formula Var(Y) = E(\(Y^2\)) - [E(Y)]\(^2\).

🎯 Exam Tip: Clearly define the random variable (winning amount Y in this case) and its probability distribution before starting calculations. Carefully distinguish between X (number of heads) and Y (winning amount).

Question 12. Let the p.m.f. of the r.v. X be
\[ p(x) = \begin{cases} \frac{3-x}{10} & \text{for } x = -1, 0, 1, 2 \\ 0 & \text{otherwise} \end{cases} \]
Calculate E(X) and Var(X).
Solution:
For x = -1,
p(-1) = \( \frac{3-(-1)}{10} \) = \( \frac{4}{10} \)
For x = 0,
p(0) = \( \frac{3-0}{10} \) = \( \frac{3}{10} \)
For x = 1,
p(1) = \( \frac{3-1}{10} \) = \( \frac{2}{10} \)
For x = 2,
p(2) = \( \frac{3-2}{10} \) = \( \frac{1}{10} \)

E(X) = \( \Sigma xp \) = 0
V(X) = \( \Sigma x^2 p \) – (\( \Sigma xp \))\(^2\)
= \( \frac{10}{10} \) - (0)\(^2\)
= 1 - 0
= 1
In simple words: First, calculate the individual probabilities P(x) for each given x value using the p.m.f. formula. Then, compute the expected value E(X) by summing \(x \cdot P(x)\). Finally, calculate the variance V(X) using E(\(X^2\)) - [E(X)]\(^2\), where E(\(X^2\)) is the sum of \(x^2 \cdot P(x)\).

🎯 Exam Tip: Always verify that your calculated probabilities sum to 1 before proceeding with E(X) and Var(X) calculations. A tabular format helps organize calculations neatly.

Question 13. Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\[ f(x) = \begin{cases} k (4-x^2) & \text{for } -2 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases} \]
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
We know that
\[ \int_{-2}^{2} k(4-x^2)dx = 1 \]
\( \implies k \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = 1 \)
\( \implies k \left[ \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) \right] = 1 \)
\( \implies k \left[ \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) \right] = 1 \)
\( \implies k \left[ 8 - \frac{8}{3} + 8 - \frac{8}{3} \right] = 1 \)
\( \implies k \left[ 16 - \frac{16}{3} \right] = 1 \)
\( \implies k \left[ \frac{48-16}{3} \right] = 1 \)
\( \implies k \left[ \frac{32}{3} \right] = 1 \)
\(\therefore\) k = \( \frac{3}{32} \)
(i) P(X > 0) = \( \int_{0}^{2} \frac{3}{32} (4-x^2)dx \)
= \( \frac{3}{32} \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} \)
= \( \frac{3}{32} \left[ \left( 4(2) - \frac{2^3}{3} \right) - \left( 0 \right) \right] \)
= \( \frac{3}{32} \left[ 8 - \frac{8}{3} \right] \)
= \( \frac{3}{32} \left[ \frac{24-8}{3} \right] \)
= \( \frac{3}{32} \cdot \frac{16}{3} \)
= \( \frac{16}{32} \)
= \( \frac{1}{2} \)
(ii) P(-1 < X < 1) = \( \int_{-1}^{1} \frac{3}{32} (4-x^2)dx \)
= \( \frac{3}{32} \left[ 4x - \frac{x^3}{3} \right]_{-1}^{1} \)
= \( \frac{3}{32} \left[ \left( 4(1) - \frac{1^3}{3} \right) - \left( 4(-1) - \frac{(-1)^3}{3} \right) \right] \)
= \( \frac{3}{32} \left[ \left( 4 - \frac{1}{3} \right) - \left( -4 + \frac{1}{3} \right) \right] \)
= \( \frac{3}{32} \left[ 4 - \frac{1}{3} + 4 - \frac{1}{3} \right] \)
= \( \frac{3}{32} \left[ 8 - \frac{2}{3} \right] \)
= \( \frac{3}{32} \left[ \frac{24-2}{3} \right] \)
= \( \frac{3}{32} \cdot \frac{22}{3} \)
= \( \frac{22}{32} \)
= \( \frac{11}{16} \)
(iii) P(X < -0.5 or X > 0.5) = P(X < -0.5) + P(X > 0.5)
= \( \int_{-2}^{-0.5} \frac{3}{32} (4-x^2)dx + \int_{0.5}^{2} \frac{3}{32} (4-x^2)dx \)
= \( \frac{3}{32} \left[ 4x - \frac{x^3}{3} \right]_{-2}^{-0.5} + \frac{3}{32} \left[ 4x - \frac{x^3}{3} \right]_{0.5}^{2} \)
= \( \frac{3}{32} \left[ \left( 4(-0.5) - \frac{(-0.5)^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) \right] + \frac{3}{32} \left[ \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(0.5) - \frac{(0.5)^3}{3} \right) \right] \)
= \( \frac{3}{32} \left[ \left( -2 - \frac{-0.125}{3} \right) - \left( -8 + \frac{8}{3} \right) \right] + \frac{3}{32} \left[ \left( 8 - \frac{8}{3} \right) - \left( 2 - \frac{0.125}{3} \right) \right] \)
= \( \frac{3}{32} \left[ \left( -2 + \frac{1}{24} \right) - \left( -8 + \frac{8}{3} \right) + \left( 8 - \frac{8}{3} \right) - \left( 2 - \frac{1}{24} \right) \right] \)
= \( \frac{3}{32} \left[ -2 + \frac{1}{24} + 8 - \frac{8}{3} + 8 - \frac{8}{3} - 2 + \frac{1}{24} \right] \)
= \( \frac{3}{32} \left[ 12 - \frac{16}{3} + \frac{2}{24} \right] \)
= \( \frac{3}{32} \left[ \frac{12 \cdot 24 - 16 \cdot 8 + 2}{24} \right] \)
= \( \frac{3}{32} \left[ \frac{288 - 128 + 2}{24} \right] \)
= \( \frac{3}{32} \left[ \frac{162}{24} \right] \)
= \( \frac{162}{32 \cdot 8} \)
= \( \frac{162}{256} \)
= \( \frac{81}{128} \)
= 0.633
In simple words: First, find the constant 'k' by integrating the probability density function over its entire range and setting it equal to 1. Then, for each probability (P(X>0), P(-10.5)), integrate the p.d.f. with the calculated 'k' over the specified intervals.

🎯 Exam Tip: Remember that for a continuous random variable, the total probability over its entire domain is 1. Pay close attention to the integration limits for each probability calculation and handle negative numbers carefully.

Question 14. The p.d.f. of the r.v. X is given by
\[ f(x) = \begin{cases} \frac{1}{2a} & \text{for } 0 < x < 2a \\ 0 & \text{otherwise} \end{cases} \]
Show that P(X < \( \frac{a}{2} \)) = P(X > \( \frac{3a}{2} \))
Solution:
P(X < \( \frac{a}{2} \)) = \( \int_{0}^{\frac{a}{2}} \frac{1}{2a} dx \)
= \( \frac{1}{2a} [x]_{0}^{\frac{a}{2}} \)
= \( \frac{1}{2a} \left[ \frac{a}{2} - 0 \right] \)
= \( \frac{1}{2a} \cdot \frac{a}{2} \)
= \( \frac{1}{4} \)
P(X > \( \frac{3a}{2} \)) = \( \int_{\frac{3a}{2}}^{2a} \frac{1}{2a} dx \)
= \( \frac{1}{2a} [x]_{\frac{3a}{2}}^{2a} \)
= \( \frac{1}{2a} \left[ 2a - \frac{3a}{2} \right] \)
= \( \frac{1}{2a} \left[ \frac{4a-3a}{2} \right] \)
= \( \frac{1}{2a} \cdot \frac{a}{2} \)
= \( \frac{1}{4} \)
\(\therefore\) P(X < \( \frac{a}{2} \)) = P(X > \( \frac{3a}{2} \))
In simple words: To show the equality, calculate the probability P(X < a/2) by integrating the given p.d.f. from 0 to a/2. Then, calculate P(X > 3a/2) by integrating the p.d.f. from 3a/2 to 2a. If both integrals yield the same result, the equality is proven.

🎯 Exam Tip: For a uniform distribution, probability over an interval is simply (length of interval) / (total length of domain). This can serve as a quick check for your integral calculations.

Question 15. Determine k if
\[ f(x) = \begin{cases} ke^{-\theta x} & \text{for } 0 \leq x < \infty, \theta > 0 \\ 0 & \text{otherwise} \end{cases} \]
is the p.d.f. of the r.v. X. Also find P(X > \( \frac{1}{\theta} \)). Find M if P(0 < X < M) = \( \frac{1}{2} \).
Solution:
We know that
\[ \int_{0}^{\infty} ke^{-\theta x} dx = 1 \]
\( \implies k \left[ \frac{e^{-\theta x}}{-\theta} \right]_{0}^{\infty} = 1 \)
\( \implies -\frac{k}{\theta} [e^{-\theta x}]_{0}^{\infty} = 1 \)
\( \implies -\frac{k}{\theta} [e^{-\infty} - e^0] = 1 \)
\( \implies -\frac{k}{\theta} [0 - 1] = 1 \)
\( \implies \frac{k}{\theta} = 1 \)
\(\therefore\) k = \( \theta \)
P(X > \( \frac{1}{\theta} \)) = \( \int_{\frac{1}{\theta}}^{\infty} \theta e^{-\theta x} dx \)
= \( \theta \left[ \frac{e^{-\theta x}}{-\theta} \right]_{\frac{1}{\theta}}^{\infty} \)
= \( -[e^{-\theta x}]_{\frac{1}{\theta}}^{\infty} \)
= \( -[e^{-\infty} - e^{-\theta \left( \frac{1}{\theta} \right)}] \)
= \( -[0 - e^{-1}] \)
= \( e^{-1} \)
P(0 < X < M) = \( \frac{1}{2} \)
\[ \int_{0}^{M} \theta e^{-\theta x} dx = \frac{1}{2} \]
\( \implies \theta \left[ \frac{e^{-\theta x}}{-\theta} \right]_{0}^{M} = \frac{1}{2} \)
\( \implies -[e^{-\theta x}]_{0}^{M} = \frac{1}{2} \)
\( \implies -[e^{-\theta M} - e^0] = \frac{1}{2} \)
\( \implies -[e^{-\theta M} - 1] = \frac{1}{2} \)
\( \implies 1 - e^{-\theta M} = \frac{1}{2} \)
\( \implies e^{-\theta M} = 1 - \frac{1}{2} \)
\( \implies e^{-\theta M} = \frac{1}{2} \)
\( \implies -\theta M = \text{log} \left( \frac{1}{2} \right) \)
\( \implies -\theta M = -\text{log} 2 \)
\( \implies M = \frac{\text{log} 2}{\theta} \)
In simple words: First, find the value of k by integrating the p.d.f. from 0 to infinity and setting it to 1. Then, calculate P(X > 1/θ) by integrating the p.d.f. from 1/θ to infinity. Finally, to find M, integrate the p.d.f. from 0 to M and set the result equal to 1/2.

🎯 Exam Tip: This problem involves an exponential distribution. Remember that \( \int e^{-ax} dx = -\frac{1}{a} e^{-ax} \) and \( e^{-\infty} = 0 \). Be careful with the signs during integration and logarithmic properties.

Question 16. The p.d.f. of the r.v. X is given by
\[ f_X(x) = \begin{cases} \frac{k}{\sqrt{x}} & \text{for } 0 < x < 4 \\ 0 & \text{otherwise} \end{cases} \]
Determine k, c.d.f. of X and hence find P(X ≤ 2) and P(X ≥ 1).
Solution:
We know that
\[ \int_{0}^{4} \frac{k}{\sqrt{x}} dx = 1 \]
\( \implies k \int_{0}^{4} x^{-\frac{1}{2}} dx = 1 \)
\( \implies k \left[ \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{0}^{4} = 1 \)
\( \implies k [2\sqrt{x}]_{0}^{4} = 1 \)
\( \implies 2k [\sqrt{4} - \sqrt{0}] = 1 \)
\( \implies 2k [2 - 0] = 1 \)
\( \implies 4k = 1 \)
\(\therefore\) k = \( \frac{1}{4} \)
c.d.f. of f(x) is
F(x) = \( \int_{0}^{x} \frac{k}{\sqrt{t}} dt \)
\(\therefore\) F(x) = \( \int_{0}^{x} \frac{1}{4} t^{-\frac{1}{2}} dt \)
= \( \frac{1}{4} \left[ \frac{t^{\frac{1}{2}}}{\frac{1}{2}} \right]_{0}^{x} \)
= \( \frac{1}{4} [2\sqrt{t}]_{0}^{x} \)
= \( \frac{1}{4} [2\sqrt{x} - 0] \)
= \( \frac{2\sqrt{x}}{4} \)
= \( \frac{\sqrt{x}}{2} \)
P(X ≤ 2) = F(2)
= \( \frac{\sqrt{2}}{2} \)
= 0.7071
P(X ≥ 1) = 1 - P(X < 1) = 1 - F(1)
= \( 1 - \frac{\sqrt{1}}{2} \)
= \( 1 - \frac{1}{2} \)
= 0.5
In simple words: First, find the constant 'k' by integrating the p.d.f. over its domain (0 to 4) and setting the result to 1. Next, determine the c.d.f. F(x) by integrating the p.d.f. from 0 to x. Finally, use the c.d.f. to find P(X ≤ 2) as F(2) and P(X ≥ 1) as 1 - F(1).

🎯 Exam Tip: Integration involving fractional exponents like \( \frac{1}{\sqrt{x}} \) often causes errors. Remember \( \int x^n dx = \frac{x^{n+1}}{n+1} \) and handle the limits of integration carefully.

Question 17. Let X denote the reaction temperature (in °C) of a certain chemical process. Let X be a continuous r.v. with p.d.f.
\[ f(x) = \begin{cases} \frac{1}{10} & \text{for } -5 \leq x \leq 5 \\ 0 & \text{otherwise} \end{cases} \]
Compute P(X < 0).
Solution:
Let its c.d.f. F(x) be given by
F(x) = \( \int_{-5}^{x} f(y)dy \), where \( f(y) = \frac{1}{10} \)
= \( \int_{-5}^{x} \frac{1}{10} dy \)
= \( \frac{1}{10} [y]_{-5}^{x} \)
= \( \frac{1}{10} [x - (-5)] \)
= \( \frac{1}{10} [x+5] \)
= \( \frac{x}{10} + \frac{5}{10} \)
= \( \frac{x}{10} + \frac{1}{2} \)
P(X < 0) = F(0)
= \( \frac{0}{10} + \frac{1}{2} \)
= \( \frac{1}{2} \)
In simple words: First, find the cumulative distribution function (c.d.f.) F(x) by integrating the given probability density function (p.d.f.) from the lower bound of the domain (-5) up to x. Then, to compute P(X < 0), simply evaluate F(0) using the derived c.d.f.

🎯 Exam Tip: For a uniform distribution, the c.d.f. is a linear function. A quick check for P(X < 0) for this symmetric distribution is that it should be 0.5, given that the mean is 0.

Part - II

Question 1. Let X ~ B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8)
Solution:
X ~ B(10, 0.2)
n = 10, p = 0.2
\(\therefore\) q = 1 – p = 1 – 0.2 = 0.8
(i) P(X = 1) = \( {}^{10}\text{C}_1 (0.2)^1 (0.8)^9 \)
= 0.2684
(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= \( 1 - {}^{10}\text{C}_0 (0.2)^0 (0.8)^{10} \)
= 1 - 0.1074
= 0.8926
(iii) P(X ≤ 8) = 1 – P(X > 8)
= 1 – [P(X=9) + P(X=10)]
= \( 1 - [{}^{10}\text{C}_9 (0.2)^9 (0.8)^1 + {}^{10}\text{C}_{10} (0.2)^{10}] \)
= 1 - 0.00000041984
= 0.9999
In simple words: This problem involves a binomial distribution. For each part, use the binomial probability formula P(X=x) = nCx * p^x * q^(n-x). For P(X ≥ 1), it's easier to calculate 1 - P(X=0). For P(X ≤ 8), calculate 1 - P(X > 8), which means 1 - [P(X=9) + P(X=10)].

🎯 Exam Tip: Remember the complementary probability rule: P(X ≥ k) = 1 - P(X < k) and P(X ≤ k) = 1 - P(X > k). This often simplifies calculations for binomial distributions.

Question 3.
If a fair coin is tossed 4 times, find the probability that it shows (i) 3 heads, (ii) head in the first 2 tosses, and tail in the last 2 tosses.
Solution:
n: No. of times a coin is tossed
\( \implies \) n = 4
X: No. of heads
P: Probability of getting heads
\( \implies p = \frac{1}{2} \)
\( \implies q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \implies X \sim B \left( 4, \frac{1}{2} \right) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
(i) P(3 heads)
P(X = 3) = \( \binom{4}{3} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^{4-3} \)
\( = 4 \times \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^1 \)
\( = 4 \times \left( \frac{1}{2} \right)^4 \)
\( = 4 \times \frac{1}{16} \)
\( = \frac{1}{4} \)
(ii) P(Head in 1st two tosses and tail in last 2 tosses)
P(X = 2) = \( \binom{4}{2} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^{4-2} \)
\( = 6 \times \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^2 \)
\( = 6 \times \left( \frac{1}{2} \right)^4 \)
\( = 6 \times \frac{1}{16} \)
\( = \frac{3}{8} \)
In simple words: This problem involves binomial distribution to calculate probabilities for coin toss outcomes, specifically for a certain number of heads and a specific sequence of heads and tails.

🎯 Exam Tip: Remember that for a fair coin, the probability of heads (p) and tails (q) is always 1/2. Binomial distribution is key for multiple independent trials.

Question 4.
The probability that a bomb will hit the target is 0.8. Find the probability that, out of 5 bombs, exactly 2 will miss the target.
Solution:
X: No. of bombs miss the target
p: Probability that bomb miss the target
\( \implies \) q = 0.8 (probability of hitting the target)
\( \implies \) p = 1 - q = 1 - 0.8 = 0.2 (probability of missing the target)
n = No. of bombs = 5
\( \implies X \sim B(5, 0.2) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
P(X = 2) = \( \binom{5}{2} (0.2)^2 (0.8)^{5-2} \)
\( = 10 \times 0.04 \times (0.8)^3 \)
\( = 10 \times 0.04 \times 0.512 \)
\( = 0.4 \times 0.512 \)
\( = 0.2048 \)
In simple words: We calculate the probability of a bomb missing the target and then use the binomial distribution to find the chance that exactly two out of five bombs will miss.

🎯 Exam Tip: Clearly define 'success' (in this case, missing the target) to correctly determine p and q. Pay attention to the exponent calculations for p and q.

Question 5.
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom can not be used on a random occasion.
Solution:
X: No. of lamps not burning
p: Probability that the lamp is not burning
\( \implies \) q = 0.3 (probability of burning)
\( \implies \) p = 1 - q = 1 - 0.3 = 0.7 (probability of not burning)
n = No. of lamps fitted = 3
\( \implies X \sim B(3, 0.7) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
P(classroom cannot be used) = P(X < 2) = p(0) + p(1)
\( = \binom{3}{0} (0.7)^0 (0.3)^{3-0} + \binom{3}{1} (0.7)^1 (0.3)^{3-1} \)
\( = 1 \times 1 \times (0.3)^3 + 3 \times 0.7 \times (0.3)^2 \)
\( = (0.3)^2 [0.3 + 3 \times 0.7] \)
\( = 0.09 [0.3 + 2.1] \)
\( = 0.09 [2.4] \)
\( = 0.216 \)
In simple words: We define the event of a lamp not burning as 'success' and use binomial distribution to find the probability that fewer than two lamps are burning, which makes the classroom unusable.

🎯 Exam Tip: Carefully read the condition for 'unusable' (less than 2 burning) to correctly identify the events (p(0) + p(1)) needed for calculation.

Question 6.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
X: No. of defective items
n: No. of items selected = 4
p: Probability of getting defective items
\( \implies \) p = 0.1
\( \implies \) q = 1 - p = 1 - 0.1 = 0.9
P(At most one defective item) = P(X \( \le \) 1) = p(0) + p(1)
\( = \binom{4}{0} (0.1)^0 (0.9)^{4-0} + \binom{4}{1} (0.1)^1 (0.9)^{4-1} \)
\( = 1 \times 1 \times (0.9)^4 + 4 \times 0.1 \times (0.9)^3 \)
\( = (0.9)^3 [0.9 + 4 \times 0.1] \)
\( = (0.9)^3 \times [0.9 + 0.4] \)
\( = 0.729 \times 1.3 \)
\( = 0.9477 \)
In simple words: Given the defective rate, we use binomial distribution to calculate the probability of finding zero or one defective item among four selected items.

🎯 Exam Tip: 'At most one' means X=0 or X=1. Ensure correct calculation of powers for p and q and binomial coefficients.

Question 7.
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Solution:
p = 0.6 (probability of survival)
q = 1 - 0.6 = 0.4 (probability of not surviving)
n = 4 (number of components tested)
x = 2 (number of components surviving)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
P(X = 2) = \( \binom{4}{2} (0.6)^2 (0.4)^2 \)
\( = 6 \times 0.36 \times 0.16 \)
\( = 0.3456 \)
In simple words: We apply the binomial probability formula with the given survival rate and number of trials to find the chance of exactly two components surviving.

🎯 Exam Tip: Identify n, p, q, and x correctly from the problem statement to apply the binomial formula accurately.

Question 8.
An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer randomly. Find the probability that this student gets 4 or more correct answers.
Solution:
n: No. of multiple-choice questions
\( \implies \) n = 5
X: No. of correct answers
p: Probability of getting correct answer
\( \implies \) There are 4 options out of which one is correct
\( \implies p = \frac{1}{4} \)
\( \implies q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \)
\( \implies X \sim B \left( 5, \frac{1}{4} \right) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
P(Four or more correct answers) = P(X \( \ge \) 4) = p(4) + p(5)
\( = \binom{5}{4} \left( \frac{1}{4} \right)^4 \left( \frac{3}{4} \right)^{5-4} + \binom{5}{5} \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^{5-5} \)
\( = 5 \times \left( \frac{1}{4} \right)^4 \left( \frac{3}{4} \right)^1 + 1 \times \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0 \)
\( = 5 \times \frac{1}{256} \times \frac{3}{4} + 1 \times \frac{1}{1024} \times 1 \)
\( = \frac{15}{1024} + \frac{1}{1024} \)
\( = \frac{16}{1024} \)
\( = \frac{1}{64} \)
In simple words: For a guessing student, the probability of a correct answer is 1/4. We use binomial distribution to find the probability of getting 4 or 5 correct answers out of 5 questions.

🎯 Exam Tip: 'Four or more' means calculating for X=4 and X=5 and summing their probabilities. Simplify fractions carefully at the end.

Question 9.
The probability that a machine will produce all bolts in a production run with in the specification is 0.9. A sample of 3 machines is taken at random. Calculate the probability that all machines will produce all bolts in a production run within the specification.
Solution:
n: No. of samples selected
\( \implies \) n = 3
X: No. of bolts produce by machines
p: Probability of getting bolts
\( \implies \) p = 0.9
\( \implies \) q = 1 - p = 1 - 0.9 = 0.1
\( \implies X \sim B(3, 0.9) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
P(Machine will produce all bolts) = P(X = 3)
\( = \binom{3}{3} (0.9)^3 (0.1)^{3-3} \)
\( = 1 \times (0.9)^3 \times (0.1)^0 \)
\( = 1 \times (0.9)^3 \times 1 \)
\( = (0.9)^3 \)
\( = 0.729 \)
In simple words: Given the probability of a machine producing bolts within specification, we use the binomial distribution to find the chance that all three sampled machines meet this specification.

🎯 Exam Tip: 'All machines' implies x=n. Remember that any non-zero number raised to the power of 0 is 1.

Question 10.
A computer installation has 3 terminals. The probability that anyone terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that (i) 0 (ii) 1 terminal requires attention during a week.
Solution:
n: No. of terminals
\( \implies \) n = 3
X: No. of terminals need attention
p: Probability of getting terminals need attention
\( \implies \) p = 0.1
\( \implies \) q = 1 - p = 1 - 0.1 = 0.9
\( \implies X \sim B(3, 0.1) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
(i) P(No attention)
\( \implies P(X = 0) = \binom{3}{0} \times (0.1)^0 (0.9)^{3-1} \)
\( = 1 \times 1 \times (0.9)^3 \)
\( = 0.729 \)
(ii) P(One terminal need attention)
\( \implies P(X = 1) = \binom{3}{1} (0.1)^1 (0.9)^{3-1} \)
\( = 3 \times 0.1 \times (0.9)^2 \)
\( = 0.3 \times 0.81 \)
\( = 0.243 \)
In simple words: We calculate the binomial probabilities for zero and one terminal needing attention, given the individual probability and total number of terminals.

🎯 Exam Tip: Clearly differentiate between 'no attention' (x=0) and 'one attention' (x=1) in your calculations. Ensure correct binomial coefficient and power calculations.

Question 11.
In a large school, 80% of the students like mathematics. A visitor asks each of 4 students, selected at random, whether they like mathematics, (i) Calculate the probabilities of obtaining an answer yes from all of the selected students, (ii) Find the probability that the visitor obtains the answer yes from at least 3 students.
Solution:
X: No. of students like mathematics
p: Probability that students like mathematics
\( \implies \) p = 0.8
\( \implies \) q = 1 - p = 1 - 0.8 = 0.2
n: No. of students selected
\( \implies \) n = 4
\( \implies X \sim B(4, 0.8) \)
\( \implies p(x) = \binom{n}{x} p^x q^{n-x} \)
(i) P(All students like mathematics)
\( \implies P(X = 4) = \binom{4}{4} (0.8)^4 (0.2)^{4-4} \)
\( = 1 \times (0.8)^4 \times (0.2)^0 \)
\( = 1 \times (0.8)^4 \times 1 \)
\( = 0.4096 \)
(ii) P(At least 3 students like mathematics)
\( \implies P(X \ge 3) = p(3) + p(4) \)
\( = \binom{4}{3} (0.8)^3 (0.2)^{4-3} + 0.4096 \)
\( = 4 \times (0.8)^3 (0.2)^1 + 0.4096 \)
\( = 0.8 \times (0.8)^3 + 0.4096 \)
\( = (0.8)^4 + 0.4096 \)
\( = 0.4096 + 0.4096 \)
\( = 0.8192 \)
In simple words: We calculate the probabilities for all 4 students liking math and for at least 3 students liking math, using the binomial distribution with the given percentage.

🎯 Exam Tip: 'All students' means x=n, and 'at least 3' means x=3 or x=4. Utilize previous calculations (like p(4)) to save time.

Question 12.
It is observed that it rains on 10 days out of 30 days. Find the probability that (i) it rains on exactly 3 days of a week. (ii) it rains at most 2 days a week.
Solution:
X: No. of days it rains in a week
p: Probability that it rains
\( \implies p = \frac{10}{30} = \frac{1}{3} \)
\( \implies q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \)
n: No. of days in a week
\( \implies \) n = 7
\( \implies X \sim B \left( 7, \frac{1}{3} \right) \)
(i) P(Rains on Exactly 3 days of a week)
\( \implies P(X = 3) = \binom{7}{3} \left( \frac{1}{3} \right)^3 \left( \frac{2}{3} \right)^{7-3} \)
\( = 35 \times \left( \frac{1}{3} \right)^3 \left( \frac{2}{3} \right)^4 \)
\( = 35 \times \frac{1}{27} \times \frac{16}{81} \)
\( = \frac{560}{2187} \)
(ii) P(Rains on at most 2 days of a week)
\( \implies P(X \le 2) = p(0) + p(1) + p(2) \)
\( = \binom{7}{0} \left( \frac{1}{3} \right)^0 \left( \frac{2}{3} \right)^{7-0} + \binom{7}{1} \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^{7-1} \)
\( + \binom{7}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^{7-2} \)
\( = 1 \times 1 \times \left( \frac{2}{3} \right)^7 + 7 \times \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^6 + 21 \times \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^5 \)
\( = \left( \frac{2}{3} \right)^5 \left[ \left( \frac{2}{3} \right)^2 + 7 \left( \frac{1}{3} \right) \left( \frac{2}{3} \right) + 21 \left( \frac{1}{3} \right)^2 \right] \)
\( = \left( \frac{2}{3} \right)^5 \left[ \frac{4}{9} + \frac{14}{9} + \frac{21}{9} \right] \)
\( = \frac{32}{243} \left[ \frac{4+14+21}{9} \right] \)
\( = \frac{32}{243} \times \frac{39}{9} \)
\( = \frac{32 \times 13}{243 \times 3} \)
\( = \frac{416}{729} \)
In simple words: We determine the probability of rain per day and then use binomial distribution to find the chance of exactly 3 rainy days and at most 2 rainy days within a week.

🎯 Exam Tip: 'At most 2' means calculating p(0) + p(1) + p(2). Factor out common terms to simplify calculations for binomial sums.

Question 13.
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
Solution:
X: Follows Poisson Distribution
\( \implies p(x) = \frac{e^{-m} m^x}{x!} \)
Given P(X = 1) = 0.4 and P(X = 2) = 0.2
\( \implies \) P(X = 1) = 2 P(X = 2)
\( \implies \frac{e^{-m} m^1}{1!} = 2 \times \frac{e^{-m} m^2}{2!} \)
\( \implies e^{-m} m = 2 \times \frac{e^{-m} m^2}{2} \)
\( \implies e^{-m} m = e^{-m} m^2 \)
Since \( e^{-m} \ne 0 \), we can divide by \( e^{-m} \).
\( \implies m = m^2 \)
\( \implies m^2 - m = 0 \)
\( \implies m(m-1) = 0 \)
\( \implies m = 0 \) or \( m = 1 \)
Since m is a parameter of Poisson distribution, m > 0.
Therefore, m = 1.
Mean = m = 1
Variance of X = m = 1
In simple words: By using the given probabilities for a Poisson distribution, we establish a relationship to solve for the parameter 'm', which represents both the mean and variance.

🎯 Exam Tip: For a Poisson distribution, the mean and variance are both equal to the parameter 'm'. Setting up the ratio or relationship between given probabilities is crucial.

Question 14.
If X has Poisson distribution with parameter m, such that \( \frac{P(X=x+1)}{P(X=x)} = \frac{m}{x+1} \), find probabilities P(X = 1) and P(X = 2), when X follows Poisson distribution with m = 2 and P(X = 0) = 0.1353.
Solution:
Given that the random variable X follows the Poisson distribution with parameter m = 2.
i.e., X \( \sim \) P(2)
Its p.m.f. is satisfying the given equation.
\( \frac{P(X=x+1)}{P(X=x)} = \frac{m}{x+1} \)
When x = 0,
\( \frac{P(X=1)}{P(X=0)} = \frac{2}{0+1} \)
P(X = 1) = 2P(X = 0)
= 2(0.1353)
= 0.2706
When x = 1,
\( \frac{P(X=2)}{P(X=1)} = \frac{2}{1+1} \)
P(X = 2) = P(X = 1)
= 0.2706
In simple words: Using the recursive property of Poisson probabilities and the given parameter 'm' and P(X=0), we calculate the probabilities for P(X=1) and P(X=2).

🎯 Exam Tip: The recursive formula for Poisson probabilities is a useful shortcut. Remember that for Poisson, P(X=x+1) = (m/(x+1)) * P(X=x).