GSEB Class 12 Maths Solutions Chapter 13 Probability Exercise 13.4

 

Question 1. State which of the following are not probability distribution of a random variable. Give reasons for your answers.
(i)
(ii)
(iii)
(iv)
Answer:
(i) Sum of probabilities \( = 0.4 + 0.4 + 0.2 = 1 \). This distribution is a probability distribution. It is valid because all probabilities are non-negative and their sum is 1.
(ii) In this distribution, \( P(X=3) = -0.1 \), which is a negative probability. Probabilities cannot be negative. Therefore, this distribution is not a probability distribution.
(iii) Sum of probabilities \( = 0.6 + 0.1 + 0.1 = 0.9 \neq 1 \). The sum of all probabilities must equal 1. Since it does not, this given distribution is not a probability distribution.
(iv) Sum of probabilities \( = 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 > 1 \). The total sum of probabilities must be exactly 1. Because the sum is greater than 1, this is not a probability distribution.

Exam Tip: To determine if a distribution is a probability distribution, always check two conditions: (1) All probabilities must be between 0 and 1 (inclusive), and (2) The sum of all probabilities must exactly equal 1.

 

Question 2. An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. What are the possible values of X, if X is a random variable? Is X a random variable?
Answer: When two balls are selected, they may be picked as Red-Red (RR), Red-Black (RB), Black-Red (BR), or Black-Black (BB). The variable X counts the number of black balls. So, X can have values 0 (no black balls), 1 (one black ball), or 2 (both black balls). Yes, X is a random variable because its value is a numerical outcome from a random process.
In simple words: You can get 0, 1, or 2 black balls when picking two balls. Yes, X is a random variable.

Exam Tip: To find possible values of a random variable, list all possible outcomes of the experiment and then count the desired characteristic (like number of black balls) for each outcome. A random variable assigns a numerical value to each outcome of a random experiment.

 

Question 3. Let X represents the difference between the number of heads and the number of tails obtained, when a coin is tossed 6 times. What are the possible values of X?
Answer: When a coin is tossed 6 times, we can determine the number of heads and tails. The variable X shows the difference between these counts.

The possible values of X are \( 0, 2, 4, 6 \). We consider the absolute difference between heads and tails, so the result is always non-negative.
In simple words: When a coin is tossed 6 times, the number of heads and tails can vary. The possible differences between the number of heads and tails are 0, 2, 4, or 6.

Exam Tip: When defining a random variable as a difference, always clarify if it's an absolute difference (always positive) or a signed difference. Here, "difference" typically implies absolute value unless otherwise specified.

 

Question 4. Find the probability distribution of:
(i) number of heads in two tosses of a coin.
(ii) number of tails in the simultaneous tosses of three coins.
(iii) number of heads in four tosses of a coin.
Answer:
(i) When a coin is tossed two times, the sample space is \( \{TT, TH, HT, HH\} \). Let X be the number of heads.
Zero success (No head) means two tails (TT). So, \( P(X=0) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
One success (1 head) means one head and one tail (TH, HT). So, \( P(X=1) = \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \).
Two successes (Both heads) means two heads (HH). So, \( P(X=2) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
The probability distribution is:

(ii) When three coins are tossed at the same time, the sample space is \( \{TTT, TTH, THT, HTT, THH, HTH, HHT, HHH\} \). Let X be the number of tails.
Zero success (No tail) means all heads (HHH). So, \( P(X=0) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
One success (1 tail) means 1 tail and 2 heads (THH, HTH, HHT). There are 3 such outcomes. So, \( P(X=1) = 3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{3}{8} \).
Two successes (2 tails) means 2 tails and 1 head (TTH, THT, HTT). There are 3 such outcomes. So, \( P(X=2) = 3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{3}{8} \).
Three successes (3 tails) means all tails (TTT). So, \( P(X=3) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
The probability distribution is:
(iii) When a coin is tossed 4 times: Let X be the number of heads.
Zero success (No head) means all tails (TTTT). So, \( P(X=0) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \).
One success (1 head) means 1 head and 3 tails (HTTT, THTT, TTHT, TTTH). There are \( 4C_1 = 4 \) such outcomes. So, \( P(X=1) = 4 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{4}{16} = \frac{1}{4} \).
Two successes (2 heads) means 2 heads and 2 tails (HHTT, HTHT, HTTH, THHT, THTH, TTHH). There are \( 4C_2 = 6 \) such outcomes. So, \( P(X=2) = 6 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{6}{16} = \frac{3}{8} \).
Three successes (3 heads) means 3 heads and 1 tail (HHHT, HHTH, HTHH, THHH). There are \( 4C_3 = 4 \) such outcomes. So, \( P(X=3) = 4 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{4}{16} = \frac{1}{4} \).
Four successes (4 heads) means all heads (HHHH). So, \( P(X=4) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \).
Thus, the probability distribution is:In simple words: For each part, list all possible outcomes, count the number of heads or tails for each, and then calculate the probability for each possible count to build the distribution table.

Exam Tip: Remember that for n coin tosses, the total number of outcomes is \( 2^n \). For calculating probabilities of exactly k heads (or tails), you can use the binomial probability formula, \( P(X=k) = nC_k p^k (1-p)^{n-k} \), where p=0.5 for a fair coin.

 

Question 5. Find the probability distribution of the number of successes in two tosses of a dice, where a success is defined as
(i) number greater than 4.
(ii) six appeared on at least one dice.
Answer: When two dice are tossed, the total number of cases in the sample space is \( 6 \times 6 = 36 \).
(i) Success is defined as getting a number greater than 4 (i.e., 5 or 6). Let X be the number of successes.
Zero success (P(0)): No number greater than 4 appears on either die. This means both dice show 1, 2, 3, or 4. Such cases are \( \{(1,1), (1,2), \dots, (4,4)\} \), which are \( 4 \times 4 = 16 \) cases. So, \( P(X=0) = \frac{16}{36} = \frac{4}{9} \).
One success (P(1)): One die shows 5 or 6, and the other shows 1, 2, 3, or 4. The number of ways for this is \( (2 \times 4) + (4 \times 2) = 8 + 8 = 16 \) cases. So, \( P(X=1) = \frac{16}{36} = \frac{4}{9} \).
Two successes (P(2)): Both dice show 5 or 6. Cases are \( \{(5,5), (5,6), (6,5), (6,6)\} \), which are \( 2 \times 2 = 4 \) cases. So, \( P(X=2) = \frac{4}{36} = \frac{1}{9} \).
The probability distribution is as follows:

The table in the OCR is partial for (i). Let's use the calculations.
(ii) Success is defined as a six appearing on at least one die. Let X be the number of successes. Zero success (P(0)): A six does not appear on any die. This means both dice show 1, 2, 3, 4, or 5. Such cases are \( 5 \times 5 = 25 \) cases. So, \( P(X=0) = \frac{25}{36} \).
One success (P(1)): A six appears on one die, and the other die shows 1, 2, 3, 4, or 5. The number of ways for this is \( (1 \times 5) + (5 \times 1) = 5 + 5 = 10 \) cases. So, \( P(X=1) = \frac{10}{36} \).
Two successes (P(2)): Sixes appear on both dice. The case is \( \{(6,6)\} \), which is 1 case. So, \( P(X=2) = \frac{1}{36} \).
The probability distribution is:In simple words: For dice rolls, list all possible outcomes for each number of successes. Then divide the number of ways to get that success by the total number of outcomes (36 for two dice).

Exam Tip: Always clearly define what a "success" means in the context of the problem. For dice rolls, it's often helpful to think about the complement event (no success) or to list combinations systematically to avoid missing cases.

 

Question 6. From a lot of 30 bulbs, which include 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of number of defective bulbs.
Answer: There are 30 bulbs in total, with 6 defective bulbs and 24 non-defective bulbs. A sample of 4 bulbs is drawn with replacement. Let X be the number of defective bulbs in the sample of 4. The probability of getting a defective bulb is \( p = \frac{6}{30} = \frac{1}{5} \).
The probability of getting a good (non-defective) bulb is \( q = 1 - \frac{1}{5} = \frac{4}{5} \).
This is a binomial distribution with \( n=4 \) (number of trials) and \( p=\frac{1}{5} \) (probability of success, i.e., defective bulb).
\( P(X=x) = nC_x p^x q^{n-x} \)
\( P(X=0) = 4C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^4 = 1 \times 1 \times \frac{256}{625} = \frac{256}{625} \)
\( P(X=1) = 4C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^3 = 4 \times \frac{1}{5} \times \frac{64}{125} = \frac{256}{625} \)
\( P(X=2) = 4C_2 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^2 = 6 \times \frac{1}{25} \times \frac{16}{25} = \frac{96}{625} \)
\( P(X=3) = 4C_3 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^1 = 4 \times \frac{1}{125} \times \frac{4}{5} = \frac{16}{625} \)
\( P(X=4) = 4C_4 \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^0 = 1 \times \frac{1}{625} \times 1 = \frac{1}{625} \)
The probability distribution of defective bulbs is:

In simple words: This is a binomial probability problem. Find the chance of drawing a defective bulb, and then use the binomial formula to find the probability of getting 0, 1, 2, 3, or 4 defective bulbs in a sample of 4.

Exam Tip: For problems "with replacement," the probability of success (p) remains constant for each trial, making it a binomial distribution. Remember to define p (probability of success) and n (number of trials) clearly.

 

Question 7. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of tails.
Answer: Let the probability of getting a tail be \( P(T) = x \). Since the head is 3 times as likely as the tail, the probability of getting a head is \( P(H) = 3x \).
We know that \( P(H) + P(T) = 1 \).
So, \( 3x + x = 1 \)
\( 4x = 1 \)
\( x = \frac{1}{4} \).
Therefore, \( P(T) = \frac{1}{4} \) and \( P(H) = 3 \times \frac{1}{4} = \frac{3}{4} \).
The coin is tossed twice. Let X be the number of tails.
The possible values for X are 0, 1, 2.
\( P(X=0) \): No tails, meaning both are heads (HH).
\( P(HH) = P(H) \times P(H) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \).
\( P(X=1) \): One tail, meaning (HT or TH).
\( P(HT) + P(TH) = \left(\frac{3}{4} \times \frac{1}{4}\right) + \left(\frac{1}{4} \times \frac{3}{4}\right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8} \).
\( P(X=2) \): Two tails, meaning both are tails (TT).
\( P(TT) = P(T) \times P(T) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \).
The probability distribution of occurring tails is:

In simple words: First, figure out the individual probabilities of getting a head or a tail using the given bias. Then, for two tosses, list all outcomes and calculate the probability for each number of tails.

Exam Tip: For biased coins, ensure the probabilities for head and tail sum to 1. When tossing a coin multiple times, assume independence for each toss to calculate combined probabilities correctly.

 

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

Determine:
(i) if it's a valid probability distribution.
(ii) P(X< 3)
(iii) P(X > 0)
(iv) P(0 < X < 3)
Answer:
(i) To check if it's a valid probability distribution, we verify two conditions: 1. All probabilities are non-negative: \( \frac{9}{16} \ge 0 \), \( \frac{3}{8} \ge 0 \), \( \frac{1}{16} \ge 0 \). This condition is satisfied. 2. The sum of probabilities equals 1: \( \frac{9}{16} + \frac{3}{8} + \frac{1}{16} = \frac{9}{16} + \frac{6}{16} + \frac{1}{16} = \frac{9+6+1}{16} = \frac{16}{16} = 1 \). This condition is also satisfied. Therefore, it is a valid probability distribution.
(ii) \( P(X < 3) \) means the probability that X is less than 3. Since the maximum value X can take is 2, \( P(X < 3) \) includes \( P(X=0) + P(X=1) + P(X=2) \), which is the sum of all probabilities. So, \( P(X < 3) = 1 \).
(iii) \( P(X > 0) \) means the probability that X is greater than 0. This includes \( P(X=1) + P(X=2) \). \( P(X > 0) = P(X=1) + P(X=2) = \frac{3}{8} + \frac{1}{16} = \frac{6}{16} + \frac{1}{16} = \frac{7}{16} \).
(iv) \( P(0 < X < 3) \) means the probability that X is greater than 0 but less than 3. This includes \( P(X=1) + P(X=2) \). \( P(0 < X < 3) = P(X=1) + P(X=2) = \frac{3}{8} + \frac{1}{16} = \frac{6}{16} + \frac{1}{16} = \frac{7}{16} \).
In simple words: First, add up all the probabilities to check if they make 1. Then, for each part, find the probabilities for the specific X values mentioned in the question and add them together.

Exam Tip: When evaluating probabilities like \( P(X < k) \) or \( P(X > k) \), carefully list all the discrete values of X that satisfy the inequality. Remember that the sum of all probabilities in a valid distribution is always 1.

 

Question 9. The random variable X has a probability distribution P(X) of the following form, where k is some number:

(a) Determine the value of k.
(b) Find \( P(X < 2) \), \( P(X \le 2) \), \( P(X \ge 0) \).
Answer:
(a) The sum of all probabilities in a probability distribution must be equal to 1. So, \( P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1 \).
\( 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2+k) = 1 \)
Combine like terms: \( (k+2k+2k+3k+k) + (k^2+2k^2+7k^2) = 1 \)
\( 9k + 10k^2 = 1 \)
\( 10k^2 + 9k - 1 = 0 \)
Factor the quadratic equation: \( (10k - 1)(k + 1) = 0 \)
This gives two possible values for k: \( k = \frac{1}{10} \) or \( k = -1 \).
Since probability values cannot be negative, k must be positive. If \( k = -1 \), then \( P(X=1) = -1 \), which is not allowed. Therefore, \( k = \frac{1}{10} \).
The probability distribution is:
(b) Calculate the required probabilities using \( k = \frac{1}{10} \):
\( P(X < 2) = P(X=0) + P(X=1) = 0 + k = 0 + \frac{1}{10} = \frac{1}{10} \).
\( P(X \le 2) = P(X=0) + P(X=1) + P(X=2) = 0 + k + 2k = 3k = 3 \times \frac{1}{10} = \frac{3}{10} \).
\( P(X \ge 0) \): This means the probability that X is greater than or equal to 0. This is the sum of all probabilities, which equals 1.
In simple words: To find 'k', add up all the probabilities and set them equal to 1. Solve the equation for 'k', remembering that probabilities can't be negative. Then, use that 'k' value to find the probabilities for X less than, less than or equal to, or greater than given numbers.

Exam Tip: When solving for 'k' in a probability distribution, always verify that the chosen value of 'k' results in all probabilities being non-negative. If a quadratic equation yields multiple solutions for 'k', choose the one that makes the probabilities valid (i.e., between 0 and 1).

 

Question 10. Find the mean number of heads in three tosses of a fair coin.
Answer: Let H denote getting a head, and T denote getting a tail. When three fair coins are tossed, the sample space is: \( \{TTT, TTH, THT, HTT, THH, HTH, HHT, HHH\} \). Let X be the random variable representing the number of heads. The possible values for X are 0, 1, 2, 3.
For a fair coin, \( P(H) = P(T) = \frac{1}{2} \).
\( P(X=0) \) (No heads, TTT) \( = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
\( P(X=1) \) (One head, TTH, THT, HTT) \( = 3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{3}{8} \).
\( P(X=2) \) (Two heads, THH, HTH, HHT) \( = 3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{3}{8} \).
\( P(X=3) \) (Three heads, HHH) \( = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \).
The probability distribution is:

The mean (Expected Value) \( E(X) = \sum X \cdot P(X) \)
\( E(X) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8}) \)
\( E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} \)
\( E(X) = \frac{3+6+3}{8} = \frac{12}{8} = \frac{3}{2} = 1.5 \).
The mean number of heads in three tosses of a fair coin is 1.5.
In simple words: Find all the possible results when you toss a coin three times. Count how many heads you get in each result. Then, use those counts and their probabilities to calculate the average number of heads.

Exam Tip: For binomial distributions (like coin tosses), the mean (expected value) can also be calculated as \( E(X) = np \), where n is the number of trials and p is the probability of success. Here, \( n=3 \) and \( p=0.5 \), so \( E(X) = 3 \times 0.5 = 1.5 \).

 

Question 11. Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Answer: When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \). Let X be the number of sixes. X can take values 0, 1, or 2.
Probability of getting a six on one die \( = \frac{1}{6} \).
Probability of not getting a six on one die \( = \frac{5}{6} \).
\( P(X=0) \): No sixes appear on either die. This means both dice show a number from 1 to 5.
\( P(X=0) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36} \).
\( P(X=1) \): One six appears. This can be (Six on first die, No six on second) or (No six on first die, Six on second).
\( P(X=1) = \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right) = \frac{5}{36} + \frac{5}{36} = \frac{10}{36} \).
\( P(X=2) \): Two sixes appear, meaning (Six on first die, Six on second).
\( P(X=2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \).
The probability distribution for X is:

The expectation of X, \( E(X) = \sum X \cdot P(X) \):
\( E(X) = \left(0 \times \frac{25}{36}\right) + \left(1 \times \frac{10}{36}\right) + \left(2 \times \frac{1}{36}\right) \)
\( E(X) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3} \).
The expectation of X (number of sixes) is \( \frac{1}{3} \).
In simple words: When you roll two dice, find the chance of getting zero, one, or two sixes. Then, multiply each number of sixes by its probability and add them all up to find the average number of sixes you'd expect.

Exam Tip: For problems involving two dice, it's often useful to visualize the \( 6 \times 6 \) grid of outcomes. The expectation is a weighted average, where each outcome's value is weighted by its probability.

 

Question 12. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Answer: The first six positive integers are 1, 2, 3, 4, 5, 6. We select two numbers without replacement. The total number of ways to select two numbers from 6 is \( 6 \times 5 = 30 \) if order matters, or \( 6C_2 = \frac{6 \times 5}{2} = 15 \) if order doesn't matter. Since X denotes the larger of the two, the order of selection doesn't affect the value of X, so we can use combinations or ordered pairs consistently. Let's use ordered pairs for clarity, so the total cases are 30.
Let X be the larger of the two selected numbers. The possible values of X are 2, 3, 4, 5, 6 (since X must be larger than at least one other selected number).
We list the favorable cases for each value of X and their probabilities:

Expected value \( E(X) = \sum X \cdot P(X) \)
\( E(X) = \left(2 \times \frac{2}{30}\right) + \left(3 \times \frac{4}{30}\right) + \left(4 \times \frac{6}{30}\right) + \left(5 \times \frac{8}{30}\right) + \left(6 \times \frac{10}{30}\right) \)
\( E(X) = \frac{4}{30} + \frac{12}{30} + \frac{24}{30} + \frac{40}{30} + \frac{60}{30} \)
\( E(X) = \frac{4 + 12 + 24 + 40 + 60}{30} = \frac{140}{30} = \frac{14}{3} \)
\( E(X) = 4 \frac{2}{3} \).
In simple words: List all pairs of numbers you can pick. For each pair, identify the larger number. Count how many times each possible larger number appears and calculate its probability. Then, multiply each larger number by its probability and add these results to find the expected value.

Exam Tip: When sampling "without replacement," the total number of ways to pick items changes with each selection. Be systematic in listing favorable outcomes to ensure accuracy. When calculating expected value, always sum the product of each possible value and its probability.

 

Question 13. Let X denotes the sum of the numbers obtained, when two fair dice are rolled. Find the variance and standard deviation of X.
Answer: When two fair dice are rolled, the number of exhaustive cases is \( 6 \times 6 = 36 \). Let X be the sum of the numbers obtained. The possible values of X range from 2 (1+1) to 12 (6+6). We first find the probability distribution for the sum X:

Now, we calculate the Expected Value \( E(X) = \sum X \cdot P(X) \) and \( E(X^2) = \sum X^2 \cdot P(X) \).First, calculate the mean \( E(X) \):
\( E(X) = \sum X \cdot P(X) = \frac{2+6+12+20+30+42+40+36+30+22+12}{36} = \frac{252}{36} = 7 \).
Next, calculate \( E(X^2) \):
\( E(X^2) = \sum X^2 \cdot P(X) = \frac{4+18+48+100+180+294+320+324+300+242+144}{36} = \frac{1974}{36} = \frac{329}{6} \approx 54.8333 \).
Now, calculate Variance \( Var(X) = E(X^2) - [E(X)]^2 \):
\( Var(X) = \frac{1974}{36} - (7)^2 = \frac{1974}{36} - 49 \)
\( Var(X) = \frac{1974 - (49 \times 36)}{36} = \frac{1974 - 1764}{36} = \frac{210}{36} = \frac{35}{6} \approx 5.83 \).
Finally, calculate Standard Deviation \( SD(X) = \sqrt{Var(X)} \):
\( SD(X) = \sqrt{\frac{35}{6}} \approx \sqrt{5.8333} \approx 2.415 \approx 2.4 \) (nearly).
In simple words: First, list all possible sums when rolling two dice and find the probability for each sum. Then, calculate the average sum (mean) and the average of the squared sums. Use these two averages to find the variance, which tells you how spread out the sums are. The square root of the variance is the standard deviation.

Exam Tip: For problems involving the sum of two dice, always remember that 7 is the most probable sum. To calculate variance, accurately compute both \( E(X) \) and \( E(X^2) \) using a structured table to avoid calculation errors. Standard deviation is always the positive square root of variance.

 

Question 14. A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 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. What is the probability distribution of the random variable X? Find the mean, variance and standard deviation of X.
Answer: There are 15 students in the class. Each student has an equal chance of being selected, so the probability of selecting any particular student is \( \frac{1}{15} \). First, let's list the distinct ages (X) and their frequencies (\( f_i \)): Ages: 14, 15, 16, 17, 18, 19, 20, 21 Count each age: 14: 2 students 15: 1 student 16: 2 students 17: 3 students 18: 1 student 19: 2 students 20: 3 students 21: 1 student Total students = 15. The probability \( P(X=x_i) = p_i = \frac{f_i}{15} \). The probability distribution is:

Mean \( E(X) = \sum X \cdot p_i = \frac{28+15+32+51+18+38+60+21}{15} = \frac{263}{15} \approx 17.5333 \).
\( E(X^2) = \sum X^2 \cdot p_i = \frac{392+225+512+867+324+722+1200+441}{15} = \frac{4683}{15} = 312.2 \).
Variance \( Var(X) = E(X^2) - [E(X)]^2 \)
\( Var(X) = 312.2 - \left(\frac{263}{15}\right)^2 = 312.2 - (17.5333)^2 \)
\( Var(X) = 312.2 - 307.41766 = 4.78234 \approx 4.78 \).
Standard Deviation \( SD(X) = \sqrt{Var(X)} = \sqrt{4.78234} \approx 2.186 \approx 2.19 \).
Thus, the mean is approximately 17.53, the variance is approximately 4.78, and the standard deviation is approximately 2.19.
In simple words: First, list all the unique ages and how many students have each age. Use this to find the probability of picking a student of a certain age. Then, calculate the average age (mean) and how much the ages spread out (variance and standard deviation).

Exam Tip: When given a list of raw data, always start by creating a frequency distribution to clarify the possible values of the random variable and their probabilities. Be careful with calculations for \( \sum X \cdot p_i \) and \( \sum X^2 \cdot p_i \) to avoid errors, especially with fractions or decimals.

 

Question 15. In a meeting, 70% of the members favour and 30% 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).
Answer: Let X be the random variable. If a member opposed, \( X = 0 \). The probability of this is \( P(X=0) = 30\% = 0.3 \).
If a member favoured, \( X = 1 \). The probability of this is \( P(X=1) = 70\% = 0.7 \).
The probability distribution is:

Expected Value \( E(X) = \sum X \cdot P(X) = (0 \times 0.30) + (1 \times 0.70) = 0 + 0.70 = 0.7 \).
To find Variance, we first need \( E(X^2) \):
\( E(X^2) = \sum X^2 \cdot P(X) = (0^2 \times 0.3) + (1^2 \times 0.7) = (0 \times 0.3) + (1 \times 0.7) = 0 + 0.7 = 0.7 \).
Variance \( Var(X) = E(X^2) - [E(X)]^2 \)
\( Var(X) = 0.7 - (0.7)^2 = 0.7 - 0.49 = 0.21 \).
The mean \( E(X) \) is 0.7 and the variance \( Var(X) \) is 0.21.
In simple words: This setup is a simple Bernoulli trial. To find the mean, multiply each outcome (0 for oppose, 1 for favour) by its probability and add them. To find the variance, first find the mean of the squared outcomes, then subtract the square of the mean.

Exam Tip: For a Bernoulli distribution (binary outcomes 0 and 1), the mean is \( p \) (probability of success) and the variance is \( p(1-p) \). In this case, success (favour) is \( p=0.7 \), so \( E(X)=0.7 \) and \( Var(X)=0.7(1-0.7) = 0.7 \times 0.3 = 0.21 \).

 

Question 16. The means of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1
(B) 2
(C) 3
(D) \( \frac{8}{3} \)
Answer: (B) 2
In simple words: First, find the probability of each number appearing. Then, multiply each number by its probability and add all the results together to find the average (mean).

Exam Tip: When a die has repeated numbers, the probability of each distinct number appearing is based on how many faces it covers divided by the total number of faces. The mean is the sum of (value * probability).

 

Question 17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then, value of E(X) is
(A) \( \frac{37}{221} \)
(B) \( \frac{27}{221} \)
(C) \( \frac{1}{13} \)
(D) \( \frac{2}{13} \)
Answer: (D) \( \frac{2}{13} \)
In simple words: Calculate the chance of getting zero, one, or two aces when drawing two cards. Then, multiply each number of aces by its probability and add them all up to find the expected number of aces.

Exam Tip: This is a hypergeometric distribution problem since cards are drawn without replacement. Alternatively, for expected value, use linearity of expectation: E(X) = E(card1 is ace) + E(card2 is ace), where each E is simply the probability of drawing an ace.

 

Question 16. The mean of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is
(A) 1
(B) 2
(C) 3
(D) \( \frac{8}{3} \)
Answer: (B) 2
In simple words: The numbers on the dice are 1, 2, and 5. The number 1 appears on three faces, 2 on two faces, and 5 on one face. To find the mean, you multiply each number by its probability and then sum these results. The calculation shows the mean to be 2.

Exam Tip: For problems involving weighted averages or means of discrete random variables, always ensure the sum of probabilities equals 1 and that you multiply each value by its respective probability before summing them up.

 

Question 17. Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then, value of E(X) is
(A) \( \frac{37}{221} \)
(B) \( \frac{27}{221} \)
(C) \( \frac{1}{13} \)
(D) \( \frac{2}{13} \)
Answer: (D) \( \frac{2}{13} \)
In simple words: When you pick two cards from a standard deck, the expected number of aces you will get is calculated by looking at the chances of getting zero aces, one ace, or two aces. The calculated expected value, or mean, for the number of aces is \( \frac{2}{13} \).

Exam Tip: When calculating expected values (E(X)) for card problems, remember to find the probabilities for each possible outcome of X (e.g., 0 aces, 1 ace, 2 aces) using combinations, and then apply the formula \( E(X) = \sum x P(x) \).