OP Malhotra Class 12 Maths Solutions Chapter 20 Theoretical Probability Distribution Exercise 20 (A)

S Chand Class 12 ICSE Maths Solutions Chapter 20 Theoretical Probability Distribution Ex 20(a)

 

Question 1. Which of the following experiments give a discrete random variable. (You are not asked to find any probabilities).
(i) A book is chosen at random from a shelf with 50 books and its author noted.
(ii) A book is chosen at random from a shelf with 50 books and the number of pages noted.
(iii) A pupil is chosen at random from a particular class and the pupil's name, is noted.
(iv) A pupil is chosen at random from a particular class and the pupil's height is recorded to the nearest cm.
(v) The number of cars passing a given point on the road between 10:00 and 11:00 hours on a particular day.
Answer:
(i) The variable here is the author's name, which is not a numerical value that can be counted. Therefore, it is not a discrete random variable.
(ii) The variable here is the number of pages, which gives numerical values that can be counted. So, it is a discrete random variable.
(iii) The variable here is the pupil's name, which is not a numerical value that can be counted. Hence, it is not a discrete random variable.
(iv) The variable here is the pupil's height recorded to the nearest cm. This gives numerical values that can be counted in whole units. Thus, it is a discrete random variable.
(v) The variable here is the number of cars, which gives numerical values that can be counted. So, it is a discrete random variable.
In simple words: A discrete random variable is something you can count, like whole numbers. If you are counting things or categories, it's discrete. If it's something you measure and can have any value in between, like height (even to the nearest cm), it's still treated as discrete if the measurement is specific.

🎯 Exam Tip: To identify a discrete random variable, check if its possible values can be counted (usually whole numbers) or if it represents categories. Continuous random variables take any value within a range.

 

Question 2. (i) Determine which of the following can be probability distribution of a random variable X:
(a)

(b)

(ii) Find the value of k: if a random variable X has the following probability distribution:

Answer:
(i)
(a) The sum of all probabilities is \( 0.4 + 0.4 + 0.2 = 1.0 \). Since the total probability is 1, this is a valid probability distribution.
(b) The sum of all probabilities is \( 0.5 + 0.2 + 0.2 = 0.9 \). Since the total probability is not 1 (it is less than 1), this is not a valid probability distribution.
(ii) For a probability distribution, the sum of all probabilities must be 1. So, we add all the given probabilities and set the sum equal to 1.
\( P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1 \)
\( \implies 0.1 + k + 0.2 + 2k + 0.3 + k = 1 \)
\( \implies 4k + 0.6 = 1 \)
\( \implies 4k = 1 - 0.6 \)
\( \implies 4k = 0.4 \)
\( \implies k = \frac{0.4}{4} \)
\( \implies k = 0.1 \)
In simple words: The sum of probabilities for all possible outcomes in any probability distribution will always add up to exactly one. We use this rule to check if a given set of probabilities forms a valid distribution or to find any missing values.

🎯 Exam Tip: Remember two key rules for a probability distribution: each probability must be between 0 and 1, and the sum of all probabilities must be exactly 1.

 

Question 3. Two cards are drawn in succession from a well shuffled deck of 52 cards, the first card being replaced, before the second is drawn. Let X denote the number of spades drawn. Find the probability distribution of X.
Answer:
Here, \(X\) represents the number of spade cards drawn. Since two cards are drawn, \(X\) can be 0 (no spades), 1 (one spade), or 2 (two spades).
First, we find the probability of drawing a spade. There are 13 spades in a deck of 52 cards.
Probability of getting a spade (\(p\)) \( = \frac{13}{52} = \frac{1}{4} \)
The probability of not getting a spade (\(q\)) is \( 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \)
Since the first card is replaced, the draws are independent.
Probability of \(X = 0\) (no spades): This means drawing a non-spade, then another non-spade.
\( P(X = 0) = q \times q = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)
Probability of \(X = 1\) (one spade): This means drawing a spade then a non-spade, or a non-spade then a spade.
\( P(X = 1) = pq + qp = \left( \frac{1}{4} \times \frac{3}{4} \right) + \left( \frac{3}{4} \times \frac{1}{4} \right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} \)
Probability of \(X = 2\) (two spades): This means drawing a spade, then another spade.
\( P(X = 2) = p \times p = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \)
The probability distribution of \(X\) is shown in the table below.

In simple words: This problem is an example of a binomial distribution because each draw is an independent Bernoulli trial (either spade or not spade) with a fixed probability of success, and the number of trials is fixed.

🎯 Exam Tip: When cards are drawn with replacement, the probability of drawing a specific card remains the same for each draw, making the events independent.

 

Question 4. Shew graphically the probability function of the discrete random variable X, where X is the number of heads appearing when an unbiased coin is tossed twice in succession.
Answer:
Here, \(X\) is the number of heads we get when an unbiased coin is tossed two times in a row.
The possible results when tossing a coin twice are: {Head-Head, Head-Tail, Tail-Head, Tail-Tail}.
From these results, the possible values for \(X\) (number of heads) are 0, 1, or 2.
Probability of \(X = 0\) (zero heads, which is Tail-Tail): \( P(X = 0) = P(TT) = \frac{1}{4} \)
Probability of \(X = 1\) (one head, which is Head-Tail or Tail-Head): \( P(X = 1) = P(HT \text{ or } TH) = \frac{2}{4} = \frac{1}{2} \)
Probability of \(X = 2\) (two heads, which is Head-Head): \( P(X = 2) = P(HH) = \frac{1}{4} \)
The probability function for \(X\) is shown in the graph below. Each coin toss is an independent event, and since the coin is unbiased, the probability of getting a head or a tail is equal (1/2).

In simple words: When a coin is tossed twice, we can get 0, 1, or 2 heads. We figure out how likely each number of heads is and then draw a picture showing these chances as bars.

🎯 Exam Tip: For small numbers of trials, listing the sample space can help accurately determine the probabilities for each value of the random variable. A bar chart is a common way to visualize discrete probability distributions.

 

Question 5. A box contains 3 red marbles and 5 green marbles. Two marbles are taken at random without replacement and X is the number of green marbles obtained. Find the probability distribution of X. Sketch the graphs.
Answer:
In the box, there are 3 red marbles and 5 green marbles, making a total of 8 marbles. We are picking two marbles without putting the first one back. \(X\) represents the number of green marbles we get. So, \(X\) can be 0, 1, or 2.
First, we find the total number of ways to choose 2 marbles from 8: \( { }^8 \mathrm{C}_2 = \frac{8 \times 7}{2 \times 1} = 28 \) ways.
Probability of \(X = 0\) (getting 0 green marbles, which means 2 red marbles):
This is choosing 2 red marbles from 3. Number of ways \( = { }^3 \mathrm{C}_2 = \frac{3 \times 2}{2 \times 1} = 3 \)
So, \( P(X = 0) = \frac{3}{28} \)
Probability of \(X = 1\) (getting 1 green marble and 1 red marble):
This is choosing 1 green from 5 AND 1 red from 3. Number of ways \( = { }^5 \mathrm{C}_1 \times { }^3 \mathrm{C}_1 = 5 \times 3 = 15 \)
So, \( P(X = 1) = \frac{15}{28} \)
Probability of \(X = 2\) (getting 2 green marbles):
This is choosing 2 green marbles from 5. Number of ways \( = { }^5 \mathrm{C}_2 = \frac{5 \times 4}{2 \times 1} = 10 \)
So, \( P(X = 2) = \frac{10}{28} \)
The probability distribution of \(X\) is shown in the table and graph below. This problem involves combinations without replacement, which is a key concept in hypergeometric probability distributions.

In simple words: We find the total ways to pick 2 marbles from 8. Then, we find ways to pick 0, 1, or 2 green marbles from 5 green ones and 3 red ones. The probability for each is the number of ways divided by total ways.

🎯 Exam Tip: When calculating probabilities without replacement, remember that the total number of items and the number of items of a specific type decrease after each pick.

 

Question 6. A fair coin has, the number ' 1' on one face and the number ' 2 ' on the other. The coin is thrown with a fair die and X is the sum of the scores. Find the probability distribution of X.
Answer:
We have a special coin with faces marked '1' and '2'. We also have a regular six-sided die (marked '1' through '6'). We roll both, and \(X\) is the sum of the numbers we get.
The total number of possible outcomes is 2 (for the coin) multiplied by 6 (for the die), which is 12 outcomes. These outcomes are:
(Coin=1, Die=1), (1,2), (1,3), (1,4), (1,5), (1,6)
(Coin=2, Die=1), (2,2), (2,3), (2,4), (2,5), (2,6)
The possible sums (\(X\) values) are from \(1+1=2\) to \(2+6=8\). So, \(X\) can be 2, 3, 4, 5, 6, 7, or 8.
Probability of \(X = 2\): Only (1,1) gives sum 2. \( P(X=2) = \frac{1}{12} \)
Probability of \(X = 3\): (1,2) or (2,1) give sum 3. \( P(X=3) = \frac{2}{12} \)
Probability of \(X = 4\): (1,3) or (2,2) give sum 4. \( P(X=4) = \frac{2}{12} \)
Probability of \(X = 5\): (1,4) or (2,3) give sum 5. \( P(X=5) = \frac{2}{12} \)
Probability of \(X = 6\): (1,5) or (2,4) give sum 6. \( P(X=6) = \frac{2}{12} \)
Probability of \(X = 7\): (1,6) or (2,5) give sum 7. \( P(X=7) = \frac{2}{12} \)
Probability of \(X = 8\): Only (2,6) gives sum 8. \( P(X=8) = \frac{1}{12} \)
The probability distribution of \(X\) is shown in the table and graph below. This scenario demonstrates how to combine outcomes from independent events (coin toss and die roll) to find the probability of their sum.

In simple words: We list all the possible sums when a special coin and a die are rolled. Then, we count how many times each sum can happen and divide by the total number of ways (12) to get the probability for each sum.

🎯 Exam Tip: When dealing with two independent events, it's helpful to list all possible outcomes in a table or as ordered pairs to ensure no combination is missed, especially for sums.

 

Question 7. An urn contains 4 white and 3 red balls. Find the probability distribution of the number of red balls in a random draw of three balls.
Answer:
An urn contains a total of 7 balls: 4 white and 3 red. We randomly pick three balls. Let \(X\) be the number of red balls drawn.
The possible number of red balls (\(X\)) can be 0, 1, 2, or 3.
First, calculate the total number of ways to choose 3 balls from 7: \( { }^7 \mathrm{C}_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \) ways.
Probability of \(X = 0\) (drawing 0 red balls, meaning 3 white balls):
Number of ways to choose 3 white balls from 4: \( { }^4 \mathrm{C}_3 = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4 \)
So, \( P(X = 0) = \frac{4}{35} \)
Probability of \(X = 1\) (drawing 1 red ball and 2 white balls):
Number of ways to choose 1 red from 3 AND 2 white from 4: \( { }^3 \mathrm{C}_1 \times { }^4 \mathrm{C}_2 = 3 \times \frac{4 \times 3}{2 \times 1} = 3 \times 6 = 18 \)
So, \( P(X = 1) = \frac{18}{35} \)
Probability of \(X = 2\) (drawing 2 red balls and 1 white ball):
Number of ways to choose 2 red from 3 AND 1 white from 4: \( { }^3 \mathrm{C}_2 \times { }^4 \mathrm{C}_1 = \frac{3 \times 2}{2 \times 1} \times 4 = 3 \times 4 = 12 \)
So, \( P(X = 2) = \frac{12}{35} \)
Probability of \(X = 3\) (drawing 3 red balls):
Number of ways to choose 3 red balls from 3: \( { }^3 \mathrm{C}_3 = 1 \)
So, \( P(X = 3) = \frac{1}{35} \)
The probability distribution of \(X\) is given in the table and graph below. This problem is an example of a hypergeometric distribution because it involves sampling without replacement from a finite population divided into two categories.

In simple words: We have a bag of red and white balls. We pull out 3 balls without putting them back. We want to know the chances of getting 0, 1, 2, or 3 red balls. We calculate this by using combinations.

🎯 Exam Tip: For problems involving combinations, always clearly define the total number of items and the number of items in each category before calculating the probabilities.

 

Question 8. Two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of aces.
Answer:
We draw two cards one after another from a 52-card deck without replacing the first card. \(X\) represents the number of aces we get.
There are 4 aces and \(52 - 4 = 48\) non-aces in the deck.
Since we draw two cards, \(X\) can be 0 (no aces), 1 (one ace), or 2 (two aces).
Probability of \(X = 0\) (getting no aces):
This means the first card is a non-ace, AND the second card is also a non-ace (from the remaining 51 cards).
\( P(X=0) = P(\text{1st non-ace}) \times P(\text{2nd non-ace } | \text{ 1st non-ace}) = \frac{48}{52} \times \frac{47}{51} = \frac{12}{13} \times \frac{47}{51} = \frac{188}{221} \)
Probability of \(X = 1\) (getting one ace):
This can happen in two ways: (Ace then Non-Ace) OR (Non-Ace then Ace).
\( P(\text{1st Ace, 2nd Non-Ace}) = \frac{4}{52} \times \frac{48}{51} \)
\( P(\text{1st Non-Ace, 2nd Ace}) = \frac{48}{52} \times \frac{4}{51} \)
\( P(X=1) = \left( \frac{4}{52} \times \frac{48}{51} \right) + \left( \frac{48}{52} \times \frac{4}{51} \right) = \frac{192}{2652} + \frac{192}{2652} = \frac{384}{2652} = \frac{32}{221} \)
Probability of \(X = 2\) (getting two aces):
This means the first card is an ace, AND the second card is also an ace (from the remaining 3 aces and 51 total cards).
\( P(X=2) = P(\text{1st Ace}) \times P(\text{2nd Ace } | \text{ 1st Ace}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \)
The probability distribution for \(X\) is shown below. When drawing without replacement, the probability of the second event depends on the outcome of the first event, as the total number of items changes.

In simple words: We draw two cards from a deck, but we don't put the first one back. We want to find the chances of getting 0, 1, or 2 aces. Because we don't replace the card, the chances change for the second draw.

🎯 Exam Tip: Always adjust the denominator (total remaining cards) and numerator (remaining cards of a specific type) for each successive draw when sampling without replacement.

 

Question 9. From a list containing 25 items, 5 of which are defective, 4 are chosen at random. Let X be the number of defectives found. Obtain the probability distribution of X, if the items are chosen without replacement.
Answer:
We have a list of 25 items, where 5 are defective and \( 25 - 5 = 20 \) are non-defective. We choose 4 items without replacing them. \(X\) is the number of defective items we find among the 4 chosen.
The possible values for \(X\) are 0, 1, 2, 3, or 4.
First, find the total number of ways to choose 4 items from 25:
\( { }^{25} \mathrm{C}_4 = \frac{25 \times 24 \times 23 \times 22}{4 \times 3 \times 2 \times 1} = 12650 \) ways.
Probability of \(X = 0\) (getting no defective items, meaning all 4 are non-defective):
Choose 4 non-defective items from 20: \( { }^{20} \mathrm{C}_4 = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845 \)
So, \( P(X = 0) = \frac{4845}{12650} = \frac{969}{2530} \)
Probability of \(X = 1\) (getting 1 defective and 3 non-defective items):
Choose 1 defective from 5 AND 3 non-defective from 20: \( { }^5 \mathrm{C}_1 \times { }^{20} \mathrm{C}_3 = 5 \times \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 5 \times 1140 = 5700 \)
So, \( P(X = 1) = \frac{5700}{12650} = \frac{114}{253} \)
Probability of \(X = 2\) (getting 2 defective and 2 non-defective items):
Choose 2 defective from 5 AND 2 non-defective from 20: \( { }^5 \mathrm{C}_2 \times { }^{20} \mathrm{C}_2 = \left( \frac{5 \times 4}{2 \times 1} \right) \times \left( \frac{20 \times 19}{2 \times 1} \right) = 10 \times 190 = 1900 \)
So, \( P(X = 2) = \frac{1900}{12650} = \frac{38}{253} \)
Probability of \(X = 3\) (getting 3 defective and 1 non-defective item):
Choose 3 defective from 5 AND 1 non-defective from 20: \( { }^5 \mathrm{C}_3 \times { }^{20} \mathrm{C}_1 = \left( \frac{5 \times 4}{2 \times 1} \right) \times 20 = 10 \times 20 = 200 \)
So, \( P(X = 3) = \frac{200}{12650} = \frac{4}{253} \)
Probability of \(X = 4\) (getting 4 defective items):
Choose 4 defective items from 5: \( { }^5 \mathrm{C}_4 = 5 \)
So, \( P(X = 4) = \frac{5}{12650} = \frac{1}{2530} \)
The probability distribution of \(X\) is shown in the table below. Quality control in manufacturing often uses probability distributions like this to estimate the likelihood of finding defective items in a sample.

In simple words: We have a group of items, some of which are faulty. We pick 4 items without putting them back. We then calculate the probability of getting 0, 1, 2, 3, or 4 faulty items in our selection.

🎯 Exam Tip: Ensure that the sum of the probabilities for all possible values of \(X\) equals 1 to verify your calculations. This is a common method for quality control sampling.

 

Question 10. A random variable A " has the following probability distribution.

Determine:
(i) The value of a.
(ii) P(X < 3)
(iii) P(X > 4)
(iv) P(0 < X < 5)
Answer:
(i) For any probability distribution, the sum of all probabilities must be 1. So, we add all the probabilities given for \(X\) and set the sum equal to 1.
\( P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8) = 1 \)
\( \implies a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1 \)
\( \implies 81a = 1 \)
\( \implies a = \frac{1}{81} \)
(ii) To find \( P(X < 3) \), we sum the probabilities for \(X=0\), \(X=1\), and \(X=2\).
\( P(X < 3) = P(X=0) + P(X=1) + P(X=2) \)
\( = a + 3a + 5a = 9a \)
Substitute the value of \(a\): \( 9 \times \frac{1}{81} = \frac{9}{81} = \frac{1}{9} \)
(iii) To find \( P(X > 4) \), we sum the probabilities for \(X=5\), \(X=6\), \(X=7\), and \(X=8\).
\( P(X > 4) = P(X=5) + P(X=6) + P(X=7) + P(X=8) \)
\( = 11a + 13a + 15a + 17a = 56a \)
Substitute the value of \(a\): \( 56 \times \frac{1}{81} = \frac{56}{81} \)
(iv) To find \( P(0 < X < 5) \), we sum the probabilities for \(X=1\), \(X=2\), \(X=3\), and \(X=4\).
\( P(0 < X < 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) \)
\( = 3a + 5a + 7a + 9a = 24a \)
Substitute the value of \(a\): \( 24 \times \frac{1}{81} = \frac{24}{81} = \frac{8}{27} \)
In simple words: First, we find the missing value 'a' by adding all probabilities and setting the sum to 1. Then, for parts (ii), (iii), and (iv), we add the 'a' values for the specific range of X to get the total probability.

🎯 Exam Tip: Carefully read the inequality signs (e.g., \( < \), \( \le \), \( > \), \( \ge \)) to correctly include or exclude boundary values when calculating probabilities for a range.

 

Question 11. A random variable X has the following probability distribution:

(i) Find the value of a;
(ii) Find P(X < 3), P(X < 4), P(0 < X < 5);
(iii) Give the smallest value of m for which P(X > m) < 0.6.
Answer:
(i) For a probability distribution, the sum of all probabilities must equal 1. So, we add all the probabilities and set the sum to 1.
\( P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1 \)
\( \implies a + 4a + 3a + 7a + 8a + 10a + 6a + 9a = 1 \)
\( \implies 48a = 1 \)
\( \implies a = \frac{1}{48} \)
(ii)
To find \( P(X < 3) \), we sum probabilities for \(X=0\), \(X=1\), and \(X=2\).
\( P(X < 3) = P(X=0) + P(X=1) + P(X=2) = a + 4a + 3a = 8a \)
Substitute \(a = \frac{1}{48}\): \( 8 \times \frac{1}{48} = \frac{8}{48} = \frac{1}{6} \) or approximately \( 0.167 \).
To find \( P(X < 4) \), we sum probabilities for \(X=0\), \(X=1\), \(X=2\), and \(X=3\).
\( P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = a + 4a + 3a + 7a = 15a \)
Substitute \(a = \frac{1}{48}\): \( 15 \times \frac{1}{48} = \frac{15}{48} = \frac{5}{16} \) or \( 0.3125 \).
To find \( P(0 < X < 5) \), we sum probabilities for \(X=1\), \(X=2\), \(X=3\), and \(X=4\).
\( P(0 < X < 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 4a + 3a + 7a + 8a = 22a \)
Substitute \(a = \frac{1}{48}\): \( 22 \times \frac{1}{48} = \frac{22}{48} = \frac{11}{24} \) or approximately \( 0.458 \).
(iii) We need to find the smallest whole number \(m\) such that \( P(X > m) < 0.6 \).
Let's check values of \(m\) starting from the largest possible \(X\) minus 1:
If \(m=7\), \(P(X > 7) = 0\). This is \( < 0.6 \).
If \(m=6\), \(P(X > 6) = P(X=7) = 9a = 9 \times \frac{1}{48} = \frac{9}{48} = \frac{3}{16} = 0.1875\). This is \( < 0.6 \).
If \(m=5\), \(P(X > 5) = P(X=6) + P(X=7) = 6a + 9a = 15a = 15 \times \frac{1}{48} = \frac{15}{48} = \frac{5}{16} = 0.3125\). This is \( < 0.6 \).
If \(m=4\), \(P(X > 4) = P(X=5) + P(X=6) + P(X=7) = 10a + 6a + 9a = 25a = 25 \times \frac{1}{48} = \frac{25}{48} \approx 0.5208\). This is \( < 0.6 \).
If \(m=3\), \(P(X > 3) = P(X=4) + P(X=5) + P(X=6) + P(X=7) = 8a + 10a + 6a + 9a = 33a = 33 \times \frac{1}{48} = \frac{33}{48} = \frac{11}{16} = 0.6875\). This is NOT \( < 0.6 \).
Therefore, the smallest value of \(m\) for which \( P(X > m) < 0.6 \) is 4.
In simple words: First, we find 'a' by making sure all probabilities add up to 1. Then, we add up the probabilities for the given ranges of X. Finally, we test different 'm' values to find the smallest one where the chance of X being bigger than 'm' is less than 0.6.

🎯 Exam Tip: When finding values based on inequalities like \( P(X > m) < \text{value} \), it is often useful to calculate cumulative probabilities step-by-step and identify where the condition is first met.

 

Question 1. Which of the following experiments give a discrete random variable. (You are not asked to find any probabilities).
(i) A book is chosen at random from a shelf with 50 books and its author noted.
(ii) A book is chosen at random from a shelf with 50 books and the number of pages noted.
(iii) A pupil is chosen at random from a particular class and the pupil's name, is noted.
(iv) A pupil is chosen at random from a particular class and the pupil's height is recorded to the nearest cm.
(v) The number of cars passing a given point on the road between 10:00 and 11:00 hours on a particular day.
Answer:
(i) This is not a discrete random variable because the values (authors' names) are not numbers that can be counted. A discrete variable must have numerical outcomes.
(ii) This is a discrete random variable because the number of pages can be counted. We can assign specific integer values like 100, 250, 500, etc., to the number of pages.
(iii) This is not a discrete random variable because the pupil's name is not a numerical value that can be counted or ordered. Random variables need quantifiable outcomes.
(iv) This is a discrete random variable. Even though height can be continuous, when it is recorded to the nearest cm, it becomes a specific, countable numerical value. Measuring to a specific unit makes it discrete.
(v) This is a discrete random variable because the number of cars passing a point can be counted as specific, whole numbers (0, 1, 2, 3, etc.). Counting whole items makes a variable discrete.
In simple words: A discrete random variable is one where you can count the possible outcomes as whole numbers. If the outcomes are names or continuous measurements not rounded to a whole number, it's not discrete.

🎯 Exam Tip: Remember that "discrete" means countable outcomes, often whole numbers. "Continuous" means outcomes can take any value within a range.

 

Question 2. Determine which of the following can be probability distribution of a random variable X:
(a)