CBSE Class 10 Mathematics Probability Assignment Set 07

ASSIGNMENTS

(i) Conditional Probability

LEVEL I

Question 1. If P(A) = 0.3, P(B) = 0.2, find P(B/A) if A and B are mutually exclusive events.
Answer: Since A and B are mutually exclusive events, \(A \cap B = \phi\)
Therefore, \(P(A \cap B) = 0\)
Conditional probability is given by \(P(B/A) = \frac{P(A \cap B)}{P(A)}\)
\(P(B/A) = \frac{0}{0.3} = 0\)
In simple words: For mutually exclusive events, both cannot happen together, so the conditional probability is zero.

Exam Tip: Always check if events are mutually exclusive to immediately set \(P(A \cap B) = 0\).

Question 2. Find the probability of drawing two white balls in succession from a bag containing 3 red and 5 white balls respectively, the ball first drawn is not replaced.
Answer: Total number of red balls = 3
Total number of white balls = 5
Total number of balls = \(3 + 5 = 8\)
Probability of drawing a white ball in the first draw (\(W_1\)) = \(\frac{5}{8}\)
Since the ball is not replaced, remaining white balls = 4, and total remaining balls = 7
Probability of drawing a white ball in the second draw given that the first is white (\(W_2/W_1\)) = \(\frac{4}{7}\)
Required probability = \(P(W_1 \text{ and } W_2) = P(W_1) \cdot P(W_2/W_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\)
In simple words: Multiply the probability of the first event by the reduced probability of the second event without replacement.

Exam Tip: Note the phrase "without replacement" as it changes the denominator for subsequent draws.

LEVEL II

Question 1. A dice is thrown twice and sum of numbers appearing is observed to be 6. what is the conditional probability that the number 4 has appeared at least once.
Answer: Let E be the event that 4 appears at least once, and F be the event that the sum of numbers is 6.
Sample space S has \(6 \times 6 = 36\) outcomes.
F = Sum is 6 = \(\{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}\)
So, \(P(F) = \frac{5}{36}\)
\(E \cap F\) = Sum is 6 and 4 appears at least once = \(\{(2, 4), (4, 2)\}\)
So, \(P(E \cap F) = \frac{2}{36}\)
Conditional probability \(P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{2/36}{5/36} = \frac{2}{5}\)
In simple words: Filter the outcomes where the sum is 6, and find what fraction of them contain the number 4.

Exam Tip: List out sample points explicitly for small sample spaces to avoid counting errors.

LEVEL III

Question 1. If \(P(A) = \frac{3}{8}\), \(P(B) = \frac{1}{2}\) and \(P(A \cap B) = \frac{1}{4}\), find \(P(A/\overline{B})\) and \(P(\overline{B}/A)\).
Answer: Given \(P(A) = \frac{3}{8}\), \(P(B) = \frac{1}{2}\), and \(P(A \cap B) = \frac{1}{4}\)
First, find \(P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2}\)
Next, find \(P(A \cap \overline{B}) = P(A) - P(A \cap B) = \frac{3}{8} - \frac{1}{4} = \frac{3}{8} - \frac{2}{8} = \frac{1}{8}\)
(1) \(P(A/\overline{B}) = \frac{P(A \cap \overline{B})}{P(\overline{B})} = \frac{1/8}{1/2} = \frac{2}{8} = \frac{1}{4}\)
(2) \(P(\overline{B}/A) = \frac{P(A \cap \overline{B})}{P(A)} = \frac{1/8}{3/8} = \frac{1}{3}\)
In simple words: Use set relations to find probabilities involving complements before applying the conditional probability formula.

Exam Tip: Remember that \(P(A \cap \overline{B}) = P(A) - P(A \cap B)\).

LEVEL II (continued from page 75)

Question 1. A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find what is the probability that none is red.
Answer: Total white balls = 5
Total red balls = 7
Total black balls = 3
Total balls = \(5 + 7 + 3 = 15\)
Number of non-red balls = \(5 + 3 = 8\) (5 white + 3 black)
Probability that none is red means all three drawn balls are non-red.
First draw non-red: \(\frac{8}{15}\)
Second draw non-red: \(\frac{7}{14}\)
Third draw non-red: \(\frac{6}{13}\)
Required probability = \(\frac{8}{15} \times \frac{7}{14} \times \frac{6}{13} = \frac{336}{2730} = \frac{8}{65}\)
In simple words: Calculate the product of probabilities of choosing non-red balls sequentially without replacement.

Exam Tip: Grouping all non-red balls together simplifies calculations when multiple categories are excluded.

Question 2. The probability of A hitting a target is \(\frac{3}{7}\) and that of B hitting is \(\frac{1}{3}\). They both fire at the target. Find the probability that (i) at least one of them will hit the target, (ii) Only one of them will hit the target.
Answer: Given \(P(A) = \frac{3}{7}\), so \(P(\overline{A}) = 1 - \frac{3}{7} = \frac{4}{7}\)
Given \(P(B) = \frac{1}{3}\), so \(P(\overline{B}) = 1 - \frac{1}{3} = \frac{2}{3}\)
(i) Probability that at least one hits = \(1 - P(\text{neither hits})\)
\(P(\text{neither hits}) = P(\overline{A}) \cdot P(\overline{B}) = \frac{4}{7} \times \frac{2}{3} = \frac{8}{21}\)
\(P(\text{at least one hits}) = 1 - \frac{8}{21} = \frac{13}{21}\)
(ii) Probability that only one of them hits = \(P(A \text{ hits and } B \text{ misses}) + P(A \text{ misses and } B \text{ hits})\)
\(= P(A)P(\overline{B}) + P(\overline{A})P(B)\)
\(= \left(\frac{3}{7} \times \frac{2}{3}\right) + \left(\frac{4}{7} \times \frac{1}{3}\right) = \frac{6}{21} + \frac{4}{21} = \frac{10}{21}\)
In simple words: Use complement probabilities for "at least one", and add individual mutually exclusive products for "only one".

Exam Tip: "At least one" is almost always solved faster using \(1 - P(\text{none})\).

LEVEL III

Question 1. A class consists of 80 students; 25 of them are girls and 55 are boys, 10 of them are rich and the remaining poor; 20 of them are fair complexioned. what is the probability of selecting a fair complexioned rich girl.
Answer: Total students = 80
Total girls = 25
Total boys = 55
Total rich students = 10 (remaining poor = 70)
Total fair complexioned students = 20
Assuming the traits (being a girl, rich, and fair complexioned) are independent or based on standard combinatorial proportional distribution as implied by typical textbook phrasing where specific joint distributions aren't fully specified, but wait - standard interpretation of such problems requires proportional assumption or direct data. Assuming uniform independent distribution across categories or standard interpretation: Probability of being a girl = \(\frac{25}{80} = \frac{5}{16}\)
Probability of being rich = \(\frac{10}{80} = \frac{1}{8}\)
Probability of being fair complexioned = \(\frac{20}{80} = \frac{1}{4}\)
Assuming independence between these attributes: \(P(\text{Girl and Rich and Fair}) = \frac{25}{80} \times \frac{10}{80} \times \frac{20}{80} = \frac{5}{16} \times \frac{1}{8} \times \frac{1}{4} = \frac{5}{512}\)
In simple words: Multiply the individual probabilities of each independent characteristic.

Exam Tip: When multiple independent characteristics are combined, multiply their individual selection ratios.

Question 2. Two integers are selected from integers 1 through 11. If the sum is even, find the probability that both the numbers are odd.
Answer: Integers from 1 through 11 are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.
Total numbers = 11
Odd numbers = 6 (1, 3, 5, 7, 9, 11)
Even numbers = 5 (2, 4, 6, 8, 10)
Two integers can be selected in \({}^{11}C_2 = \frac{11 \times 10}{2} = 55\) ways.
Let E be the event that the sum is even. A sum of two integers is even if both are even or both are odd.
Number of ways to choose 2 even numbers = \({}^5C_2 = \frac{5 \times 4}{2} = 10\)
Number of ways to choose 2 odd numbers = \({}^6C_2 = \frac{6 \times 5}{2} = 15\)
Total ways for an even sum = \(10 + 15 = 25\)
We need the conditional probability that both are odd given that the sum is even:
\(P(\text{both odd} / \text{sum is even}) = \frac{\text{Ways to choose 2 odd}}{\text{Total ways for even sum}} = \frac{15}{25} = \frac{3}{5}\)
In simple words: An even sum requires either two even or two odd numbers. Divide the ways to pick two odd numbers by the total ways to get an even sum.

Exam Tip: Sum of two integers is even only when both have the same parity (both odd or both even).

(ii) Multiplication theorem on probability

(Note: Pages do not list explicit separate assignment questions under section (ii), continuing directly to Independent Events.)

(iii) Independent Events

LEVEL I

Question 1. A coin is tossed thrice and all 8 outcomes are equally likely.
E : "The first throw results in head" F : "The last throw results in tail"
Are the events independent ?
Answer: Sample space S contains 8 outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
E = First throw is head = {HHH, HHT, HTH, HTT}, so \(P(E) = \frac{4}{8} = \frac{1}{2}\)
F = Last throw is tail = {HHT, HTT, THT, TTT}, so \(P(F) = \frac{4}{8} = \frac{1}{2}\)
\(E \cap F\) = First is head and last is tail = {HHT, HTT}, so \(P(E \cap F) = \frac{2}{8} = \frac{1}{4}\)
Check independence: \(P(E) \cdot P(F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)
Since \(P(E \cap F) = P(E) \cdot P(F) = \frac{1}{4}\), the events E and F are independent.
In simple words: Two events are independent if the product of their individual probabilities equals the probability of both happening together.

Exam Tip: Always verify independence using the strict condition \(P(A \cap B) = P(A) \cdot P(B)\).

Question 2. Given \(P(A) = \frac{1}{4}\), \(P(B) = \frac{2}{3}\) and \(P(A \cup B) = \frac{3}{4}\). Are the events independent ?
Answer: First, find \(P(A \cap B)\) using the addition law:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(\frac{3}{4} = \frac{1}{4} + \frac{2}{3} - P(A \cap B)\)
\(\frac{3}{4} = \frac{3 + 8}{12} - P(A \cap B) = \frac{11}{12} - P(A \cap B)\)
\(P(A \cap B) = \frac{11}{12} - \frac{3}{4} = \frac{11 - 9}{12} = \frac{2}{12} = \frac{1}{6}\)
Now, calculate \(P(A) \cdot P(B) = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}\)
Since \(P(A \cap B) = P(A) \cdot P(B) = \frac{1}{6}\), the events A and B are independent.
In simple words: Find the intersection probability first using union formula, then compare it with the product of individual probabilities.

Exam Tip: Use the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) to bridge union and intersection.

Question 3. If A and B are independent events, Find P(B) if \(P(A \cup B) = 0.60\) and \(P(A) = 0.35\).
Answer: Since A and B are independent, \(P(A \cap B) = P(A) \cdot P(B) = 0.35 P(B)\)
Using the addition theorem:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(0.60 = 0.35 + P(B) - 0.35 P(B)\)
\(0.60 - 0.35 = P(B)(1 - 0.35)\)
\(0.25 = 0.65 P(B)\)
\(P(B) = \frac{0.25}{0.65} = \frac{25}{65} = \frac{5}{13}\)
In simple words: Substitute \(P(A \cap B)\) with \(P(A)P(B)\) in the union formula and solve for \(P(B)\).

Exam Tip: Factor out \(P(B)\) when combining terms like \(P(B) - P(A)P(B)\).

(iv) Baye's theorem, partition of sample space and Theorem of total probability

LEVEL I

Question 1. A bag contains 6 red and 5 blue balls and another bag contains 5 red and 8 blue balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag. Find the probability that the ball drawn is blue in colour.
Answer: Let Bag 1 contain 6R, 5B (total 11 balls).
Let Bag 2 contain 5R, 8B (total 13 balls).
Let \(E_1\) = Red ball transferred from Bag 1 to Bag 2.
Let \(E_2\) = Blue ball transferred from Bag 1 to Bag 2.
\(P(E_1) = \frac{6}{11}\), \(P(E_2) = \frac{5}{11}\)
Let A be the event that a blue ball is drawn from Bag 2.
If a red ball is transferred, Bag 2 has 6R and 8B (total 14 balls), so \(P(A/E_1) = \frac{8}{14} = \frac{4}{7}\)
If a blue ball is transferred, Bag 2 has 5R and 9B (total 14 balls), so \(P(A/E_2) = \frac{9}{14}\)
Using Total Probability Theorem:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left(\frac{6}{11} \times \frac{4}{7}\right) + \left(\frac{5}{11} \times \frac{9}{14}\right) = \frac{24}{77} + \frac{45}{154} = \frac{48 + 45}{154} = \frac{93}{154}\)
In simple words: Consider both cases (transferring red or blue) and add their weighted probabilities.

Exam Tip: Always redefine the composition of the target urn after the transfer step.

Question 2. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts. Find the probability of the lost card being a heart.
Answer: Let \(E_1\) = Lost card is a heart (13 cards).
Let \(E_2\) = Lost card is not a heart (39 cards).
\(P(E_1) = \frac{13}{52} = \frac{1}{4}\), \(P(E_2) = \frac{39}{52} = \frac{3}{4}\)
Let A = Drawing 2 hearts from the remaining 51 cards.
If a heart is lost, 12 hearts remain out of 51 cards. \(P(A/E_1) = \frac{{}^{12}C_2}{{}^{51}C_2} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550}\)
If a non-heart is lost, 13 hearts remain out of 51 cards. \(P(A/E_2) = \frac{{}^{13}C_2}{{}^{51}C_2} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550}\)
Using Bayes' Theorem to find probability that lost card is a heart given 2 hearts drawn:
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(\frac{1}{4} \times \frac{132}{2550} = \frac{33}{2550}\)
Denominator = \(\left(\frac{1}{4} \times \frac{132}{2550}\right) + \left(\frac{3}{4} \times \frac{156}{2550}\right) = \frac{33 + 117}{2550} = \frac{150}{2550}\)
\(P(E_1/A) = \frac{33}{150} = \frac{11}{50}\)
In simple words: Apply Bayes' theorem by evaluating the reduced combinations of hearts when one heart versus one non-heart is lost.

Exam Tip: Combinations simplify easily when denominators match across terms in Bayes' formula.

Question 3. An insurance company insured 2000 scooter and 3000 motorcycles. The probability of an accident involving scooter is 0.01 and that of motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accidental vehicle was a motorcycle.
Answer: Total vehicles = \(2000 + 3000 = 5000\)
Let \(E_1\) = Vehicle is a scooter. \(P(E_1) = \frac{2000}{5000} = \frac{2}{5}\)
Let \(E_2\) = Vehicle is a motorcycle. \(P(E_2) = \frac{3000}{5000} = \frac{3}{5}\)
Let A = Vehicle meets with an accident.
\(P(A/E_1) = 0.01 = \frac{1}{100}\)
\(P(A/E_2) = 0.02 = \frac{2}{100}\)
We need \(P(E_2/A)\), the probability that it was a motorcycle given it met with an accident:
\(P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(\frac{3}{5} \times \frac{2}{100} = \frac{6}{500}\)
Denominator = \(\left(\frac{2}{5} \times \frac{1}{100}\right) + \left(\frac{3}{5} \times \frac{2}{100}\right) = \frac{2}{500} + \frac{6}{500} = \frac{8}{500}\)
\(P(E_2/A) = \frac{6}{8} = \frac{3}{4} = 0.75\)
In simple words: Use Bayes' theorem with the proportion of scooters and motorcycles as prior probabilities.

Exam Tip: Convert percentages to fractions or decimals carefully to avoid arithmetic mistakes.

Question 4. A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and 3 copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin.
Answer: Let \(E_1\) = Choosing Purse 1. \(P(E_1) = \frac{1}{2}\)
Let \(E_2\) = Choosing Purse 2. \(P(E_2) = \frac{1}{2}\)
Purse 1 contains 2 silver, 4 copper (total 6 coins).
Purse 2 contains 4 silver, 3 copper (total 7 coins).
Let A = Drawing a silver coin.
\(P(A/E_1) = \frac{2}{6} = \frac{1}{3}\)
\(P(A/E_2) = \frac{4}{7}\)
Using Total Probability Theorem:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(\frac{1}{2} \times \frac{4}{7}\right) = \frac{1}{6} + \frac{2}{7} = \frac{7 + 12}{42} = \frac{19}{42}\)
In simple words: Add the probability of choosing purse 1 and drawing silver to the probability of choosing purse 2 and drawing silver.

Exam Tip: When purses/bags are chosen at random, their prior probabilities are usually equal (\(\frac{1}{2}\)).

Question 5. Two thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy is getting a first class is 0.28. Find the probability that a student chosen at random will get first class marks in the subject.
Answer: Let \(E_1\) = Student is a boy. \(P(E_1) = \frac{2}{3}\)
Let \(E_2\) = Student is a girl. \(P(E_2) = 1 - \frac{2}{3} = \frac{1}{3}\)
Let A = Student gets first class marks.
\(P(A/E_1) = 0.28 = \frac{28}{100} = \frac{7}{25}\)
\(P(A/E_2) = 0.25 = \frac{25}{100} = \frac{1}{4}\)
Using Total Probability Theorem:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left(\frac{2}{3} \times \frac{7}{25}\right) + \left(\frac{1}{3} \times \frac{1}{4}\right) = \frac{14}{75} + \frac{1}{12}\)
Find common denominator (300):
\(P(A) = \frac{14 \times 4}{300} + \frac{1 \times 25}{300} = \frac{56 + 25}{300} = \frac{81}{300} = \frac{27}{100} = 0.27\)
In simple words: Calculate the weighted average of first-class probabilities for boys and girls.

Exam Tip: Convert decimals to fractions when finding common denominators makes arithmetic cleaner.

LEVEL II (continued from page 76)

Question 1. Find the probability of drawing a one-rupee coin from a purse with two compartments one of which contains 3 fifty-paise coins and 2 one-rupee coins and other contains 2 fifty-paise coins and 3 one-rupee coins.
Answer: Let \(E_1\) = Compartment 1 chosen. \(P(E_1) = \frac{1}{2}\)
Let \(E_2\) = Compartment 2 chosen. \(P(E_2) = \frac{1}{2}\)
Compartment 1: 3 fifty-paise coins, 2 one-rupee coins (total 5 coins).
Compartment 2: 2 fifty-paise coins, 3 one-rupee coins (total 5 coins).
Let A = Drawing a one-rupee coin.
\(P(A/E_1) = \frac{2}{5}\)
\(P(A/E_2) = \frac{3}{5}\)
Using Total Probability Theorem:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left(\frac{1}{2} \times \frac{2}{5}\right) + \left(\frac{1}{2} \times \frac{3}{5}\right) = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2} = 0.5\)
In simple words: Sum the probabilities of getting a one-rupee coin from either compartment.

Exam Tip: Equal compartment selection probabilities simplify total probability calculations.

Question 2. Suppose 5 men out of 100 and 25 women out of 1000 are good orator. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal number of men and women.
Answer: Let M = Selected person is male. \(P(M) = \frac{1}{2}\)
Let W = Selected person is female. \(P(W) = \frac{1}{2}\)
Let G = Person is a good orator.
\(P(G/M) = \frac{5}{100} = \frac{1}{20}\)
\(P(G/W) = \frac{25}{1000} = \frac{1}{40}\)
We need the conditional probability that a male person is selected given that the chosen orator is good: \(P(M/G)\).
Using Bayes' Theorem:
\(P(M/G) = \frac{P(M)P(G/M)}{P(M)P(G/M) + P(W)P(G/W)}\)
Numerator = \(\frac{1}{2} \times \frac{1}{20} = \frac{1}{40}\)
Denominator = \(\left(\frac{1}{2} \times \frac{1}{20}\right) + \left(\frac{1}{2} \times \frac{1}{40}\right) = \frac{1}{40} + \frac{1}{80} = \frac{2 + 1}{80} = \frac{3}{80}\)
\(P(M/G) = \frac{1/40}{3/80} = \frac{1}{40} \times \frac{80}{3} = \frac{2}{3}\)
In simple words: Apply Bayes' theorem using equal prior probabilities for men and women.

Exam Tip: Read carefully whether the question asks for unconditional probability or conditional probability via Bayes' theorem.

Question 3. A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant 40%. Out of that 80% of the bicycles are rated of standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be standard quality. Find the probability that it comes from the second plant.
Answer: Let \(E_1\) = Bicycle manufactured by Plant 1. \(P(E_1) = 0.60 = \frac{3}{5}\)
Let \(E_2\) = Bicycle manufactured by Plant 2. \(P(E_2) = 0.40 = \frac{2}{5}\)
Let A = Bicycle is of standard quality.
\(P(A/E_1) = 0.80 = \frac{4}{5}\)
\(P(A/E_2) = 0.90 = \frac{9}{10}\)
We need \(P(E_2/A)\), the probability that it came from the second plant given it is standard quality.
Using Bayes' Theorem:
\(P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(\frac{2}{5} \times \frac{9}{10} = \frac{18}{50}\)
Denominator = \(\left(\frac{3}{5} \times \frac{4}{5}\right) + \left(\frac{2}{5} \times \frac{9}{10}\right) = \frac{12}{25} + \frac{18}{50} = \frac{24 + 18}{50} = \frac{42}{50}\)
\(P(E_2/A) = \frac{18/50}{42/50} = \frac{18}{42} = \frac{3}{7}\)
In simple words: Use Bayes' theorem with plant manufacturing percentages as priors and standard quality rates as conditional probabilities.

Exam Tip: Converting percentage shares directly into fractions like \(\frac{3}{5}\) and \(\frac{2}{5}\) makes computation cleaner.

LEVEL III

Question 1. A letter is known to have come either from LONDON or CLIFTON. On the envelope just has two consecutive letters ON are visible. What is the probability that the letter has come from (i) LONDON (ii) CLIFTON ?
Answer: Let \(E_1\) = Letter came from LONDON. \(P(E_1) = \frac{1}{2}\)
Let \(E_2\) = Letter came from CLIFTON. \(P(E_2) = \frac{1}{2}\)
LONDON has 6 letters: L, O, N, D, O, N. Consecutive letter pairs (length 5): LO, ON, ND, DO, OS... wait, pairs are: LO, ON, ND, DO, ON. Total 5 pairs. How many 'ON'? Pairs: (L,O), (O,N), (N,D), (D,O), (O,N). Exactly 2 pairs of 'ON' out of 5 total pairs. So \(P(A/E_1) = \frac{2}{5}\).
CLIFTON has 7 letters: C, L, I, F, T, O, N. Consecutive letter pairs (length 6): CL, LI, IF, FT, TO, ON. Total 6 pairs. Exactly 1 pair of 'ON'. So \(P(A/E_2) = \frac{1}{6}\).
Total probability of getting 'ON':
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2) = \left(\frac{1}{2} \times \frac{2}{5}\right) + \left(\frac{1}{2} \times \frac{1}{6}\right) = \frac{1}{5} + \frac{1}{12} = \frac{12 + 5}{60} = \frac{17}{60}\)
(i) Probability that it came from LONDON given 'ON':
\(P(E_1/A) = \frac{\frac{1}{2} \times \frac{2}{5}}{\frac{17}{60}} = \frac{1/5}{17/60} = \frac{1}{5} \times \frac{60}{17} = \frac{12}{17}\)
(ii) Probability that it came from CLIFTON given 'ON':
\(P(E_2/A) = 1 - \frac{12}{17} = \frac{5}{17}\)
In simple words: Count total adjacent letter pairs in each word, find how many are 'ON', and apply Bayes' theorem.

Exam Tip: Carefully list out adjacent letter pairs to correctly determine conditional probabilities in word-based problems.

Question 2. A test detection of a particular disease is not fool proof. The test will correctly detect the disease 90% of the time, but will incorrectly detect the disease 1% of the time. For a large population of which an estimated 0.2% have the disease, a person is selected at random, given the test, and told that he has the disease. What are the chances that the person actually have the disease.
Answer: Let \(E_1\) = Person has the disease. \(P(E_1) = 0.2\% = \frac{0.2}{100} = 0.002 = \frac{2}{1000}\)
Let \(E_2\) = Person does not have the disease. \(P(E_2) = 1 - 0.002 = 0.998 = \frac{998}{1000}\)
Let A = Test result is positive (tells he has the disease).
\(P(A/E_1) = \text{Probability test is positive given person has disease} = 90\% = 0.90 = \frac{90}{100}\)
\(P(A/E_2) = \text{Probability test is positive given person does not have disease (false positive)} = 1\% = 0.01 = \frac{1}{100}\)
We need \(P(E_1/A)\), the probability the person actually has the disease given a positive test.
Using Bayes' Theorem:
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(\frac{2}{1000} \times \frac{90}{100} = \frac{180}{100000} = \frac{18}{10000}\)
Denominator = \(\frac{18}{10000} + \left(\frac{998}{1000} \times \frac{1}{100}\right) = \frac{18}{10000} + \frac{998}{100000} = \frac{180 + 998}{100000} = \frac{1178}{100000}\)
\(P(E_1/A) = \frac{180/100000}{1178/100000} = \frac{180}{1178} = \frac{90}{589} \approx 0.1528 \text{ (or about } 15.3\% \text{)}\)
In simple words: Use Bayes' theorem to balance true positives against false positives in a low-prevalence population.

Exam Tip: Keep denominators uniform (like powers of 10) when adding multi-term probabilities in Bayes' formula.

Question 3. Given three identical boxes I, II and III each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold ? [CBSE 2011]
Answer: Let \(E_1, E_2, E_3\) be the events of choosing Box I, Box II, and Box III respectively.
\(P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}\)
Let A be the event that the drawn coin is gold.
Box I has 2 gold coins: \(P(A/E_1) = \frac{2}{2} = 1\)
Box II has 0 gold coins (both silver): \(P(A/E_2) = 0\)
Box III has 1 gold and 1 silver coin: \(P(A/E_3) = \frac{1}{2}\)
We need the probability that the other coin is also gold, which means the chosen box is Box I (since only Box I has two gold coins).
Using Bayes' Theorem to find \(P(E_1/A)\):
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)}\)
Numerator = \(\frac{1}{3} \times 1 = \frac{1}{3}\)
Denominator = \(\left(\frac{1}{3} \times 1\right) + \left(\frac{1}{3} \times 0\right) + \left(\frac{1}{3} \times \frac{1}{2}\right) = \frac{1}{3} + 0 + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2}\)
\(P(E_1/A) = \frac{1/3}{1/2} = \frac{2}{3}\)
In simple words: The other coin is gold only if Box I was chosen. Apply Bayes' formula considering all boxes containing gold coins.

Exam Interpretation Tip: "Other coin is also gold" is equivalent to "Box I was selected given a gold coin was drawn".

(v) Random variables & probability distribution Mean & variance of random variables

LEVEL I

Question 1. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of spades.
Answer: Total cards = 52, Spades = 13, Non-spades = 39.
Probability of getting a spade in a single draw, \(p = \frac{13}{52} = \frac{1}{4}\)
Probability of not getting a spade, \(q = 1 - \frac{1}{4} = \frac{3}{4}\)
Let X denote the number of spades in 2 draws. X can take values 0, 1, 2.
This follows a binomial distribution with \(n = 2\), \(p = \frac{1}{4}\), \(q = \frac{3}{4}\):
\(P(X = x) = {}^{n}C_x p^x q^{n-x}\)
- For \(X = 0\):
\(P(X = 0) = {}^{2}C_0 \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^2 = 1 \times 1 \times \frac{9}{16} = \frac{9}{16}\)
- For \(X = 1\):
\(P(X = 1) = {}^{2}C_1 \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^1 = 2 \times \frac{1}{4} \times \frac{3}{4} = \frac{6}{16} = \frac{3}{8}\)
- For \(X = 2\):
\(P(X = 2) = {}^{2}C_2 \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^0 = 1 \times \frac{1}{16} \times 1 = \frac{1}{16}\)
Probability distribution table:

In simple words: Use binomial formula with probability of success \(p = \frac{1}{4}\) for 2 independent trials.

Exam Tip: Verify that the sum of all probabilities in the distribution equals 1 (\(\frac{9}{16} + \frac{6}{16} + \frac{1}{16} = 1\)).

Question 2. 4 defective apples are accidentally mixed with 16 good ones. Three apples are drawn at random from the mixed lot. Find the probability distribution of the number of defective apples.
Answer: Defective apples = 4, Good apples = 16, Total apples = 20.
Three apples are drawn, so X (number of defective apples) can take values 0, 1, 2, 3.
Total ways of drawing 3 apples = \({}^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140\)
- For \(X = 0\) (0 defective, 3 good):
\(P(X = 0) = \frac{{}^4C_0 \times {}^{16}C_3}{{}^{20}C_3} = \frac{1 \times \frac{16 \times 15 \times 14}{6}}{1140} = \frac{560}{1140} = \frac{56}{114} = \frac{28}{57}\)
- For \(X = 1\) (1 defective, 2 good):
\(P(X = 1) = \frac{{}^4C_1 \times {}^{16}C_2}{{}^{20}C_3} = \frac{4 \times \frac{16 \times 15}{2}}{1140} = \frac{4 \times 120}{1140} = \frac{480}{1140} = \frac{48}{114} = \frac{8}{19}\)
- For \(X = 2\) (2 defective, 1 good):
\(P(X = 2) = \frac{{}^4C_2 \times {}^{16}C_1}{{}^{20}C_3} = \frac{6 \times 16}{1140} = \frac{96}{1140} = \frac{8}{95}\)
- For \(X = 3\) (3 defective, 0 good):
\(P(X = 3) = \frac{{}^4C_3 \times {}^{16}C_0}{{}^{20}C_3} = \frac{4 \times 1}{1140} = \frac{4}{1140} = \frac{1}{285}\)
Probability distribution table:

In simple words: Use combinations (hypergeometric distribution) to find the probability of drawing each possible number of defective apples.

Exam Tip: Always double check that combination numerators sum up to total possible selections.

Question 3. A random variable X is specified by the following distribution.
Find the variance of the distribution.
X: 2, 3, 4
P(X): 0.3, 0.4, 0.3
Answer: Given distribution:

First, calculate Mean \(E(X) = \sum x_i p_i\):
\(E(X) = (2 \times 0.3) + (3 \times 0.4) + (4 \times 0.3) = 0.6 + 1.2 + 1.2 = 3.0\)
Next, calculate \(E(X^2) = \sum x_i^2 p_i\):
\(E(X^2) = (2^2 \times 0.3) + (3^2 \times 0.4) + (4^2 \times 0.3) = (4 \times 0.3) + (9 \times 0.4) + (16 \times 0.3)\)
\(E(X^2) = 1.2 + 3.6 + 4.8 = 9.6\)
Now, calculate Variance \(Var(X) = E(X^2) - [E(X)]^2\):
\(Var(X) = 9.6 - (3.0)^2 = 9.6 - 9.0 = 0.6\)
In simple words: Compute mean \(E(X)\) and expected square \(E(X^2)\), then use the variance formula \(Var(X) = E(X^2) - [E(X)]^2\).

Exam Tip: The variance formula \(E(X^2) - [E(X)]^2\) is much faster than computing \(\sum (x_i - \mu)^2 p_i\).

LEVEL III (continued from page 77)

Question 1. A coin is biased so that the head is 3 times as likely to occur as a tail. If the coin is tossed twice. Find the probability distribution of the number of tails.
Answer: Let probability of tail be \(P(T) = p\). Then probability of head is \(P(H) = 3p\).
Since \(P(H) + P(T) = 1\), we have \(3p + p = 1 \implies 4p = 1 \implies p = \frac{1}{4}\)
So, \(P(T) = \frac{1}{4}\) and \(P(H) = \frac{3}{4}\)
Let X be the number of tails in 2 tosses. X can take values 0, 1, 2.
- For \(X = 0\) (0 tails, 2 heads):
\(P(X = 0) = P(HH) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\)
- For \(X = 1\) (1 tail, 1 head: HT or TH):
\(P(X = 1) = P(HT) + P(TH) = \left(\frac{3}{4} \times \frac{1}{4}\right) + \left(\frac{1}{4} \times \frac{3}{4}\right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}\)
- For \(X = 2\) (2 tails: TT):
\(P(X = 2) = P(TT) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\)
Probability distribution table:

In simple words: Find individual probabilities for biased coin tosses and determine probabilities for 0, 1, and 2 tails.

Exam Tip: Always solve for individual single-trial probabilities first when dealing with biased coins.

Question 2. The sum of mean and variance of a binomial distribution for 5 trials be 1.8. Find the probability distribution.
Answer: Given number of trials \(n = 5\).
For binomial distribution, Mean = \(np\) and Variance = \(npq\).
Given sum of mean and variance = 1.8:
\(np + npq = 1.8\)
\(5p + 5pq = 1.8\)
Since \(q = 1 - p\):
\(5p + 5p(1 - p) = 1.8\)
\(5p + 5p - 5p^2 = 1.8\)
\(10p - 5p^2 = 1.8\)
Multiply by 5 to simplify:
\(50p - 25p^2 = 9 \implies 25p^2 - 50p + 9 = 0\)
Solve quadratic equation for p using factorisation (\(-45p - 5p = -50\)):
\(25p^2 - 45p - 5p + 9 = 0\)
\(5p(5p - 9) - 1(5p - 9) = 0\)
\((5p - 1)(5p - 9) = 0\)
Thus, \(p = \frac{1}{5} = 0.2\) (since \(p = \frac{9}{5} > 1\) is not possible).
Therefore, \(q = 1 - \frac{1}{5} = \frac{4}{5} = 0.8\)
Now find probability distribution for \(X = 0, 1, 2, 3, 4, 5\) using \(P(X = x) = {}^{5}C_x \left(\frac{1}{5}\right)^x \left(\frac{4}{5}\right)^{5-x}\):
- \(P(X = 0) = {}^{5}C_0 \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^5 = 1 \times 1 \times \frac{1024}{3125} = \frac{1024}{3125}\)
- \(P(X = 1) = {}^{5}C_1 \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^4 = 5 \times \frac{1}{5} \times \frac{256}{625} = \frac{256}{625}\)
- \(P(X = 2) = {}^{5}C_2 \left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^3 = 10 \times \frac{1}{25} \times \frac{64}{125} = \frac{640}{3125} = \frac{128}{625}\)
- \(P(X = 3) = {}^{5}C_3 \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^2 = 10 \times \frac{1}{125} \times \frac{16}{25} = \frac{160}{3125} = \frac{32}{625}\)
- \(P(X = 4) = {}^{5}C_4 \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^1 = 5 \times \frac{1}{625} \times \frac{4}{5} = \frac{20}{3125} = \frac{4}{625}\)
- \(P(X = 5) = {}^{5}C_5 \left(\frac{1}{5}\right)^5 \left(\frac{4}{5}\right)^0 = 1 \times \frac{1}{3125} \times 1 = \frac{1}{3125}\)
In simple words: Set up the binomial mean and variance equation to find p, then compute probabilities for each value of X.

Exam Tip: Remember standard binomial formulas: Mean = \(np\) and Variance = \(npq\).

Question 3. The mean and variance of a binomial distribution are \(\frac{4}{3}\) and \(\frac{8}{9}\) respectively. Find \(P(X \ge 1)\).
Answer: Mean \(np = \frac{4}{3}\) --- (1)
Variance \(npq = \frac{8}{9}\) --- (2)
Divide (2) by (1):
\(\frac{npq}{np} = \frac{8/9}{4/3} \implies q = \frac{8}{9} \times \frac{3}{4} = \frac{2}{3}\)
Since \(p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}\)
Substitute p into (1) to find n:
\(n \left(\frac{1}{3}\right) = \frac{4}{3} \implies n = 4\)
We need \(P(X \ge 1) = 1 - P(X = 0)\):
\(P(X = 0) = {}^{4}C_0 p^0 q^4 = 1 \times 1 \times \left(\frac{2}{3}\right)^4 = \frac{16}{81}\)
\(P(X \ge 1) = 1 - \frac{16}{81} = \frac{65}{81}\)
In simple words: Divide variance by mean to find q, then find n and calculate \(1 - P(X = 0)\).

Exam Tip: Dividing variance by mean (\(\frac{npq}{np} = q\)) is the fastest way to find q in binomial problems.

(vi) Bernoulli's trials and Binomial Distribution

LEVEL II

Question 1. If a die is thrown 5 times, what is the chance that an even number will come up exactly 3 times.
Answer: Number of trials \(n = 5\).
Probability of getting an even number in a single throw of a die, \(p = \frac{3}{6} = \frac{1}{2}\)
Probability of failure (not even), \(q = 1 - \frac{1}{2} = \frac{1}{2}\)
We need probability of exactly 3 successes (\(X = 3\)):
\(P(X = 3) = {}^{5}C_3 p^3 q^{5-3} = {}^{5}C_3 \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \times \left(\frac{1}{2}\right)^5 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}\)
In simple words: Use binomial probability formula with \(n = 5\) and \(p = \frac{1}{2}\).

Exam Tip: For symmetric outcomes like even/odd on a die, \(p = q = \frac{1}{2}\).

Question 2. An experiment succeeds twice as often it fails. Find the probability that in the next six trials, there will be at least 4 success.
Answer: Let probability of success be p and failure be q.
Given success occurs twice as often as failure: \(p = 2q\)
Since \(p + q = 1\), we have \(2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3}\)
Therefore, \(p = 1 - \frac{1}{3} = \frac{2}{3}\)
Number of trials \(n = 6\).
We need probability of at least 4 successes: \(P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6)\)
- \(P(X = 4) = {}^{6}C_4 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{240}{729}\)
- \(P(X = 5) = {}^{6}C_5 \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{192}{729}\)
- \(P(X = 6) = {}^{6}C_6 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^0 = 1 \times \frac{64}{729} \times 1 = \frac{64}{729}\)
Summing these probabilities:
\(P(X \ge 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}\)
In simple words: Determine p and q from the ratio, then sum binomial probabilities for 4, 5, and 6 successes.

Exam Tip: When calculating "at least" with high trial counts, check if summing directly or using complement is faster.

Question 3. A pair of dice is thrown 200 times. If getting a sum 9 is considered a success, find the mean and variance of the number of success.
Answer: Number of trials \(n = 200\)
When a pair of dice is thrown, total outcomes = 36.
Outcomes giving a sum of 9: {(3, 6), (4, 5), (5, 4), (6, 3)}, total 4 outcomes.
Probability of success in a single trial, \(p = \frac{4}{36} = \frac{1}{9}\)
Probability of failure, \(q = 1 - \frac{1}{9} = \frac{8}{9}\)
Mean = \(np = 200 \times \frac{1}{9} = \frac{200}{9}\)
Variance = \(npq = 200 \times \frac{1}{9} \times \frac{8}{9} = \frac{1600}{81}\)
In simple words: Find single-trial success probability p, then compute mean as \(np\) and variance as \(npq\).

Exam Tip: Remember standard binomial parameters: Mean is \(np\) and Variance is \(npq\).

Questions for self evaluation

Question 1. A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Find the probability that the number is divisible by 5.
Answer: Available digits: 1, 2, 3, 5 (4 digits).
Total 4-digit numbers that can be formed without repetition = \(4! = 24\)
A number is divisible by 5 if its units digit is 5.
Fixing 5 at the units place, the remaining 3 digits (1, 2, 3) can be arranged in the first three places in \(3! = 6\) ways.
Required probability = \(\frac{6}{24} = \frac{1}{4}\)
In simple words: Divide the number of 4-digit numbers ending in 5 by the total possible 4-digit numbers.

Exam Tip: Divisibility by 5 requires the last digit to be 0 or 5. Here, only 5 is available.

Question 2. The probability that an event happens in one trial of an experiment is 0.4. Three independent trials of an experiment are performed. Find the probability that the event happens at least once.
Answer: Probability of success in one trial \(p = 0.4\)
Probability of failure in one trial \(q = 1 - 0.4 = 0.6\)
Number of trials \(n = 3\)
Probability that the event happens at least once = \(1 - P(\text{happens zero times})\)
\(P(X = 0) = {}^{3}C_0 p^0 q^3 = 1 \times 1 \times (0.6)^3 = 0.216\)
\(P(\text{at least once}) = 1 - 0.216 = 0.784\)
In simple words: Subtract the probability of zero successes from 1 to find the probability of at least one success.

Exam Tip: "At least once" problems are almost always easiest using the complement rule \(1 - P(\text{none})\).

Question 3. A football match is either won, draw or lost by the host country's team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.
Answer: For each match, probability of correct forecast \(p = \frac{1}{3}\)
Probability of incorrect forecast \(q = \frac{2}{3}\)
Number of matches \(n = 4\)
We need probability of at least 3 correct results: \(P(X \ge 3) = P(X = 3) + P(X = 4)\)
- \(P(X = 3) = {}^{4}C_3 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 = 4 \times \frac{1}{27} \times \frac{2}{3} = \frac{8}{81}\)
- \(P(X = 4) = {}^{4}C_4 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^0 = 1 \times \frac{1}{81} \times 1 = \frac{1}{81}\)
Total probability = \(\frac{8}{81} + \frac{1}{81} = \frac{9}{81} = \frac{1}{9}\)
In simple words: Use binomial distribution with \(p = \frac{1}{3}\) and \(n = 4\) for 3 and 4 successes.

Exam Tip: "At least 3" for 4 trials means you only calculate for \(X = 3\) and \(X = 4\).

Question 4. A candidate has to reach the examination center in time. Probability of him going by bus or scooter or by other means of transport are \(\frac{3}{5}, \frac{1}{10}, \frac{3}{10}\) respectively. The probability that he will be late is \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively (for bus and scooter). But he reaches in time if he uses other mode of transport. He reached late at the centre. Find the probability that he traveled by bus.
Answer: Let \(E_1\) = Goes by bus. \(P(E_1) = \frac{3}{5}\)
Let \(E_2\) = Goes by scooter. \(P(E_2) = \frac{1}{10}\)
Let \(E_3\) = Goes by other means. \(P(E_3) = \frac{3}{10}\)
Let A = Candidate reaches late.
\(P(A/E_1) = \frac{1}{3}\) (late probability by bus)
\(P(A/E_2) = \frac{1}{4}\) (late probability by scooter)
\(P(A/E_3) = 0\) (reaches in time by other means)
We need \(P(E_1/A)\), the probability that he traveled by bus given that he reached late.
Using Bayes' Theorem:
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)}\)
Numerator = \(\frac{3}{5} \times \frac{1}{3} = \frac{3}{15} = \frac{1}{5} = \frac{12}{60}\)
Denominator = \(\left(\frac{3}{5} \times \frac{1}{3}\right) + \left(\frac{1}{10} \times \frac{1}{4}\right) + \left(\frac{3}{10} \times 0\right) = \frac{3}{15} + \frac{1}{40} + 0\)
Common denominator (120):
Numerator = \(\frac{24}{120}\)
Denominator = \(\frac{24}{120} + \frac{3}{120} = \frac{27}{120}\)
\(P(E_1/A) = \frac{24/120}{27/120} = \frac{24}{27} = \frac{8}{9}\)
In simple words: Apply Bayes' theorem with three transport options, noting that the probability of being late via other transport is zero.

Exam Tip: Zero conditional probability terms in the denominator can be omitted safely.

Question 5. Let X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that
\(P(X = x) = \begin{cases} kx, & \text{if } x = 0, \text{ or } 1 \\ 2kx, & \text{if } x = 2 \\ k(5 - x), & \text{if } x = 3 \text{ or } 4 \end{cases}\), where k is a +ve constant.
Find the mean and variance of the probability distribution.
Answer: First, find k using the property that the sum of all probabilities equals 1:
\(\sum P(X = x) = 1\)
\(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1\)
\((k \times 0) + (k \times 1) + (2k \times 2) + k(5 - 3) + k(5 - 4) = 1\)
\(0 + k + 4k + 2k + k = 1\)
\(8k = 1 \implies k = \frac{1}{8}\)
Probability distribution table:
- \(P(X = 0) = 0\)
- \(P(X = 1) = k = \frac{1}{8}\)
- \(P(X = 2) = 4k = \frac{4}{8} = \frac{1}{2}\)
- \(P(X = 3) = 2k = \frac{2}{8} = \frac{1}{4}\)
- \(P(X = 4) = k = \frac{1}{8}\)
Let's compute Mean \(E(X) = \sum x_i p_i\):
\(E(X) = \left(0 \times 0\right) + \left(1 \times \frac{1}{8}\right) + \left(2 \times \frac{4}{8}\right) + \left(3 \times \frac{2}{8}\right) + \left(4 \times \frac{1}{8}\right)\)
\(E(X) = 0 + \frac{1}{8} + \frac{8}{8} + \frac{6}{8} + \frac{4}{8} = \frac{19}{8} = 2.375\)
Next, compute \(E(X^2) = \sum x_i^2 p_i\):
\(E(X^2) = \left(0^2 \times 0\right) + \left(1^2 \times \frac{1}{8}\right) + \left(2^2 \times \frac{4}{8}\right) + \left(3^2 \times \frac{2}{8}\right) + \left(4^2 \times \frac{1}{8}\right)\)
\(E(X^2) = 0 + \frac{1}{8} + \frac{16}{8} + \frac{18}{8} + \frac{16}{8} = \frac{51}{8} = 6.375\)
Now, compute Variance \(Var(X) = E(X^2) - [E(X)]^2\):
\(Var(X) = \frac{51}{8} - \left(\frac{19}{8}\right)^2 = \frac{51}{8} - \frac{361}{64} = \frac{408 - 361}{64} = \frac{47}{64}\)
In simple words: Find constant k by setting the sum of probabilities to 1, then calculate mean and variance using standard expectation formulas.

Exam Tip: Always verify that \(\sum P(X=x) = 1\) to correctly determine unknown constants like k.

Question 6. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
Answer: Probability of getting a six in a single throw, \(p = \frac{1}{6}\)
Probability of not getting a six, \(q = \frac{5}{6}\)
We need the probability of obtaining the 3rd six in the 6th throw.
This means exactly two sixes must appear in the first 5 throws, and the 6th throw must be a six.
Probability of getting 2 sixes in the first 5 throws is given by binomial probability:
\(P(2 \text{ sixes in 5 throws}) = {}^{5}C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 = 10 \times \frac{1}{36} \times \frac{125}{216} = \frac{1250}{7776}\)
Probability of the 6th throw being a six = \(\frac{1}{6}\)
Required probability = \(\frac{1250}{7776} \times \frac{1}{6} = \frac{1250}{46656} = \frac{625}{23328}\)
In simple words: Multiply the probability of getting 2 successes in 5 trials by the probability of success on the 6th trial.

Exam Tip: "Nth success on the k-th trial" means \(n-1\) successes occur in the first \(k-1\) trials, followed by a success on the k-th trial.

Question 7. On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?
Answer: Number of questions \(n = 5\)
Probability of a correct answer by guessing, \(p = \frac{1}{3}\)
Probability of an incorrect answer, \(q = \frac{2}{3}\)
We need probability of 4 or more correct answers: \(P(X \ge 4) = P(X = 4) + P(X = 5)\)
- \(P(X = 4) = {}^{5}C_4 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^1 = 5 \times \frac{1}{81} \times \frac{2}{3} = \frac{10}{243}\)
- \(P(X = 5) = {}^{5}C_5 \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^0 = 1 \times \frac{1}{243} \times 1 = \frac{1}{243}\)
Total probability = \(\frac{10}{243} + \frac{1}{243} = \frac{11}{243}\)
In simple words: Use binomial formula to calculate probabilities for 4 and 5 correct answers and add them together.

Exam Tip: "Four or more" out of 5 trials means evaluating exactly for \(X = 4\) and \(X = 5\).

Question 8. Two cards are drawn simultaneously (or successively) from a well shuffled pack of 52 cards. Find the mean and variance of the number of red cards.
Answer: Total cards = 52, Red cards = 26, Black cards = 26.
Number of cards drawn \(n = 2\)
Probability of getting a red card in a single draw, \(p = \frac{26}{52} = \frac{1}{2}\)
Probability of not getting a red card, \(q = 1 - \frac{1}{2} = \frac{1}{2}\)
Since cards are drawn simultaneously (or successively without replacement, wait - standard deck drawing without replacement follows hypergeometric distribution, but for 2 draws from 52 cards, mean and variance can be computed via binomial approximation or exact hypergeometric formulas).
Exact hypergeometric / binomial parameters for 2 draws:
X can take values 0, 1, 2.
- \(P(X = 0) = \frac{{}^{26}C_0 \times {}^{26}C_2}{{}^{52}C_2} = \frac{1 \times \frac{26 \times 25}{2}}{\frac{52 \times 51}{2}} = \frac{325}{1326} = \frac{25}{102}\)
- \(P(X = 1) = \frac{{}^{26}C_1 \times {}^{26}C_1}{{}^{52}C_2} = \frac{26 \times 26}{1326} = \frac{676}{1326} = \frac{52}{102} = \frac{26}{51}\)
- \(P(X = 2) = \frac{{}^{26}C_2 \times {}^{26}C_0}{{}^{52}C_2} = \frac{325}{1326} = \frac{25}{102}\)
Let's compute Mean \(E(X) = \sum x_i p_i\):
\(E(X) = \left(0 \times \frac{25}{102}\right) + \left(1 \times \frac{52}{102}\right) + \left(2 \times \frac{25}{102}\right) = \frac{52 + 50}{102} = \frac{102}{102} = 1\)
Next, compute \(E(X^2) = \sum x_i^2 p_i\):
\(E(X^2) = \left(0^2 \times \frac{25}{102}\right) + \left(1^2 \times \frac{52}{102}\right) + \left(2^2 \times \frac{25}{102}\right) = \frac{52 + (4 \times 25)}{102} = \frac{52 + 100}{102} = \frac{152}{102} = \frac{76}{51}\)
Now, compute Variance \(Var(X) = E(X^2) - [E(X)]^2\):
\(Var(X) = \frac{76}{51} - (1)^2 = \frac{76}{51} - \frac{51}{51} = \frac{25}{51}\)
In simple words: Find the probability distribution of red cards using combinations, then compute mean and variance using expectation formulas.

Exam Tip: For drawing without replacement, use hypergeometric probabilities (\({n}C_x\) ratios) rather than standard binomial formulas.