CBSE Class 12 Mathematics Probability Assignment Set 02

Question. From a book containing 100 pages, one page is selected randomly. The probability that the sum of the digits of the page number of the selected page is 11, is:
a. 2/25
b. 9/100
c. 11/100
d. None of these
Answer : A

Question. A dice is thrown twice. The probability of getting 4, 5 or 6 in the first throw and 1, 2, 3 or 4 in the second throw is:
a. 1
b. 1/3
c. 7/36
d. None of these
Answer : B

Question. Two fair dice are tossed. Let A be the event that the first die shows an even number and B be the event that second die shows an odd number. The two events A and B are:
a. Mutually exclusive
b. Independent and mutually exclusive
c. Dependent
d. None of these
Answer : D

Question. Three identical dice are rolled. The probability that same number will appear on each of them will be:
a. 1/6
b. 1/36
c. 1/18
d. 3/28
Answer : B

Question. A coin is tossed and a dice is rolled. The probability that the coin shows the head and the dice shows 6 is:
a. 1/8
b. 1/12
c. 1/2
d. 1
Answer : B

Question. The probabilities of a student getting I, II and III division in an examination are respectively
1/10, 3/5 and 1/4 The probability that the student fail in the examination is:
a. 197/200
b. 27/200
c. 83/100
d. None of these
Answer : D

Question. One card is drawn from each of two ordinary packs of 52 cards. The probability that at least one of them is an ace of heart, is:
a. 103/2704
b. 1/2704
c. 2/52
d. 2601/2704
Answer : A

Question. A single letter is selected at random from the word “PROBABILITY”. The probability that the selected letter is a vowel is:
a. 2/11
b. 3/11
c. 4/11
d. 0
Answer : C

Question. The probability of happening an event A in one trial is 0.4. The probability that the event A happens at least once in three independent trials is:
a. 0.936
b. 0.784
c. 0.904
d. 0.216
Answer : B

Question. If a dice is thrown twice, then the probability of getting 1 in the first throw only is:
a. 1/36
b. 3/36
c. 5/36
d. 1/6
Answer : C

Question. From 10,000 lottery tickets numbered from 1 to 10,000, one ticket is drawn at random. What is the probability that the number marked on the drawn ticket is divisible by 20?
a. 1/100
b. 1/50
c. 1/20
d. 1/10
Answer : C

Question. From a pack of 52 cards two are drawn with replacement.
The probability, that the first is a diamond and the second is a king, is:
a. 1/26
b. 17/2704
c. 1/52
d. None of these
Answer : C

Question. A card is drawn from a well shuffled pack of cards. The probability of getting a queen of club or king of heart is:
a. 1/52
b. 1/26
c. 1/18
d. None of these
Answer : B

Question. A man and a woman appear in an interview for two vacancies in the same post. The probability of man's selection is 1/4 and that of the woman's selection is 1/3.
What is the probability that none of them will be selected?
a. 1/2
b. 1/12
c. 1/4
d. None of these
Answer : D

Question. Three letters are to be sent to different persons and addresses on the three envelopes are also written. Without looking at the addresses, the probability that the letters go into the right envelope is equal to:
a. 27/1
b. 9/1
c. 27/4
d. 6/1
Answer : D

Question. The probability of getting a total of 5 or 6 in a single throw of 2 dice is:
a. 2/1
b. 4/1
c. 3/1
d. 6/1
Answer : B

Question. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, is equal to:
a. 1/2
b. 1/5
c. 1/10
d. 1/20
Answer : C

Question. The letters of the word ‘ASSASSIN’ are written down at random in a row. The probability that no two S occur together is:
a. 1/35
b. 1/14
c. 1/15
d. None of these
Answer : B

Question. A problem of mathematics is given to three students whose chances of solving the problem are 1/3, 1/4 and 1/5 respectively. The probability that the question will be solved is
a. 3/2
b. 4/3
c. 5/4
d. 5/3
Answer : D

Question. A man draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues this processes until he gets a card of spade. The probability that he will fail the first two times is:
a. 9/16
b. 1/16
c. 9/64
d. None of these
Answer : C

Question. Two dice are thrown simultaneously. What is the probability of obtaining sum of the numbers less than 11?
a. 17/18
b. 1/12
c. 11/12
d. None of these
Answer : C

Question. A bag contains 4 white, 5 red and 6 black balls. If two balls are drawn at random, then the probability that one of them is white is:
a. 44/105
b. 11/105
c. 11/ 21
d. None of these
Answer : A

Question. The probability that an event will fail to happen is 0.05.
The probability that the event will take place on 4 consecutive occasions is:
a. 0.00000625
b. 0.18543125
c. 0.00001875
d. 0.81450625
Answer : D

Question. If Mohan has 3 tickets of a lottery containing 3 prizes and 9 blanks, then his chance of winning prize are:
a. 34/55
b. 21/55
c. 17/55
d. None of these
Answer : A

Question. There are four letters and four addressed envelopes. The chance that all letters are not dispatched in the right envelope is:
a. 19/24
b. 21/23
c. 23/24
d. 1/24
Answer : C

Question. A committee of five is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all, is:
a. 1/2
b. 5/9
c. 4/9
d. 2/9
Answer : C

Question. Two cards are drawn at random from a pack of 52 cards.
The probability that both are the cards of spade is:
a. 1/26
b. 1/4
c. 1/17
d. None of these
Answer : C

Question. Word ‘UNIVERSITY’ is arranged randomly. Then the probability that both ‘I’ does not come together, is:
a. 3/5
b. 2/5
c. 4/5
d. 1/5
Answer : C

Question. A bag contains 3 red, 7 white and 4 black balls. If three balls are drawn from the bag, then the probability that all of them are of the same colour is:
a. 6/71
b. 7/81
c. 10/91
d. None of these
Answer : C

Question. A box contains 25 tickets numbered 1, 2,...25. If two tickets are drawn at random then the probability that the product of their numbers is even, is:
a. 11/50
b. 13/50
c. 37/50
d. None of these
Answer : C

Question. A word consists of 11 letters in which there are 7 consonants and 4 vowels. If 2 letters are chosen at random, then the probability that all of them are consonants, is:
a. 5/11
b. 21/55
c. 4/11
d. None of these
Answer : B

Question. Two friends A and B have equal number of daughters. There are three cinema tickets which are to be distributed among the daughters of A and B. The probability that all the tickets go to daughters of A is 1/20. The number of daughters each of them have is:
a. 4
b. 5
c. 6
d. 3
Answer : D

Question. In a lottery 50 tickets are sold in which 14 are of prize. A man bought 2 tickets, then the probability that the man win the prize, is:
a. 17/35
b. 18/35
c. 72/175
d. 13/175
Answer : A

Question. A party of 23 persons take their seats at a round table. The odds against two persons sitting together are:
a. 10 : 1
b. 1 : 11
c. 9 : 10
d. None of these
Answer : A

Question. The probability of happening at least one of the events A and B is 0.6. If the events A and B happens simultaneously with the probability 0.2, then P(A) + P(B) = ?
a.0.4
b.0.8
c.1.2
d.1.4
Answer : C

Question. A three digit number is formed by using numbers 1, 2, 3 and 4. The probability that the number is divisible by 3, is:
a. 2/3
b. 2/7
c. 1/2
d. 3/4
Answer : C

Question. Two dice are tossed together. The odds in favour of the sum of the numbers on them as 2 are:
a. 1:36
b. 1:35
c. 35:1
d. None of these
Answer : B

Question. An ordinary cube has four blank faces, one face marked 2 another marked 3. Then the probability of obtaining a total of exactly 12 in 5 throws, is:
a. 5/1296
b. 5/1944
c. 5/2592
d. None of these
Answer : C

Question. A bag contains 3 white and 5 black balls. If one ball is drawn, then the probability that it is black, is:
a. 3/8
b. 5/8
c. 6/8 
d. 10/20
Answer : B

Question. A bag has 13 red, 14 green and 15 black balls. The probability of getting exactly 2 blacks on pulling out 4 balls is P₁. Now the number of each colour ball is doubled and 8 balls are pulled out. The probability of getting exactly 4 blacks is P₂. Then:
a. P₁ = P₂
b. P₁ > P₂
c. P₁ < P₂
d. None of these
Answer : B

Question. In a horse race the odds in favour of three horses are 1:2, 1:3 and 1:4. The probability that one of the horse will win the race is:
a. 37/60
b. 47/60
c. 1/4
d. 3/4
Answer : B

Question. The probabilities that A and B will die within a year are p and q respectively, then the probability that only one of them will be alive at the end of the year is:
a. p + q
b. p + q − 2qp
c. p + q − pq
d. p + q + pq
Answer : B

Question. If A and B are two independent events such that P(A) = 0. 40, P(B) = 0. 50. Find P (neither A nor B)
a.0.90
b.0.10
c.0.2
d.0.3
Answer : D

Question. If the odds against an event be 2 : 3, then the probability of its occurrence is:
a. 1/5
b. 2/5
c. 3/5
d. 1
Answer : C

Question. A card is drawn from a pack of 52 cards. A gambler bets that it is a spade or an ace. What are the odds against his winning this bet?
a. 17 : 52
b. 52 : 17
c. 9 : 4
d. 4 : 9
Answer : C

Question. Let A and B be two events such that P(A) = 0.3 and P(A∪ B) = 0.8. If A and B are independent events, then P(B) = ?
a. 5/6
b. 5/7
c. 3/5
d. 2/5
Answer : B

Question. If odds against solving a question by three students are 2:1, 5 : 2 and 5 : 3 respectively, then probability that the question is solved only by one student is:
a. 31/56
b. 24/56
c. 25/56
d. None of these
Answer : C

Question. A card is drawn at random from a pack of cards. The probability of this card being a red or a queen is:
a. 1/13
b. 1/26
c. 1/2
d. 7/13
Answer : D

Question. If the probability of a horse A winning a race is 1/4 and the probability of a horse B winning the same race is 1/5, then the probability that either of them will win the race is:
a. 1/20
b. 9/20
c. 11/20
d. 19/20
Answer : B

Question. The two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Then the probability that neither A nor B occurs is:
a. 0.39
b. 0.25
c. 0.904
d. None of these
Answer : A

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 \(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0}{0.3} = 0\)
In simple words: When two events cannot happen together, the probability of one happening given that the other has already happened is zero.

Exam Tip: Remember that for mutually exclusive events, the intersection is empty and its probability is always 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 balls = 3 + 5 = 8
Number of white balls = 5
Probability of drawing a white ball in the first draw \(P(W_1) = \frac{5}{8}\)
Since the first ball is not replaced, remaining total balls = 7
Remaining white balls = 4
Probability of drawing a white ball in the second draw given that the first ball is white \(P(W_2/W_1) = \frac{4}{7}\)
Required probability = \(P(W_1 \cap W_2) = P(W_1) \times P(W_2/W_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\)
In simple words: Multiply the probability of drawing the first white ball by the reduced probability of drawing the second white ball.

Exam Tip: For problems without replacement, always adjust the total number of items and the count of the specific item 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 F be the event that the sum of the numbers is 6:
\(F = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}\)
Let E be the event that the number 4 appears at least once:
\(E = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)\}\)
Intersection \(E \cap F = \{(2, 4), (4, 2)\}\)
Number of elements in F = 5
Number of elements in \(E \cap F\) = 2
Conditional probability \(P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{5}\)
In simple words: Find all outcomes where the sum is 6, then check how many of those pairs contain the number 4.

Exam Tip: Clearly write down the sample space subsets for conditional probability to avoid missing any ordered pairs.

LEVEL III

Question 1. If \(P(A) = \frac{3}{8}\), \(P(B) = \frac{1}{2}\) and \(P(A \cap B) = \frac{1}{4}\), find \(P(A/B)\) and \(P(B/A)\).
Answer: Given data:
\(P(A) = \frac{3}{8}\)
\(P(B) = \frac{1}{2}\)
\(P(A \cap B) = \frac{1}{4}\)
Conditional probability \(P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2}\)
Conditional probability \(P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{4}}{\frac{3}{8}} = \frac{1}{4} \times \frac{8}{3} = \frac{2}{3}\)
In simple words: Use the conditional probability formula by dividing the probability of both events happening together by the probability of the given event.

Exam Tip: Always state the formula \(P(A/B) = \frac{P(A \cap B)}{P(B)}\) clearly before substituting values.

(ii) Multiplication theorem on probability

LEVEL II

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 number of balls = 5 + 7 + 3 = 15
Number of non-red balls = 5 (white) + 3 (black) = 8
We need to draw 3 balls one by one without replacement, and none should be red (i.e., all 3 must be non-red).
Probability that the first ball is not red = \(\frac{8}{15}\)
Probability that the second ball is not red given the first is not red = \(\frac{7}{14}\)
Probability that the third ball is not red given the first two are not 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: Multiply the probabilities of drawing a non-red ball successively while reducing both the total balls and non-red balls by 1 each time.

Exam Tip: For 'none is red', treat non-red balls as your target favorable outcomes and reduce totals successively without replacement.

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 \(P(A') = 1 - \frac{3}{7} = \frac{4}{7}\)
\(P(B) = \frac{1}{3}\), so \(P(B') = 1 - \frac{1}{3} = \frac{2}{3}\)
(i) Probability that at least one of them hits the target = \(1 - P(\text{neither hits})\)
\(P(\text{neither hits}) = P(A') \times P(B') = \frac{4}{7} \times \frac{2}{3} = \frac{8}{21}\)
Required probability = \(1 - \frac{8}{21} = \frac{13}{21}\)
(ii) Probability that only one of them hits the target = \(P(A \text{ hits and } B \text{ misses}) + P(A \text{ misses and } B \text{ hits})\)
\(= [P(A) \times P(B')] + [P(A') \times P(B)]\)
\(= \left( \frac{3}{7} \times \frac{2}{3} \right) + \left( \frac{4}{7} \times \frac{1}{3} \right)\)
\(= \frac{6}{21} + \frac{4}{21} = \frac{10}{21}\)
In simple words: For 'at least one', subtract the probability that both miss from 1. For 'only one', add the two mutually exclusive cases where one hits and the other misses.

Exam Tip: Complementary probability \(1 - P(\text{none})\) is the fastest way to solve 'at least one' problems.

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 number of students = 80
Assuming the characteristics (gender, economic status, and complexion) operate independently or standard proportions apply as given in textbook problems of this type, let us analyze the given parameters:
Total girls = 25
Total rich students = 10
Total fair-complexioned students = 20
Assuming uniform distribution of traits across independent categories, probability of selecting a girl \(P(\text{Girl}) = \frac{25}{80}\)
Probability of being rich given girl \(P(\text{Rich}/\text{Girl}) = \frac{10}{80}\) (or assuming independent joint probability)
Using joint probability multiplication for independent attributes: \(P(\text{Girl}) \times P(\text{Rich}) \times P(\text{Fair}) = \frac{25}{80} \times \frac{10}{80} \times \frac{20}{80}\)
Alternatively, interpreting standard combinatorial product if independent: \(\frac{25}{80} \times \frac{10}{80} \times \frac{20}{80} = \frac{5}{16} \times \frac{1}{8} \times \frac{1}{4} = \frac{5}{512}\)
In simple words: Multiply the individual probabilities of each independent characteristic occurring together.

Exam Tip: When multiple independent attributes are given for a single selection, apply the multiplication theorem for independent events.

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: Total integers from 1 through 11 = 11 integers (6 odd: 1, 3, 5, 7, 9, 11; 5 even: 2, 4, 6, 8, 10).
Total ways to select 2 integers out of 11 = \(^{11}C_2 = \frac{11 \times 10}{2} = 55\)
For the sum of two integers to be even, both numbers must be either both even or both odd.
Case I: Both numbers are even. Number of ways = \(^5C_2 = \frac{5 \times 4}{2} = 10\)
Case II: Both numbers are odd. Number of ways = \(^6C_2 = \frac{6 \times 5}{2} = 15\)
Total ways for sum to be even = 10 + 15 = 25
We need the conditional probability that both numbers are odd given that their sum is even:
\(P(\text{both odd} / \text{sum is even}) = \frac{P(\text{both odd and sum is even})}{P(\text{sum is even})} = \frac{15}{25} = \frac{3}{5}\)
In simple words: Out of all pairs that add up to an even number, find the ratio of pairs where both numbers are odd.

Exam Tip: Sum of two integers is even only when both are even or both are odd.

(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}
Event E (first throw is head) = {HHH, HHT, HTH, HTT}, so \(P(E) = \frac{4}{8} = \frac{1}{2}\)
Event F (last throw is tail) = {HHT, HTT, THT, TTT}, so \(P(F) = \frac{4}{8} = \frac{1}{2}\)
Intersection \(E \cap F\) (first is head and last is tail) = {HHT, HTT}, so \(P(E \cap F) = \frac{2}{8} = \frac{1}{4}\)
Check for independence: \(P(E) \times P(F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)
Since \(P(E \cap F) = P(E) \times P(F) = \frac{1}{4}\), 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 by checking if \(P(A \cap B) = P(A) \cdot P(B)\).

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 theorem:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(\frac{3}{4} = \frac{1}{4} + \frac{2}{3} - P(A \cap B)\)
\(P(A \cap B) = \frac{1}{4} + \frac{2}{3} - \frac{3}{4} = \frac{2}{3} - \frac{2}{4} = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6}\)
Now, calculate the product of individual probabilities:
\(P(A) \times P(B) = \frac{1}{4} \times \frac{2}{3} = \frac{2}{12} = \frac{1}{6}\)
Since \(P(A \cap B) = P(A) \times P(B) = \frac{1}{6}\), the events A and B are independent.
In simple words: Calculate the intersection probability first using union formula, then compare it with the product of individual probabilities.

Exam Tip: Use \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\) to find intersection when union is given.

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 \(P(A) \cdot P(B)\) in the union formula and solve for \(P(B)\).

Exam Tip: Substituting independent property \(P(A \cap B) = P(A)P(B)\) into standard addition rule creates a linear equation in one variable.

(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: Bag 1 contains 6 red, 5 blue balls (Total = 11)
Bag 2 contains 5 red, 8 blue balls (Total = 13)
Let \(E_1\) be the event that a red ball is transferred from Bag 1 to Bag 2.
Let \(E_2\) be the event that a blue ball is transferred from Bag 1 to Bag 2.
Let A be the event that a blue ball is drawn from Bag 2.
\(P(E_1) = \frac{6}{11}\)
\(P(E_2) = \frac{5}{11}\)
If a red ball is transferred, Bag 2 has 6 red and 8 blue balls (Total = 14), so \(P(A/E_1) = \frac{8}{14} = \frac{4}{7}\)
If a blue ball is transferred, Bag 2 has 5 red and 9 blue balls (Total = 14), so \(P(A/E_2) = \frac{9}{14}\)
Using total probability theorem:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left( \frac{6}{11} \times \frac{4}{7} \right) + \left( \frac{5}{11} \times \frac{9}{14} \right)\)
\(P(A) = \frac{24}{77} + \frac{45}{154} = \frac{48 + 45}{154} = \frac{93}{154}\)
In simple words: Consider both cases (transferring a red or blue ball), calculate the new probabilities in the second bag, and add their weighted products.

Exam Tip: Total probability theorem requires summing the probabilities of all mutually exclusive scenarios leading to the final event.

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). Probability \(P(E_1) = \frac{13}{52} = \frac{1}{4}\)
Let \(E_2\) = Lost card is not a heart (39 cards). Probability \(P(E_2) = \frac{39}{52} = \frac{3}{4}\)
Let A = Two cards drawn from remaining 51 cards are both hearts.
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 are 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 = \(\frac{33}{2550} + \left( \frac{3}{4} \times \frac{156}{2550} \right) = \frac{33}{2550} + \frac{468}{4 \times 2550} = \frac{33}{2550} + \frac{117}{2550} = \frac{150}{2550}\)
\(P(E_1/A) = \frac{33}{150} = \frac{11}{50} = 0.22\)
In simple words: Apply Bayes theorem by considering whether the lost card was a heart or not, and compute the ratio of favorable combinations.

Exam Tip: Combinations (like \(^nC_r\)) simplify calculations when drawing multiple cards from remaining decks.

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 insured = 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 = The insured 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 to find \(P(E_2/A)\), the probability that it was a motorcycle given that 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 to find the proportion of accidents caused by motorcycles out of all total accidents.

Exam Tip: Convert vehicle counts to prior probabilities by dividing by the total population before applying Bayes formula.

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 the first purse. \(P(E_1) = \frac{1}{2}\)
Let \(E_2\) = Choosing the second purse. \(P(E_2) = \frac{1}{2}\)
Let A = Drawing a silver coin.
First purse has 2 silver and 4 copper coins (Total = 6). \(P(A/E_1) = \frac{2}{6} = \frac{1}{3}\)
Second purse has 4 silver and 3 copper coins (Total = 7). \(P(A/E_2) = \frac{4}{7}\)
Using the Theorem of Total Probability:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left( \frac{1}{2} \times \frac{1}{3} \right) + \left( \frac{1}{2} \times \frac{4}{7} \right)\)
\(P(A) = \frac{1}{6} + \frac{2}{7} = \frac{7 + 12}{42} = \frac{19}{42}\)
In simple words: Add the probability of choosing the first purse and drawing silver to the probability of choosing the second purse and drawing silver.

Exam Tip: Whenever an experiment involves selection among multiple containers followed by drawing an item, use the total probability theorem.

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 \(E_1\) = The student is a boy. \(P(E_1) = \frac{2}{3}\)
Let \(E_2\) = The student is a girl. \(P(E_2) = 1 - \frac{2}{3} = \frac{1}{3}\)
Let A = The student gets a first class.
\(P(A/E_1) = 0.28 = \frac{28}{100} = \frac{7}{25}\)
\(P(A/E_2) = 0.25 = \frac{25}{100} = \frac{1}{4}\)
Using the Theorem of Total Probability:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left( \frac{2}{3} \times \frac{7}{25} \right) + \left( \frac{1}{3} \times \frac{1}{4} \right)\)
\(P(A) = \frac{14}{75} + \frac{1}{12} = \frac{56 + 25}{300} = \frac{81}{300} = \frac{27}{100} = 0.27\)
In simple words: Combine the weighted probabilities of boys and girls getting a first class.

Exam Tip: Fractions and decimals can be interchanged to simplify addition; convert to a common denominator carefully.

LEVEL II

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 \(E_1\) = Choosing compartment 1. \(P(E_1) = \frac{1}{2}\)
Let \(E_2\) = Choosing compartment 2. \(P(E_2) = \frac{1}{2}\)
Let A = Drawing a one-rupee coin.
Compartment 1 has 3 fifty-paise and 2 one-rupee coins (Total = 5). \(P(A/E_1) = \frac{2}{5}\)
Compartment 2 has 2 fifty-paise and 3 one-rupee coins (Total = 5). \(P(A/E_2) = \frac{3}{5}\)
Using the Theorem of Total Probability:
\(P(A) = P(E_1)P(A/E_1) + P(E_2)P(A/E_2)\)
\(P(A) = \left( \frac{1}{2} \times \frac{2}{5} \right) + \left( \frac{1}{2} \times \frac{3}{5} \right) = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2}\)
In simple words: Find the probability of drawing a one-rupee coin from each compartment and take their average since compartments are chosen equally.

Exam Tip: Equal compartment choice means multiplying by \(\frac{1}{2}\) for each partition branch.

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 \(E_1\) = Selected orator is a man. \(P(E_1) = \frac{1}{2}\)
Let \(E_2\) = Selected orator is a woman. \(P(E_2) = \frac{1}{2}\)
Let A = The chosen person is a good orator.
\(P(A/E_1) = \frac{5}{100} = \frac{1}{20}\)
\(P(A/E_2) = \frac{25}{1000} = \frac{1}{40}\)
We need to find \(P(E_1/A)\), the probability that a male person is selected given that the chosen orator is good:
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(\frac{1}{2} \times \frac{1}{20} = \frac{1}{40}\)
Denominator = \(\left( \frac{1}{2} \times \frac{1}{20} \right) + \left( \frac{1}{2} \times \frac{1}{40} \right) = \frac{1}{40} + \frac{1}{80} = \frac{3}{80}\)
\(P(E_1/A) = \frac{\frac{1}{40}}{\frac{3}{80}} = \frac{1}{40} \times \frac{80}{3} = \frac{2}{3}\)
In simple words: Use Bayes theorem to find the proportion of male orators out of all good orators.

Exam Tip: Carefully identify whether the question asks for unconditional total probability or conditional Bayes probability.

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. \(P(E_1) = \frac{60}{100} = 0.6\)
Let \(E_2\) = Bicycle manufactured by Plant 2. \(P(E_2) = \frac{40}{100} = 0.4\)
Let A = Bicycle is of standard quality.
\(P(A/E_1) = \frac{80}{100} = 0.8\)
\(P(A/E_2) = \frac{90}{100} = 0.9\)
We need to find \(P(E_2/A)\), the probability that it came from the second plant given it is standard quality:
\(P(E_2/A) = \frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(0.4 \times 0.9 = 0.36\)
Denominator = \((0.6 \times 0.8) + (0.4 \times 0.9) = 0.48 + 0.36 = 0.84\)
\(P(E_2/A) = \frac{0.36}{0.84} = \frac{36}{84} = \frac{3}{7}\)
In simple words: Apply Bayes theorem using plant production percentages and standard quality percentages.

Exam Tip: Percentage figures convert directly to decimals or fractions for calculating Bayes theorem fractions.

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. \(P(E_2) = \frac{1}{2}\)
Let A = The visible consecutive letters are ON.
Word LONDON has 6 letters: L, O, N, D, O, N. Consecutive pairs of letters: LO, ON, ND, DO, ON (Total 5 pairs, 'ON' appears 2 times). So \(P(A/E_1) = \frac{2}{5}\)
Word CLIFTON has 7 letters: C, L, I, F, T, O, N. Consecutive pairs of letters: CL, LI, IF, FT, TO, ON (Total 6 pairs, 'ON' appears 1 time). So \(P(A/E_2) = \frac{1}{6}\)
(i) Probability that it came from LONDON given 'ON' is visible: \(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(\frac{1}{2} \times \frac{2}{5} = \frac{1}{5}\)
Denominator = \(\frac{1}{5} + \left( \frac{1}{2} \times \frac{1}{6} \right) = \frac{1}{5} + \frac{1}{12} = \frac{12 + 5}{60} = \frac{17}{60}\)
\(P(E_1/A) = \frac{\frac{1}{5}}{\frac{17}{60}} = \frac{1}{5} \times \frac{60}{17} = \frac{12}{17}\)
(ii) Probability that it came from CLIFTON given 'ON' is visible: \(P(E_2/A) = 1 - \frac{12}{17} = \frac{5}{17}\)
In simple words: Count how many times the pair 'ON' appears out of all possible consecutive letter pairs in each word, then use Bayes theorem.

Exam Tip: For words, total consecutive pairs equal (number of letters - 1).

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. \(P(E_1) = \frac{0.2}{100} = 0.002\)
Let \(E_2\) = Person does not have the disease. \(P(E_2) = 1 - 0.002 = 0.998\)
Let A = Test result is positive (tells that he has the disease).
\(P(A/E_1) = \frac{90}{100} = 0.9\)
\(P(A/E_2) = \frac{1}{100} = 0.01\)
We need to find \(P(E_1/A)\), the probability that the person actually has the disease given a positive test:
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2)}\)
Numerator = \(0.002 \times 0.9 = 0.0018\)
Denominator = \((0.002 \times 0.9) + (0.998 \times 0.01) = 0.0018 + 0.00998 = 0.01178\)
\(P(E_1/A) = \frac{0.0018}{0.01178} = \frac{1800}{11780} = \frac{90}{589} \approx 0.1528\) (or about 15.3%)
In simple words: Use Bayes theorem to find the true positive probability among all positive test results.

Exam Tip: Pay close attention to false positive rates in medical testing problems as they significantly affect conditional probabilities.

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 = The coin drawn is gold.
- Box I has 2 gold coins, so \(P(A/E_1) = \frac{2}{2} = 1\)
- Box II has 0 gold coins, so \(P(A/E_2) = 0\)
- Box III has 1 gold and 1 silver coin, so \(P(A/E_3) = \frac{1}{2}\)
We need to find the probability that the chosen box is Box I given that the drawn coin is gold (since Box I contains two gold coins, the other coin is also gold):
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)}\)
Numerator = \(\frac{1}{3} \times 1 = \frac{1}{3}\)
Denominator = \(\left( \frac{1}{3} \times 1 \right) + \left( \frac{1}{3} \times 0 \right) + \left( \frac{1}{3} \times \frac{1}{2} \right) = \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{2}\)
\(P(E_1/A) = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}\)
In simple words: Find the probability that the gold coin came from Box I using Bayes theorem.

Exam Tip: 'Other coin is also gold' means the chosen box must be Box I.

(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: Let X denote the number of spades when 2 cards are drawn with replacement.
Probability of getting a spade in a single draw \(p = \frac{13}{52} = \frac{1}{4}\)
Probability of not getting a spade \(q = 1 - \frac{1}{4} = \frac{3}{4}\)
X can take values 0, 1, 2.
\(P(X = 0) = ^2C_0 q^2 p^0 = 1 \times \left(\frac{3}{4}\right)^2 \times 1 = \frac{9}{16}\)
\(P(X = 1) = ^2C_1 q^1 p^1 = 2 \times \frac{3}{4} \times \frac{1}{4} = \frac{6}{16} = \frac{3}{8}\)
\(P(X = 2) = ^2C_2 q^0 p^2 = 1 \times 1 \times \left(\frac{1}{4}\right)^2 = \frac{1}{16}\)
Probability distribution table:
X: 0, 1, 2
P(X): \(\frac{9}{16}, \frac{3}{8}, \frac{1}{16}\)
In simple words: Use binomial distribution formula since cards are drawn with replacement.

Exam Tip: 'With replacement' trials are independent Bernoulli trials, perfect for binomial distribution.

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: Total apples = 4 (defective) + 16 (good) = 20 apples.
Three apples are drawn at random. Let X be the number of defective apples. X can take values 0, 1, 2, 3.
Total ways of drawing 3 apples = \(^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140\)
\(P(X = 0) = \frac{^{16}C_3 \times ^4C_0}{^{20}C_3} = \frac{\frac{16 \times 15 \times 14}{6} \times 1}{1140} = \frac{560}{1140} = \frac{56}{114} = \frac{28}{57}\)
\(P(X = 1) = \frac{^{16}C_2 \times ^4C_1}{^{20}C_3} = \frac{\frac{16 \times 15}{2} \times 4}{1140} = \frac{120 \times 4}{1140} = \frac{480}{1140} = \frac{24}{57} = \frac{8}{19}\)
\(P(X = 2) = \frac{^{16}C_1 \times ^4C_2}{^{20}C_3} = \frac{16 \times \frac{4 \times 3}{2}}{1140} = \frac{16 \times 6}{1140} = \frac{96}{1140} = \frac{8}{95}\)
\(P(X = 3) = \frac{^{16}C_0 \times ^4C_3}{^{20}C_3} = \frac{1 \times 4}{1140} = \frac{4}{1140} = \frac{1}{285}\)
In simple words: Use combinations to find probabilities for each possible count of defective apples drawn without replacement.

Exam Tip: Check that the sum of all probabilities in the distribution equals 1.

Question 3. A random variable X is specified by the following distribution:
X: 2, 3, 4
P(X): 0.3, 0.4, 0.3
Find the variance of the distribution.
Answer: Given distribution:
Mean \(E(X) = \sum x_i p_i = (2 \times 0.3) + (3 \times 0.4) + (4 \times 0.3)\)
\(E(X) = 0.6 + 1.2 + 1.2 = 3.0\)
\(E(X^2) = \sum x_i^2 p_i = (2^2 \times 0.3) + (3^2 \times 0.4) + (4^2 \times 0.3)\)
\(E(X^2) = (4 \times 0.3) + (9 \times 0.4) + (16 \times 0.3)\)
\(E(X^2) = 1.2 + 3.6 + 4.8 = 9.6\)
Variance \(\text{Var}(X) = E(X^2) - [E(X)]^2\)
\(\text{Var}(X) = 9.6 - (3.0)^2 = 9.6 - 9.0 = 0.6\)
In simple words: Calculate the mean and mean of squares, then use the variance formula \(E(X^2) - [E(X)]^2\).

Exam Tip: Variance is always non-negative; double-check your arithmetic when squaring values.

LEVEL III

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 P(Tail) = p, then P(Head) = 3p
Since \(p + 3p = 1\), we get \(4p = 1 \implies p = \frac{1}{4}\)
So, P(Tail) = \(\frac{1}{4}\) and P(Head) = \(\frac{3}{4}\)
Let X be the number of tails in 2 tosses. X can take values 0, 1, 2.
\(P(X = 0) = P(\text{both heads}) = \left(\frac{3}{4}\right)^2 = \frac{9}{16}\)
\(P(X = 1) = P(\text{one tail, one head}) = 2 \times \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) = \frac{6}{16} = \frac{3}{8}\)
\(P(X = 2) = P(\text{both tails}) = \left(\frac{1}{4}\right)^2 = \frac{1}{16}\)
Probability distribution:
X: 0, 1, 2
P(X): \(\frac{9}{16}, \frac{3}{8}, \frac{1}{16}\)
In simple words: Determine individual coin probabilities first, then use binomial probabilities for 2 independent tosses.

Exam Tip: Assign probabilities carefully when coins are biased using ratio given in the problem.

Question 2. The sum of mean and variance of a binomial distribution for 5 trials be 1.8. Find the probability distribution.
Answer: Number of trials \(n = 5\)
Mean = \(np\), Variance = \(npq\)
Given: Mean + 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: \(25p^2 - 50p + 9 = 0\)
Factorizing: \(25p^2 - 45p - 5p + 9 = 0\)
\(5p(5p - 9) - 1(5p - 9) = 0\)
\((5p - 1)(5p - 9) = 0\)
Since \(p \le 1\), \(p = \frac{1}{5} = 0.2\). Thus \(q = 1 - 0.2 = 0.8\)
Binomial probability formula: \(P(X = x) = ^5C_x \left(\frac{1}{5}\right)^x \left(\frac{4}{5}\right)^{5-x}\)
Calculating for each x = 0, 1, 2, 3, 4, 5:
\(P(X = 0) = ^5C_0 \left(\frac{4}{5}\right)^5 = \frac{1024}{3125}\)
\(P(X = 1) = ^5C_1 \left(\frac{1}{5}\right)\left(\frac{4}{5}\right)^4 = 5 \times \frac{256}{3125} = \frac{1280}{3125}\)
\(P(X = 2) = ^5C_2 \left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^3 = 10 \times \frac{64}{3125} = \frac{640}{3125}\)
\(P(X = 3) = ^5C_3 \left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^2 = 10 \times \frac{16}{3125} = \frac{160}{3125}\)
\(P(X = 4) = ^5C_4 \left(\frac{1}{5}\right)^4\left(\frac{4}{5}\right)^1 = 5 \times \frac{4}{3125} = \frac{20}{3125}\)
\(P(X = 5) = ^5C_5 \left(\frac{1}{5}\right)^5 = \frac{1}{3125}\)
In simple words: Set up the quadratic equation using mean \(np\) and variance \(npq\) to find p, then substitute into the binomial formula.

Exam Tip: Remember formulas: Mean of binomial = \(np\), Variance = \(npq\).

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

Exam Tip: Dividing variance by mean is a shortcut to instantly find q in binomial distribution 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 \(p = \frac{3}{6} = \frac{1}{2}\)
Probability of failure \(q = 1 - \frac{1}{2} = \frac{1}{2}\)
We need the probability of exactly 3 successes: \(P(X = 3)\)
\(P(X = 3) = ^5C_3 p^3 q^{5-3} = 10 \times \left(\frac{1}{2}\right)^3 \times \left(\frac{1}{2}\right)^2 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}\)
In simple words: Use the binomial formula with n = 5 and p = 1/2 for getting an even number 3 times.

Exam Tip: Even numbers on a standard die are 2, 4, 6 (3 favorable out of 6).

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 that success is twice as likely as failure: \(p = 2q\)
Since \(p + q = 1\), we have \(2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3}\)
Therefore, \(p = \frac{2}{3}\) and \(q = \frac{1}{3}\), with number of trials \(n = 6\).
We need the probability of at least 4 successes: \(P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6)\)
\(P(X = 4) = ^6C_4 \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{240}{729}\)
\(P(X = 5) = ^6C_5 \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{192}{729}\)
\(P(X = 6) = ^6C_6 \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^0 = 1 \times \frac{64}{729} \times 1 = \frac{64}{729}\)
\(P(X \ge 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729}\)
In simple words: Find individual probabilities for 4, 5, and 6 successes using binomial distribution and add them up.

Exam Tip: 'At least 4' means you sum probabilities for 4, 5, and 6.

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 \(n = 200\)
Sample space for a pair of dice = 36 outcomes.
Favorable outcomes for a sum of 9: (3,6), (4,5), (5,4), (6,3) - total 4 outcomes.
Probability of success in a single throw \(p = \frac{4}{36} = \frac{1}{9}\)
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: Calculate the probability of getting a sum of 9 in one throw, then use the formulas \(np\) for mean and \(npq\) for variance.

Exam Tip: Mean and variance formulas for binomial distribution are independent of complex summation.

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: Total 4-digit numbers formed using digits 1, 2, 3, 5 without repetition = \(4! = 24\)
A number is divisible by 5 if its unit digit is 5.
Fixing 5 at the unit place, the remaining 3 digits (1, 2, 3) can be arranged in the first 3 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.

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 = \(1 - P(\text{happens zero times})\)
\(P(X = 0) = ^3C_0 p^0 q^3 = 1 \times 1 \times (0.6)^3 = 0.216\)
Required probability = \(1 - 0.216 = 0.784\)
In simple words: Subtract the probability of zero successes from 1 to find the probability of at least one success.

Exam Tip: 'At least one' is always solved most efficiently using the complement rule \(1 - P(0)\).

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 a correct forecast \(p = \frac{1}{3}\)
Probability of incorrect forecast \(q = \frac{2}{3}\)
Number of matches \(n = 4\)
We need the probability of forecasting at least 3 correct results: \(P(X \ge 3) = P(X = 3) + P(X = 4)\)
\(P(X = 3) = ^4C_3 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 = 4 \times \frac{1}{27} \times \frac{2}{3} = \frac{8}{81}\)
\(P(X = 4) = ^4C_4 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^0 = 1 \times \frac{1}{81} \times 1 = \frac{1}{81}\)
\(P(X \ge 3) = \frac{8}{81} + \frac{1}{81} = \frac{9}{81} = \frac{1}{9}\)
In simple words: Use the binomial formula to find probabilities of getting exactly 3 and 4 correct forecasts, then add them.

Exam Tip: Each match forecast acts as an independent trial with success probability \(\frac{1}{3}\).

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. 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.
\(P(A/E_1) = \frac{1}{3}\)
\(P(A/E_2) = \frac{1}{4}\)
\(P(A/E_3) = 0\) (reaches in time if he uses other mode)
We need to find \(P(E_1/A)\), the probability that he traveled by bus given that he reached late:
\(P(E_1/A) = \frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)}\)
Numerator = \(\frac{3}{5} \times \frac{1}{3} = \frac{1}{5} = \frac{3}{15}\)
Denominator = \(\left(\frac{3}{5} \times \frac{1}{3}\right) + \left(\frac{1}{10} \times \frac{1}{4}\right) + \left(\frac{3}{10} \times 0\right) = \frac{1}{5} + \frac{1}{40} + 0 = \frac{8 + 1}{40} = \frac{9}{40}\)
\(P(E_1/A) = \frac{\frac{1}{5}}{\frac{9}{40}} = \frac{1}{5} \times \frac{40}{9} = \frac{8}{9}\)
In simple words: Apply Bayes theorem using the different transport modes and their corresponding late arrival probabilities.

Exam Tip: A zero probability for late arrival via other means simplifies the denominator calculation.

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:

\[ P(X=x)= \begin{cases} kx, & x=0,1\\ 2kx, & x=2\\ k(5-x), & x=3,4 \end{cases} \]

Answer: Since the sum of all probabilities in a distribution equals 1:
\(\sum P(X = x) = 1\)
\(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1\)
\(k(0) + k(1) + 2k(2) + k(5-3) + k(5-4) = 1\)
\(0 + k + 4k + 2k + k = 1\)
\(8k = 1 \implies k = \frac{1}{8}\)
Probability distribution table:
- For \(x = 0\): \(P(0) = 0\)
- For \(x = 1\): \(P(1) = \frac{1}{8}\)
- For \(x = 2\): \(P(2) = \frac{4}{8} = \frac{1}{2}\)
- For \(x = 3\): \(P(3) = \frac{2}{8} = \frac{1}{4}\)
- For \(x = 4\): \(P(4) = \frac{1}{8}\)
Mean \(E(X) = \sum x_i P(x_i) = (0 \times 0) + \left(1 \times \frac{1}{8}\right) + \left(2 \times \frac{4}{8}\right) + \left(3 \times \frac{2}{8}\right) + \left(4 \times \frac{1}{8}\right)\)
\(E(X) = 0 + \frac{1}{8} + \frac{8}{8} + \frac{6}{8} + \frac{4}{8} = \frac{19}{8} = 2.375\)
\(E(X^2) = \sum x_i^2 P(x_i) = (0^2 \times 0) + \left(1^2 \times \frac{1}{8}\right) + \left(2^2 \times \frac{4}{8}\right) + \left(3^2 \times \frac{2}{8}\right) + \left(4^2 \times \frac{1}{8}\right)\)
\(E(X^2) = 0 + \frac{1}{8} + \frac{16}{8} + \frac{18}{8} + \frac{16}{8} = \frac{51}{8} = 6.375\)
Variance \(\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{51}{8} - \left(\frac{19}{8}\right)^2 = \frac{51}{8} - \frac{361}{64} = \frac{408 - 361}{64} = \frac{47}{64}\)
In simple words: Find constant k by equating the sum of all probabilities to 1, then calculate mean and variance.

Exam Tip: Always verify that probabilities sum up to 1 before calculating mean and variance.

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: Success (getting a 6) probability \(p = \frac{1}{6}\)
Failure (not getting a 6) probability \(q = \frac{5}{6}\)
We need the third six to appear on the exact 6th throw. This means in the first 5 throws, exactly 2 sixes must have appeared, followed by a six on the 6th throw.
Probability of getting 2 sixes in the first 5 throws = \(^5C_2 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3\)
\(= 10 \times \frac{1}{36} \times \frac{125}{216} = \frac{1250}{7776}\)
Probability of getting a six on the 6th throw = \(\frac{1}{6}\)
Required probability = \(\frac{1250}{7776} \times \frac{1}{6} = \frac{1250}{46656} = \frac{625}{23328}\)
In simple words: Multiply the probability of getting 2 sixes in 5 throws by the probability of getting a six on the 6th throw.

Exam Tip: For 'nth success on the rth trial' problems, find successes in preceding trials and multiply by the final success probability.

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: Probability of a correct answer in a single question by guessing \(p = \frac{1}{3}\)
Probability of an incorrect answer \(q = \frac{2}{3}\)
Number of questions \(n = 5\)
We need the probability of getting four or more correct answers: \(P(X \ge 4) = P(X = 4) + P(X = 5)\)
\(P(X = 4) = ^5C_4 \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^1 = 5 \times \frac{1}{81} \times \frac{2}{3} = \frac{10}{243}\)
\(P(X = 5) = ^5C_5 \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^0 = 1 \times \frac{1}{243} \times 1 = \frac{1}{243}\)
\(P(X \ge 4) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243}\)
In simple words: Use binomial probability to calculate the chance of getting exactly 4 and exactly 5 correct answers, then add them.

Exam Tip: 'Four or more' in 5 trials means you evaluate 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 drawing a red card in a single draw \(p = \frac{26}{52} = \frac{1}{2}\)
Probability of failure \(q = 1 - \frac{1}{2} = \frac{1}{2}\)
Since cards are drawn simultaneously (or successively without replacement, modeled as binomial or hypergeometric; for standard binomial approximation in curriculum):
Mean = \(np = 2 \times \frac{1}{2} = 1\)
Variance for drawing without replacement using finite population correction factor or standard binomial formula: Variance = \(npq \left(\frac{N - n}{N - 1}\right)\)
\(\text{Variance} = 2 \times \frac{1}{2} \times \frac{1}{2} \times \left(\frac{52 - 2}{52 - 1}\right) = \frac{1}{2} \times \frac{50}{51} = \frac{25}{51}\)
In simple words: Use binomial mean and hypergeometric variance formulas since cards are drawn without replacement.

Exam Tip: For simultaneous drawing without replacement, apply the finite population correction multiplier to the binomial variance formula.