OP Malhotra Class 11 Maths Solutions Chapter 22 Probability Exercise 22 (F)

Question 1.
(i) In a single throw of two coins, find the probability of getting both heads or both tails.
(ii) A dice is thrown twice. Find the probability that the sum of the two numbers obtain is 5 or 7?
(iii) Two dice are tossed once. Find the probability of getting an even number on first die or a total of 8.
Answer:
(i) When two coins are thrown, the total possible outcomes are (HH, HT, TH, TT). So, there are 4 total outcomes.
The event of getting both heads is {HH}, which is 1 outcome.
The event of getting both tails is {TT}, which is 1 outcome.
Since these two events cannot happen at the same time (they are mutually exclusive), we can add their probabilities.
The probability of both heads is \( \frac{1}{4} \).
The probability of both tails is \( \frac{1}{4} \).
So, the required probability is \( \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \). This means there's a 50% chance of getting either both heads or both tails.
(ii) When a die is thrown twice, the total number of possible outcomes is \( 6^2 = 36 \).
Let A be the event that the sum of the two numbers is 5. The pairs that sum to 5 are {(1, 4), (2, 3), (3, 2), (4, 1)}. So, there are 4 such outcomes, \( n(A) = 4 \).
Let B be the event that the sum of the two numbers is 7. The pairs that sum to 7 are {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}. So, there are 6 such outcomes, \( n(B) = 6 \).
Events A and B are mutually exclusive because a sum cannot be both 5 and 7 at the same time.
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{6}{36} \).
The required probability of getting either a sum of 5 or 7 is \( P(A \text{ or } B) = P(A) + P(B) = \frac{4}{36} + \frac{6}{36} = \frac{10}{36} = \frac{5}{18} \). Understanding mutual exclusivity helps simplify calculations in probability.
(iii) When two dice are tossed once, the total number of possible outcomes is \( 6^2 = 36 \).
Let A be the event of getting an even number on the first die. The outcomes are: {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. So, there are 18 such outcomes, \( n(A) = 18 \).
Let B be the event of getting a total of 8. The outcomes are: {(2,6), (3,5), (4,4), (5,3), (6,2)}. So, there are 5 such outcomes, \( n(B) = 5 \).
Now, we need to find the outcomes that are common to both A and B (the intersection). These are the pairs where the first die is even AND the sum is 8: {(2,6), (4,4), (6,2)}. So, \( n(A \cap B) = 3 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{18}{36} \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{5}{36} \).
The probability of both A and B is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{36} \).
The required probability of getting an even number on the first die OR a total of 8 is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( \implies P(A \cup B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} \)
\( \implies P(A \cup B) = \frac{18 + 5 - 3}{36} = \frac{20}{36} = \frac{5}{9} \). This calculation accounts for outcomes that satisfy both conditions, ensuring they are not double-counted.
In simple words: For two coins, count how many ways you get two heads or two tails and divide by all ways. For two dice, list pairs that sum to 5 and to 7, then add their chances. For an even first die OR a sum of 8, list all ways for each, see what is common, and use the formula to avoid counting common parts twice.

🎯 Exam Tip: For problems involving two dice, always list all 36 possible outcomes or visualize them in a 6x6 grid to accurately count favorable outcomes and their intersections.

Question 2. If the probability of a horse A winning a race is \( \frac { 1 }{ 5 } \) and the probability of horse B winning the same race is \( \frac { 1 }{ 4 } \), what is the probability that one of the horses will win?
Answer: Let P(A) be the probability that horse A wins, so \( P(A) = \frac{1}{5} \).
Let P(B) be the probability that horse B wins, so \( P(B) = \frac{1}{4} \).
Since only one horse can win the same race, the events of horse A winning and horse B winning are mutually exclusive (they cannot happen at the same time).
To find the probability that one of the horses will win, we add their individual probabilities.
Required probability \( = P(A) + P(B) \)
\( = \frac{1}{5} + \frac{1}{4} \)
To add these fractions, we find a common denominator, which is 20.
\( = \frac{4}{20} + \frac{5}{20} = \frac{9}{20} \). This means there's a 9 out of 20 chance that either horse A or horse B will win the race.
In simple words: If horse A has a 1/5 chance to win and horse B has a 1/4 chance to win, and only one can win, then the total chance for either of them to win is found by adding their separate chances.

🎯 Exam Tip: When events are mutually exclusive (cannot happen at the same time), their combined probability is simply the sum of their individual probabilities.

Question 3. In a single throw of two dice, what is the probability of obtaining a total of 9 or 11?
Answer: When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \). This is our sample space, \( n(S) = 36 \).
Let A be the event of obtaining a total of 9. The pairs that sum to 9 are {(3, 6), (4, 5), (5, 4), (6, 3)}. So, there are 4 such outcomes, \( n(A) = 4 \).
Let B be the event of obtaining a total of 11. The pairs that sum to 11 are {(5, 6), (6, 5)}. So, there are 2 such outcomes, \( n(B) = 2 \).
The events A and B are mutually exclusive because the sum of two dice cannot be both 9 and 11 at the same time.
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{4}{36} \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{2}{36} \).
The required probability of obtaining a total of 9 or 11 is \( P(A \text{ or } B) = P(A) + P(B) - P(A \cap B) \). Since A and B are mutually exclusive, \( P(A \cap B) = 0 \).
So, \( P(A \cup B) = P(A) + P(B) = \frac{4}{36} + \frac{2}{36} = \frac{6}{36} = \frac{1}{6} \). This shows there's a 1 in 6 chance for the sum to be either 9 or 11.
In simple words: With two dice, count how many ways you can get a sum of 9, and how many ways you can get a sum of 11. Since these cannot happen together, just add their chances and divide by the total possible outcomes (36).

🎯 Exam Tip: Always list all favorable outcomes for each event when dealing with dice problems to ensure accuracy, especially when checking for mutual exclusivity or overlap.

Question 4. In a group there are 2 men and 3 women. 3 persons are selected at random from the group. Find the probability that 1 man and 2 women or 2 men and 1 woman are selected.
Answer: The total number of people in the group is 2 men + 3 women = 5 persons.
We are selecting 3 persons from these 5. The total number of ways to do this is \( ^5C_3 \).
\( ^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \). So, \( n(S) = 10 \).
Let A be the event of selecting 1 man and 2 women.
Ways to select 1 man from 2 men = \( ^2C_1 = 2 \).
Ways to select 2 women from 3 women = \( ^3C_2 = 3 \).
So, \( n(A) = ^2C_1 \times ^3C_2 = 2 \times 3 = 6 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{6}{10} \).
Let B be the event of selecting 2 men and 1 woman.
Ways to select 2 men from 2 men = \( ^2C_2 = 1 \).
Ways to select 1 woman from 3 women = \( ^3C_1 = 3 \).
So, \( n(B) = ^2C_2 \times ^3C_1 = 1 \times 3 = 3 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{3}{10} \).
The events A and B are mutually exclusive because we cannot select both 1 man and 2 women AND 2 men and 1 woman at the same time when selecting only 3 people.
The required probability is \( P(A \text{ or } B) = P(A) + P(B) \).
\( = \frac{6}{10} + \frac{3}{10} = \frac{9}{10} \). This means there's a very high chance (90%) of selecting one of these two combinations.
In simple words: First, find all possible ways to choose 3 people from 5. Then, find ways to pick 1 man and 2 women, and separately, ways to pick 2 men and 1 woman. Since these two selections are different, add up their chances and divide by the total number of ways.

🎯 Exam Tip: For "or" probability problems with selections, always check if the events are mutually exclusive. If they are, simply add their probabilities; if not, use the inclusion-exclusion principle.

Question 5. In a class of 25 students with roll numbers 1 to 25, a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of 5 or 7?
Answer: The total number of students is 25, so the total number of possible outcomes (sample space) is \( n(S) = 25 \).
Let A be the event that the selected student's roll number is a multiple of 5.
The multiples of 5 between 1 and 25 are {5, 10, 15, 20, 25}. So, \( n(A) = 5 \).
Let B be the event that the selected student's roll number is a multiple of 7.
The multiples of 7 between 1 and 25 are {7, 14, 21}. So, \( n(B) = 3 \).
We need to check for any overlap between events A and B (numbers that are multiples of both 5 and 7). Numbers that are multiples of both 5 and 7 are multiples of 35. There are no multiples of 35 between 1 and 25. So, \( A \cap B = \emptyset \) (empty set), which means \( n(A \cap B) = 0 \). These events are mutually exclusive.
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{5}{25} \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{3}{25} \).
The required probability that the roll number is either a multiple of 5 or 7 is \( P(A \cup B) = P(A) + P(B) \) (since \( P(A \cap B) = 0 \)).
\( = \frac{5}{25} + \frac{3}{25} = \frac{8}{25} \). This indicates that 8 out of 25 students would have a roll number satisfying the condition.
In simple words: From students numbered 1 to 25, we want to find the chance of picking one whose number can be divided by 5 or by 7. We count the numbers divisible by 5, then by 7. Since no number is divisible by both 5 and 7 in this range, we just add their chances together.

🎯 Exam Tip: Always list out the favorable outcomes when the sample space is small to avoid errors, especially when identifying common elements between events.

Question 6. If chance of A, winning a certain race be \( \frac { 1 }{ 6 } \) and the chance of B winning it is \( \frac { 1 }{ 3 } \), what is the chance that neither should win?
Answer: Let P(A) be the probability that A wins, so \( P(A) = \frac{1}{6} \).
Let P(B) be the probability that B wins, so \( P(B) = \frac{1}{3} \).
In a race, if A wins, B cannot win, and vice versa. Therefore, these events are mutually exclusive.
The probability that either A or B wins is \( P(A \cup B) = P(A) + P(B) \).
\( = \frac{1}{6} + \frac{1}{3} \)
To add these, find a common denominator, which is 6.
\( = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \).
The probability that neither A nor B wins is the complement of "A or B wins". This is \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - \frac{1}{2} = \frac{1}{2} \). So, there is a 50% chance that neither of these two horses will win the race.
In simple words: We know the chances of horse A winning and horse B winning. Since only one can win, we add their chances together. Then, to find the chance that neither of them wins, we subtract this total chance from 1.

🎯 Exam Tip: Remember that the probability of an event NOT happening is 1 minus the probability of the event happening. This is crucial for "neither/nor" questions.

Question 7. Discuss and criticise the following : P(A) = \( \frac { 2 }{ 3 } \), P(B) = \( \frac { 1 }{ 4 } \), P (C) = \( \frac { 1 }{ 3 } \) are the probabilities of three mutually exclusive events A, B and C.
Answer: The statement says that A, B, and C are three mutually exclusive events. If events are mutually exclusive, the probability of any of them occurring is the sum of their individual probabilities: \( P(A \cup B \cup C) = P(A) + P(B) + P(C) \).
Let's calculate the sum of the given probabilities:
\( P(A \cup B \cup C) = \frac{2}{3} + \frac{1}{4} + \frac{1}{3} \)
To add these fractions, find a common denominator, which is 12.
\( = \frac{8}{12} + \frac{3}{12} + \frac{4}{12} \)
\( = \frac{8 + 3 + 4}{12} = \frac{15}{12} \).
We know that the probability of any event cannot be greater than 1. However, here, \( \frac{15}{12} \) is greater than 1.
\( \implies \frac{15}{12} = 1.25 \)
Since \( P(A \cup B \cup C) > 1 \), the given statement is false. It is not possible for three mutually exclusive events to have probabilities that sum up to more than 1. This fundamental rule ensures that probabilities remain within a sensible range.
In simple words: When events cannot happen at the same time (mutually exclusive), you add their chances to find the chance that any of them happens. If this total chance is more than 1, it's wrong, because probability can never be greater than 1.

🎯 Exam Tip: A key property of probability is that for any event E, \( 0 \le P(E) \le 1 \). If a calculation results in a probability outside this range, it indicates an error in the problem statement or calculation.

Question 8. E and F are two events associated with a random experiment for which P (F) = 0.35, P (E or F) = 0.85, P (E and F) = 0.15. Find P (E).
Answer: We are given the following probabilities:
Probability of event F, \( P(F) = 0.35 \).
Probability of event E or F (union), \( P(E \cup F) = 0.85 \).
Probability of event E and F (intersection), \( P(E \cap F) = 0.15 \).
We use the formula for the probability of the union of two events:
\( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)
Now, substitute the given values into the formula:
\( 0.85 = P(E) + 0.35 - 0.15 \)
First, simplify the right side:
\( 0.85 = P(E) + (0.35 - 0.15) \)
\( 0.85 = P(E) + 0.20 \)
To find \( P(E) \), subtract 0.20 from both sides:
\( P(E) = 0.85 - 0.20 \)
\( P(E) = 0.65 \). This probability means event E has a 65% chance of occurring.
In simple words: We have the chance of F, the chance of E or F, and the chance of both E and F. We use a special rule that connects these three numbers to find the chance of E alone. We just put the numbers into the rule and solve for E.

🎯 Exam Tip: The formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) is fundamental for problems involving the union of two events that may or may not overlap.

Question 9.
(i) 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. Find the probability that neither A nor B occurs.
(ii) The probability of an event A occurring is 0.5 and of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.
Answer:
(i) We are given:
Probability of event A, \( P(A) = 0.25 \).
Probability of event B, \( P(B) = 0.50 \).
Probability that both A and B occur (intersection), \( P(A \cap B) = 0.14 \).
First, find the probability that A or B occurs, \( P(A \cup B) \), using the formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.25 + 0.50 - 0.14 \)
\( = 0.75 - 0.14 = 0.61 \).
The probability that neither A nor B occurs is the complement of \( P(A \cup B) \). This means \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - 0.61 = 0.39 \). So, there is a 39% chance that neither event A nor event B will happen.
(ii) We are given:
Probability of event A, \( P(A) = 0.5 \).
Probability of event B, \( P(B) = 0.3 \).
A and B are mutually exclusive events, which means they cannot happen at the same time. Therefore, the probability of both A and B occurring is 0, i.e., \( P(A \cap B) = 0 \).
First, find the probability that A or B occurs, \( P(A \cup B) \):
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.5 + 0.3 - 0 \)
\( = 0.8 \).
The probability that neither A nor B occurs is the complement of \( P(A \cup B) \):
\( P(\overline{A \cup B}) = 1 - P(A \cup B) \)
\( = 1 - 0.8 = 0.2 \). This implies a 20% chance for neither event A nor event B to occur, a lower chance than when there was an overlap.
In simple words: For the first part, we find the chance of A or B happening by adding their chances and taking away the chance they both happen. Then, we subtract this from 1 to find the chance that neither happens. For the second part, since A and B cannot happen together, we just add their chances, and then subtract that from 1.

🎯 Exam Tip: Understanding the concepts of 'union' (\(A \cup B\)), 'intersection' (\(A \cap B\)), 'complement' (\(\overline{A}\)), and 'mutually exclusive' is essential for solving probability problems efficiently.

Question 10.
(i) A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is even ?
(ii) A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Answer:
(i) Total tickets = 25 (numbered 1 to 25).
Total number of ways to draw 2 tickets from 25 is \( ^{25}C_2 \).
\( ^{25}C_2 = \frac{25 \times 24}{2 \times 1} = 25 \times 12 = 300 \). So, \( n(S) = 300 \).
The product of two numbers is even if:
Case 1: Both tickets drawn are even numbers.
Case 2: One ticket is even and the other is odd.
(The product is odd only if both are odd).
Numbers from 1 to 25:
Even numbers: {2, 4, 6, ..., 24}. There are 12 even numbers.
Odd numbers: {1, 3, 5, ..., 25}. There are 13 odd numbers.
Let A be the event that both tickets drawn are even.
Number of ways to choose 2 even tickets from 12 = \( ^{12}C_2 \).
\( ^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66 \). So, \( n(A) = 66 \).
Let B be the event that one ticket is even and one is odd.
Number of ways to choose 1 even ticket from 12 = \( ^{12}C_1 = 12 \).
Number of ways to choose 1 odd ticket from 13 = \( ^{13}C_1 = 13 \).
So, \( n(B) = ^{12}C_1 \times ^{13}C_1 = 12 \times 13 = 156 \).
The events A and B are mutually exclusive. We can add their favorable outcomes.
Total favorable outcomes \( = n(A) + n(B) = 66 + 156 = 222 \).
Required probability \( = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{222}{300} \).
This can be simplified by dividing by 6: \( \frac{37}{50} \). This is a practical application of probability to number theory.
(ii) Total number of balls in the bag = 7 white + 5 black + 4 red = 16 balls.
We are drawing 4 balls at random without replacement.
Total number of ways to draw 4 balls from 16 = \( ^{16}C_4 \).
\( ^{16}C_4 = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820 \). So, \( n(S) = 1820 \).
We need the probability that "at least three balls are black". This means either exactly 3 black balls or exactly 4 black balls are drawn.
Case 1: Exactly 3 black balls and 1 non-black ball.
Number of ways to choose 3 black balls from 5 = \( ^5C_3 = \frac{5 \times 4}{2 \times 1} = 10 \).
Number of non-black balls = 7 white + 4 red = 11 balls.
Number of ways to choose 1 non-black ball from 11 = \( ^{11}C_1 = 11 \).
Ways for Case 1 = \( ^5C_3 \times ^{11}C_1 = 10 \times 11 = 110 \).
Case 2: Exactly 4 black balls.
Number of ways to choose 4 black balls from 5 = \( ^5C_4 = 5 \).
Ways for Case 2 = \( ^5C_4 = 5 \).
These two cases are mutually exclusive. Total favorable outcomes = \( 110 + 5 = 115 \).
Required probability \( = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{115}{1820} \).
This can be simplified by dividing by 5: \( \frac{23}{364} \). This calculation demonstrates how to handle 'at least' conditions in probability by considering all possible scenarios.
In simple words: For the tickets, we count all ways to pick 2 tickets. Then, we find ways where the product is even (both even, or one even and one odd). We add these ways and divide by the total. For the balls, we count all ways to pick 4 balls. Then, we find ways to pick at least 3 black balls (meaning 3 black and 1 other, or all 4 black). We add these ways and divide by the total.

🎯 Exam Tip: When dealing with "at least" or "at most" scenarios, remember to break down the problem into individual cases and sum their probabilities (if mutually exclusive) or use the complementary event approach.

Question 11.
(i) A and B are two mutually exclusive events of an experiment. If P(not A) = 0.75, P(A ∪ B) = 0.65 and P(B) = p, find the value of p.
(ii) A and B are two mutually exclusive events. If P (A) = 0.5 and P (\(\overline{\mathbf{B}}\)) = 0.6, find P (A or B).
Answer:
(i) We are given that A and B are mutually exclusive events, which means \( P(A \cap B) = 0 \).
We know \( P(\text{not A}) = P(\overline{A}) = 0.75 \).
The probability of A is \( P(A) = 1 - P(\overline{A}) = 1 - 0.75 = 0.25 \).
We are also given \( P(A \cup B) = 0.65 \) and \( P(B) = p \).
For mutually exclusive events, the formula for union is \( P(A \cup B) = P(A) + P(B) \).
Substitute the known values:
\( 0.65 = 0.25 + p \)
Now, solve for p:
\( p = 0.65 - 0.25 \)
\( p = 0.40 \). So, the probability of event B is 0.40. This demonstrates how to find an unknown probability using the properties of mutually exclusive events.
(ii) We are given that A and B are mutually exclusive events, so \( P(A \cap B) = 0 \).
We have \( P(A) = 0.5 \).
We are given \( P(\overline{B}) = 0.6 \).
The probability of B is \( P(B) = 1 - P(\overline{B}) = 1 - 0.6 = 0.4 \).
We need to find \( P(A \text{ or } B) \), which is \( P(A \cup B) \).
For mutually exclusive events, \( P(A \cup B) = P(A) + P(B) \).
\( = 0.5 + 0.4 = 0.9 \). This means there's a 90% chance that either A or B will occur.
In simple words: For the first part, if A and B can't happen together, we know A's chance from "not A". Then we use the rule that the chance of A or B is just the sum of their individual chances to find the unknown 'p'. For the second part, we use the same idea: find B's chance from "not B", then add the chances of A and B because they don't overlap.

🎯 Exam Tip: Always remember that \( P(\text{event}) + P(\text{not event}) = 1 \). This relationship is frequently used to find missing probabilities, especially in combination with union formulas.

Question 12.
(i) A, B and C are three mutually exclusive and exhaustive events associated with a random experiment. Find P (A) given that P (B) = \( \frac { 3 }{ 2 } \) P (A) and P (C) = \( \frac { 1 }{ 2 } \) P (B).
Answer:
(i) Since A, B, and C are mutually exclusive (cannot happen at the same time) and exhaustive (cover all possible outcomes), their probabilities must sum to 1.
\( P(A) + P(B) + P(C) = 1 \) ...(1)
We are given two relationships:
1. \( P(B) = \frac{3}{2} P(A) \)
2. \( P(C) = \frac{1}{2} P(B) \)
Substitute the first relationship into the second one to express P(C) in terms of P(A):
\( P(C) = \frac{1}{2} \left( \frac{3}{2} P(A) \right) = \frac{3}{4} P(A) \)
Now, substitute the expressions for P(B) and P(C) in terms of P(A) into equation (1):
\( P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1 \)
To solve for P(A), find a common denominator for the coefficients (which is 4):
\( \frac{4}{4} P(A) + \frac{6}{4} P(A) + \frac{3}{4} P(A) = 1 \)
Combine the terms:
\( \left( \frac{4 + 6 + 3}{4} \right) P(A) = 1 \)
\( \frac{13}{4} P(A) = 1 \)
Now, isolate P(A):
\( P(A) = \frac{4}{13} \). This value ensures all probabilities sum up to one, as expected for exhaustive events.
In simple words: If A, B, and C cover all possible outcomes and don't overlap, their chances add up to 1. We are given how B's chance relates to A's, and how C's chance relates to B's. We change everything to be about A's chance, then add them up to 1, and solve for A's chance.

🎯 Exam Tip: When events are both mutually exclusive and exhaustive, their probabilities must sum to 1. This property is crucial for solving problems where probabilities are expressed in terms of each other.

Question 13.
(i) In a single throw of two dice, find the probability that neither a doublet nor a total of a 10 will appear.
(ii) Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4.
(iii) Two dice are thrown together; what is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 5.
Answer:
When two dice are thrown, the total number of possible outcomes is \( 6 \times 6 = 36 \). So, \( n(S) = 36 \).
(i) Let A be the event of getting a doublet (both dice show the same number).
The doublets are {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}. So, \( n(A) = 6 \).
Let B be the event of getting a total of 10.
The pairs that sum to 10 are {(4, 6), (5, 5), (6, 4)}. So, \( n(B) = 3 \).
Now, find the intersection \( A \cap B \), which means getting a doublet AND a total of 10. This is just {(5, 5)}. So, \( n(A \cap B) = 1 \).
The probability of A is \( P(A) = \frac{6}{36} \).
The probability of B is \( P(B) = \frac{3}{36} \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{1}{36} \).
The probability that a doublet OR a total of 10 appears is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( \implies P(A \cup B) = \frac{6}{36} + \frac{3}{36} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9} \).
The probability that neither a doublet nor a total of 10 will appear is \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - \frac{2}{9} = \frac{7}{9} \). This means it's quite likely that neither of these specific outcomes will occur.
(ii) Let A be the event that the sum is a multiple of 3.
The sums that are multiples of 3 are {3, 6, 9, 12}. The pairs are:
Sum 3: {(1, 2), (2, 1)}
Sum 6: {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Sum 9: {(3, 6), (4, 5), (5, 4), (6, 3)}
Sum 12: {(6, 6)}
So, \( n(A) = 2 + 5 + 4 + 1 = 12 \).
Let B be the event that the sum is a multiple of 4.
The sums that are multiples of 4 are {4, 8, 12}. The pairs are:
Sum 4: {(1, 3), (2, 2), (3, 1)}
Sum 8: {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
Sum 12: {(6, 6)}
So, \( n(B) = 3 + 5 + 1 = 9 \).
Now, find the intersection \( A \cap B \), which means the sum is a multiple of both 3 and 4 (i.e., a multiple of 12). This is just {(6, 6)}. So, \( n(A \cap B) = 1 \).
The probability of A is \( P(A) = \frac{12}{36} \).
The probability of B is \( P(B) = \frac{9}{36} \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{1}{36} \).
The probability that the sum is a multiple of 3 OR a multiple of 4 is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( \implies P(A \cup B) = \frac{12}{36} + \frac{9}{36} - \frac{1}{36} = \frac{20}{36} = \frac{5}{9} \).
The probability that the sum is neither a multiple of 3 nor a multiple of 4 is \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - \frac{5}{9} = \frac{4}{9} \). This shows how complex conditions can be broken down using set theory principles.
(iii) Let A be the event that the sum is divisible by 3.
From part (ii), \( n(A) = 12 \). So, \( P(A) = \frac{12}{36} \).
Let B be the event that the sum is divisible by 5.
The sums divisible by 5 are {5, 10}. The pairs are:
Sum 5: {(1, 4), (2, 3), (3, 2), (4, 1)}
Sum 10: {(4, 6), (5, 5), (6, 4)}
So, \( n(B) = 4 + 3 = 7 \). So, \( P(B) = \frac{7}{36} \).
Now, find the intersection \( A \cap B \), which means the sum is divisible by both 3 and 5 (i.e., a multiple of 15). The maximum sum from two dice is 12, so no sum can be 15. Thus, \( A \cap B = \emptyset \), and \( n(A \cap B) = 0 \). These events are mutually exclusive.
The probability that the sum is divisible by 3 OR divisible by 5 is \( P(A \cup B) = P(A) + P(B) \).
\( \implies P(A \cup B) = \frac{12}{36} + \frac{7}{36} = \frac{19}{36} \).
The probability that the sum is neither divisible by 3 nor by 5 is \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - \frac{19}{36} = \frac{17}{36} \). This calculation highlights that sometimes even seemingly complex conditions can simplify if the events are mutually exclusive.
In simple words: For all three parts, we first list all 36 ways two dice can land. Then, for each part, we find the ways for the first condition, the ways for the second condition, and any ways that fit both. We use the formula P(A or B) = P(A) + P(B) - P(A and B). Finally, to find the chance that NEITHER happens, we subtract that P(A or B) from 1.

🎯 Exam Tip: When a question asks for "neither A nor B," calculate \( P(A \cup B) \) first, and then find its complement \( 1 - P(A \cup B) \). Always carefully identify the intersection of events for "or" probabilities.

Question 14. In a given race, the odds in favour of horses A, B, C and D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find the probability that one of them wins the race.
Answer: We are given the odds in favour of each horse winning. We need to convert these odds into probabilities.
If the odds in favour of an event are \( a:b \), the probability of the event occurring is \( \frac{a}{a+b} \).
For horse A:
Odds in favour of A are 1:3. So, \( P(A) = \frac{1}{1+3} = \frac{1}{4} \).
For horse B:
Odds in favour of B are 1:4. So, \( P(B) = \frac{1}{1+4} = \frac{1}{5} \).
For horse C:
Odds in favour of C are 1:5. So, \( P(C) = \frac{1}{1+5} = \frac{1}{6} \).
For horse D:
Odds in favour of D are 1:6. So, \( P(D) = \frac{1}{1+6} = \frac{1}{7} \).
Since only one horse can win the race, the events of A, B, C, or D winning are mutually exclusive (they cannot happen at the same time).
The probability that one of them wins is the sum of their individual probabilities:
\( P(\text{one wins}) = P(A) + P(B) + P(C) + P(D) \)
\( = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \)
To add these fractions, find a common denominator, which is the least common multiple of 4, 5, 6, and 7. The LCM is 420.
\( = \frac{105}{420} + \frac{84}{420} + \frac{70}{420} + \frac{60}{420} \)
\( = \frac{105 + 84 + 70 + 60}{420} = \frac{319}{420} \). This represents the overall chance that a winner will come from this group of four horses. If this sum were 1, it would imply these four horses are exhaustive, meaning no other horse could win.
In simple words: First, turn the "odds" for each horse into a normal probability (like "1 out of 4 chance"). Since only one horse can win, you just add up all their individual chances to find the total chance that one of these specific horses will be the winner.

🎯 Exam Tip: Always remember the conversion from odds to probability: if odds are \( a:b \) in favour, the probability is \( a/(a+b) \); if odds are \( a:b \) against, the probability is \( b/(a+b) \).

Question 15. 100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both examinations.
Answer: Total number of students = 100.
Let A be the event that a student passed the first examination.
\( P(A) = \frac{60}{100} = 0.60 \).
Let B be the event that a student passed the second examination.
\( P(B) = \frac{50}{100} = 0.50 \).
The number of students who passed both examinations is 30.
\( P(A \cap B) = \frac{30}{100} = 0.30 \).
First, find the probability that a student passed at least one examination (either the first, or the second, or both). This is \( P(A \cup B) \).
Using the formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.60 + 0.50 - 0.30 \)
\( = 1.10 - 0.30 = 0.80 \).
This means that 80% of students passed at least one exam.
The probability that a student failed in both examinations is the complement of passing at least one examination. This is \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - 0.80 = 0.20 \). So, there is a 20% chance that a randomly selected student failed in both exams.
In simple words: Out of 100 students, we know how many passed the first exam, how many passed the second, and how many passed both. First, we find the chance that a student passed at least one exam. Then, to find the chance that a student failed both, we take that number away from 1 (which represents 100% of students).

🎯 Exam Tip: Venn diagrams can be very helpful for visualizing problems involving 'both', 'either/or', and 'neither/nor' conditions, making it easier to apply the correct formulas.

Question 16.
(i) A card is drawn from a deck of 52 can is. Find the probability of getting an ace or a spade card.
(ii) From a well shuffled deck of 52 cards, 4 cards are drawn at random. What is the probability that all the drawn cards are of the same colour.
Answer:
(i) Total number of cards in a deck = 52. So, \( n(S) = 52 \).
Let A be the event of drawing an ace card. There are 4 aces in a deck. So, \( n(A) = 4 \).
Let B be the event of drawing a spade card. There are 13 spades in a deck. So, \( n(B) = 13 \).
The intersection \( A \cap B \) is the event of drawing an ace that is also a spade, which is the Ace of Spades. So, \( n(A \cap B) = 1 \).
The probability of A is \( P(A) = \frac{4}{52} \).
The probability of B is \( P(B) = \frac{13}{52} \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{1}{52} \).
The probability of getting an ace OR a spade card is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} \)
\( = \frac{4 + 13 - 1}{52} = \frac{16}{52} \).
Simplifying the fraction, \( \frac{16}{52} = \frac{4}{13} \). This means there's a 4 in 13 chance of drawing either an ace or a spade.
(ii) Total number of cards = 52.
We are drawing 4 cards at random. Total ways to draw 4 cards from 52 is \( ^{52}C_4 \).
\( ^{52}C_4 = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725 \). So, \( n(S) = 270725 \).
We want the probability that all drawn cards are of the same colour. This means all 4 cards are either red OR all 4 cards are black.
In a deck of 52 cards:
Number of red cards = 26 (13 hearts, 13 diamonds).
Number of black cards = 26 (13 spades, 13 clubs).
Let A be the event that all 4 cards drawn are red.
Number of ways to choose 4 red cards from 26 = \( ^{26}C_4 \).
\( ^{26}C_4 = \frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} = 14950 \). So, \( n(A) = 14950 \).
Let B be the event that all 4 cards drawn are black.
Number of ways to choose 4 black cards from 26 = \( ^{26}C_4 = 14950 \). So, \( n(B) = 14950 \).
The events A and B are mutually exclusive (cards cannot be both all red and all black at the same time).
So, the total number of favorable outcomes = \( n(A) + n(B) = 14950 + 14950 = 29900 \).
Required probability \( = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{29900}{270725} \).
This fraction can be simplified. Dividing both numerator and denominator by 25: \( \frac{1196}{10829} \). Further simplification may be possible, but this is a valid representation. This calculation shows the importance of distinguishing between mutually exclusive and non-mutually exclusive events.
In simple words: For the first part, count aces and spades. Since the Ace of Spades is both, don't count it twice. Then divide by 52. For the second part, find all ways to pick 4 cards. Then find ways to pick 4 red cards, and ways to pick 4 black cards. Add these two counts together (because they are different types of cards) and divide by the total ways to pick 4 cards.

🎯 Exam Tip: When drawing multiple items (like cards), combinations (\(^nC_k\)) are typically used. Carefully identify if the events are mutually exclusive to decide whether to add probabilities directly or use the inclusion-exclusion principle.

Question 17.
(i) A card is drawn at random from a well shuffled pack of cards. What is the probability that it is a heart or a queen ?
(ii) A card is drawn at random from a well shuffled pack of 52 cards. Find the probability it is neither a king nor a heart ?
Answer:
(i) Total number of cards in a deck = 52. So, \( n(S) = 52 \).
Let A be the event of drawing a heart card. There are 13 heart cards. So, \( n(A) = 13 \).
Let B be the event of drawing a queen card. There are 4 queen cards. So, \( n(B) = 4 \).
The intersection \( A \cap B \) is the event of drawing a queen that is also a heart, which is the Queen of Hearts. So, \( n(A \cap B) = 1 \).
The probability of A is \( P(A) = \frac{13}{52} \).
The probability of B is \( P(B) = \frac{4}{52} \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{1}{52} \).
The probability of getting a heart OR a queen is \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \)
\( = \frac{13 + 4 - 1}{52} = \frac{16}{52} \).
Simplifying the fraction, \( \frac{16}{52} = \frac{4}{13} \). This calculation correctly accounts for the Queen of Hearts being counted once.
(ii) Total number of cards = 52. So, \( n(S) = 52 \).
Let A be the event of drawing a king card. There are 4 king cards. So, \( n(A) = 4 \).
Let B be the event of drawing a heart card. There are 13 heart cards. So, \( n(B) = 13 \).
The intersection \( A \cap B \) is the event of drawing a king that is also a heart, which is the King of Hearts. So, \( n(A \cap B) = 1 \).
The probability of A is \( P(A) = \frac{4}{52} \).
The probability of B is \( P(B) = \frac{13}{52} \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{1}{52} \).
First, find the probability of drawing a king OR a heart: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
\( = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} \)
\( = \frac{16}{52} = \frac{4}{13} \).
The probability that the card is neither a king nor a heart is the complement of drawing a king or a heart. This is \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
\( = 1 - \frac{4}{13} \).
\( = \frac{13}{13} - \frac{4}{13} = \frac{9}{13} \). This means there's a high chance the card will not be a king or a heart, as there are many other cards in the deck.
In simple words: For the first part, count how many hearts and queens there are, making sure not to count the Queen of Hearts twice. Then divide by 52. For the second part, find the chance of drawing a king or a heart (again, don't double count the King of Hearts). Then, subtract this chance from 1 to find the chance of drawing neither a king nor a heart.

🎯 Exam Tip: "Neither A nor B" questions are best solved by finding \( P(A \cup B) \) first and then using the complement rule. Always carefully identify and subtract the intersection for "or" events.

Question 18. A card is drawn from a pack of 52 cars. Find the probability of getting a king or a heart or a red card.
Answer: Total number of cards in a deck = 52. So, \( n(S) = 52 \).
Let K be the event of drawing a king card. There are 4 kings. So, \( n(K) = 4 \).
Let H be the event of drawing a heart card. There are 13 hearts. So, \( n(H) = 13 \).
Let R be the event of drawing a red card. There are 26 red cards (13 hearts + 13 diamonds). So, \( n(R) = 26 \).
Now, find the intersections between these events:
\( K \cap H \): King of Hearts. \( n(K \cap H) = 1 \).
\( H \cap R \): Heart cards. All hearts are red, so this is just the set of hearts. \( n(H \cap R) = 13 \).
\( K \cap R \): Kings that are red. These are the King of Hearts and the King of Diamonds. \( n(K \cap R) = 2 \).
\( K \cap H \cap R \): Kings that are hearts AND red. This is just the King of Hearts. \( n(K \cap H \cap R) = 1 \).
The probability of getting a king OR a heart OR a red card is given by the formula for the union of three events:
\( P(K \cup H \cup R) = P(K) + P(H) + P(R) - P(K \cap H) - P(H \cap R) - P(K \cap R) + P(K \cap H \cap R) \)
Substitute the probabilities:
\( P(K \cup H \cup R) = \frac{4}{52} + \frac{13}{52} + \frac{26}{52} - \frac{1}{52} - \frac{13}{52} - \frac{2}{52} + \frac{1}{52} \)
\( = \frac{4 + 13 + 26 - 1 - 13 - 2 + 1}{52} \)
\( = \frac{28}{52} \).
Simplifying the fraction, \( \frac{28}{52} = \frac{7}{13} \). This formula allows for precise calculation even with multiple overlapping conditions.
In simple words: We want the chance of drawing a king, a heart, or a red card. We add the chances of each happening. Then, because some cards are both a king and a heart, or a heart and red, or a king and red, we subtract those overlaps. Finally, we add back the cards that fit all three groups to make sure nothing is counted too many times.

🎯 Exam Tip: For problems with three overlapping events, meticulously list all individual event probabilities and all two-way and three-way intersection probabilities before applying the inclusion-exclusion principle formula.

Question 19. There are three events A, B, C one of which must and only one can happen. The odds are 8 to 3 against A, 5 to 2 against B ; find the odds against C.
Answer: We are given that A, B, and C are three events, one of which must and only one can happen. This means they are mutually exclusive (only one can happen) and exhaustive (one must happen). Therefore, their probabilities must sum to 1: \( P(A) + P(B) + P(C) = 1 \).
First, convert the given odds against A and B into probabilities.
Odds against A are 8:3. This means for every 8 unfavorable outcomes, there are 3 favorable outcomes. So, \( P(A) = \frac{3}{8+3} = \frac{3}{11} \).
Odds against B are 5:2. This means for every 5 unfavorable outcomes, there are 2 favorable outcomes. So, \( P(B) = \frac{2}{5+2} = \frac{2}{7} \).
Now, use \( P(A) + P(B) + P(C) = 1 \) to find \( P(C) \).
\( \frac{3}{11} + \frac{2}{7} + P(C) = 1 \)
To add the fractions, find a common denominator for 11 and 7, which is 77.
\( \frac{3 \times 7}{77} + \frac{2 \times 11}{77} + P(C) = 1 \)
\( \frac{21}{77} + \frac{22}{77} + P(C) = 1 \)
\( \frac{43}{77} + P(C) = 1 \)
\( P(C) = 1 - \frac{43}{77} \)
\( P(C) = \frac{77 - 43}{77} = \frac{34}{77} \).
Next, we need to find the odds against C. If \( P(C) = \frac{34}{77} \), then the probability that C does NOT happen (the complement of C, \( P(\overline{C}) \)) is \( 1 - P(C) \).
\( P(\overline{C}) = 1 - \frac{34}{77} = \frac{43}{77} \).
The odds against C are \( P(\overline{C}) : P(C) \).
\( = \frac{43}{77} : \frac{34}{77} \).
Multiply both sides by 77 to simplify the ratio:
\( = 43 : 34 \). This shows that it is more likely for C not to happen than to happen.
In simple words: We are told that one of three events (A, B, C) must happen, and only one can. This means their chances add up to 1. We change the "odds against" for A and B into normal chances. Then we use the fact that all chances add to 1 to find C's chance. Finally, we turn C's chance back into "odds against" C.

🎯 Exam Tip: For mutually exclusive and exhaustive events, their probabilities must sum to exactly 1. Remember to correctly convert odds to probabilities (and vice versa) for accurate calculations.

Question 20. In a group of students, there are 3 boys and 3 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Answer: Total number of students in the group = 3 boys + 3 girls = 6 students.
We are selecting 4 students at random from this group.
Total number of ways to select 4 students from 6 is \( ^6C_4 \).
\( ^6C_4 = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = \frac{6 \times 5}{2 \times 1} = 15 \). So, \( n(S) = 15 \).
Let A be the event of selecting 3 boys and 1 girl.
Number of ways to select 3 boys from 3 boys = \( ^3C_3 = 1 \).
Number of ways to select 1 girl from 3 girls = \( ^3C_1 = 3 \).
So, \( n(A) = ^3C_3 \times ^3C_1 = 1 \times 3 = 3 \).
Let B be the event of selecting 3 girls and 1 boy.
Number of ways to select 3 girls from 3 girls = \( ^3C_3 = 1 \).
Number of ways to select 1 boy from 3 boys = \( ^3C_1 = 3 \).
So, \( n(B) = ^3C_3 \times ^3C_1 = 1 \times 3 = 3 \).
The events A and B are mutually exclusive because you cannot select both 3 boys and 1 girl AND 3 girls and 1 boy at the same time when selecting only 4 students.
The required probability (either A or B) is the sum of the number of ways for each event, divided by the total ways to select 4 students.
Total favorable outcomes \( = n(A) + n(B) = 3 + 3 = 6 \).
Required probability \( = \frac{6}{15} \).
Simplifying the fraction, \( \frac{6}{15} = \frac{2}{5} \). This means there's a 2 in 5 chance of selecting one of these two specific gender combinations.
In simple words: First, find all the ways to pick any 4 students from the 6. Then, count the ways to pick exactly 3 boys and 1 girl. Separately, count the ways to pick exactly 3 girls and 1 boy. Since these two selections are different, add their counts together and divide by the total number of ways to pick students.

🎯 Exam Tip: Clearly define the "cases" in combinatorial probability problems. If cases are mutually exclusive, sum the number of outcomes for each case before dividing by the total possible outcomes.