Samacheer Kalvi Class 10 Maths Solutions Chapter 8 Statistics and Probability Exercise 8.4

Question 1. If \( P(A) = \frac{2}{3} \), \( P(B) = \frac{2}{5} \), \( P(A \cup B) = \frac{1}{3} \), then find \( P(A \cap B) \).
Answer: We use the formula for the probability of the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Now, we put the given values into the formula:
\( \frac{1}{3} = \frac{2}{3} + \frac{2}{5} - P(A \cap B) \)
Next, we rearrange the equation to find \( P(A \cap B) \):
\( P(A \cap B) = \frac{2}{3} + \frac{2}{5} - \frac{1}{3} \)
To add and subtract these fractions, we find a common denominator, which is 15:
\( P(A \cap B) = \frac{10}{15} + \frac{6}{15} - \frac{5}{15} \)
\( P(A \cap B) = \frac{10 + 6 - 5}{15} \)
\( P(A \cap B) = \frac{11}{15} \)
In simple words: We used a special formula to connect the probabilities of events A, B, and their union. By putting in the numbers we knew, we could figure out the probability of both events A and B happening together.

🎯 Exam Tip: Remember the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) and practice rearranging it to find any missing term. Make sure to find a common denominator when adding or subtracting fractions.

Question 2. A and B are two events such that, \( P(A) = 0.42 \), \( P(B) = 0.48 \), and \( P(A \cap B) = 0.16 \). Find
(i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Answer:
Given:
\( P(A) = 0.42 \)
\( P(B) = 0.48 \)
\( P(A \cap B) = 0.16 \)

(i) To find the probability of 'not A', we subtract \( P(A) \) from 1:
\( P(\text{not A}) = P(\overline{A}) = 1 - P(A) \)
\( P(\text{not A}) = 1 - 0.42 = 0.58 \)

(ii) To find the probability of 'not B', we subtract \( P(B) \) from 1:
\( P(\text{not B}) = P(\overline{B}) = 1 - P(B) \)
\( P(\text{not B}) = 1 - 0.48 = 0.52 \)

(iii) To find the probability of 'A or B', we use the union formula:
\( P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = 0.42 + 0.48 - 0.16 \)
\( P(A \cup B) = 0.90 - 0.16 \)
\( P(A \cup B) = 0.74 \)
In simple words: We found the chance of A not happening and B not happening by taking 1 minus their individual probabilities. Then, we found the chance of A or B happening by adding their individual chances and subtracting the chance of both happening together.

🎯 Exam Tip: Remember that \( P(\text{not event}) = 1 - P(\text{event}) \). This is a basic rule in probability. For 'A or B', always use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) unless the events are mutually exclusive (meaning \( P(A \cap B) = 0 \)).

Question 3. If A and B are two mutually exclusive events of a random experiment and \( P(\text{not A}) = 0.45 \), \( P(A \cup B) = 0.65 \), then find \( P(B) \).
Answer: Given that A and B are mutually exclusive events, this means their intersection is zero: \( P(A \cap B) = 0 \).
We are given:
\( P(\text{not A}) = 0.45 \)
\( P(A \cup B) = 0.65 \)

First, find \( P(A) \) from \( P(\text{not A}) \):
\( P(A) = 1 - P(\text{not A}) \)
\( P(A) = 1 - 0.45 = 0.55 \)

Since A and B are mutually exclusive, the formula for \( P(A \cup B) \) simplifies to:
\( P(A \cup B) = P(A) + P(B) \)
Now substitute the known values:
\( 0.65 = 0.55 + P(B) \)
To find \( P(B) \), subtract 0.55 from 0.65:
\( P(B) = 0.65 - 0.55 \)
\( P(B) = 0.10 \)
So, \( P(B) = 0.1 \).
In simple words: We first found the chance of event A happening by subtracting 'not A' from 1. Since A and B cannot happen at the same time, we could simply subtract the chance of A from the chance of A or B to find the chance of B.

🎯 Exam Tip: The key phrase "mutually exclusive events" simplifies the union formula to \( P(A \cup B) = P(A) + P(B) \). Always look for this phrase or similar ones (like "disjoint events") to make calculations easier.

Question 4. The probability that at least one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find \( P(\overline{A}) + P(\overline{B}) \).
Answer: We are given:
Probability that at least one of A and B occur is \( P(A \cup B) = 0.6 \).
Probability that A and B occur simultaneously is \( P(A \cap B) = 0.2 \).

We need to find \( P(\overline{A}) + P(\overline{B}) \).
We know that \( P(\overline{A}) = 1 - P(A) \) and \( P(\overline{B}) = 1 - P(B) \).
So, \( P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B)) \).

First, let's find \( P(A) + P(B) \) using the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Rearranging the formula:
\( P(A) + P(B) = P(A \cup B) + P(A \cap B) \)
Substitute the given values:
\( P(A) + P(B) = 0.6 + 0.2 = 0.8 \)

Now substitute this sum back into the expression for \( P(\overline{A}) + P(\overline{B}) \):
\( P(\overline{A}) + P(\overline{B}) = 2 - (P(A) + P(B)) \)
\( P(\overline{A}) + P(\overline{B}) = 2 - 0.8 \)
\( P(\overline{A}) + P(\overline{B}) = 1.2 \)
In simple words: We know the chance of A or B happening, and the chance of both A and B happening. We used a probability rule to find the sum of their individual chances. Then, we used another rule to find the sum of the chances of A not happening and B not happening.

🎯 Exam Tip: This question tests your understanding of both the union formula and the complement rule. Remember that \( P(\overline{A}) + P(\overline{B}) \) is not simply \( P(\overline{A \cup B}) \) or \( P(\overline{A \cap B}) \), but rather related to \( P(A) + P(B) \).

Question 5. The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen.
Answer: Given:
\( P(A) = 0.5 \)
\( P(B) = 0.3 \)
A and B are mutually exclusive events, which means \( P(A \cap B) = 0 \).

First, find the probability that A or B happens (union of A and B):
Since A and B are mutually exclusive:
\( P(A \cup B) = P(A) + P(B) \)
\( P(A \cup B) = 0.5 + 0.3 \)
\( P(A \cup B) = 0.8 \)

Now, we need to find the probability of neither A nor B happening. This is the complement of the event (A or B).
\( P(\text{neither A nor B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) \)
\( P(\text{neither A nor B}) = 1 - 0.8 \)
\( P(\text{neither A nor B}) = 0.2 \)
In simple words: Since A and B cannot happen at the same time, we just added their chances to find the chance of either one happening. Then, to find the chance of *neither* of them happening, we subtracted that sum from 1.

🎯 Exam Tip: The phrase "neither A nor B" means \( P(\overline{A \cup B}) \), which can be calculated as \( 1 - P(A \cup B) \). Always combine the concepts of mutually exclusive events and complements correctly.

Question 6. Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Answer: When two dice are rolled once, the total number of possible outcomes (sample space) is \( 6 \times 6 = 36 \).
The sample space S is:
\( S = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), \)
\( (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \)
\( (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), \)
\( (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \)
\( (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \)
\( (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} \)
So, \( n(S) = 36 \).

Let A be the event of getting an even number on the first die.
\( A = \{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \)
\( (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), \)
\( (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} \)
The number of outcomes in event A is \( n(A) = 18 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{18}{36} \).

Let B be the event of getting a total face sum of 8.
\( B = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \)
The number of outcomes in event B is \( n(B) = 5 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{5}{36} \).

Now, find the intersection of A and B, which means getting an even number on the first die AND a total sum of 8.
\( A \cap B = \{(2, 6), (4, 4), (6, 2)\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 3 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{3}{36} \).

Finally, find the probability of getting an even number on the first die OR a total face sum of 8, using the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} \)
\( P(A \cup B) = \frac{18 + 5 - 3}{36} \)
\( P(A \cup B) = \frac{20}{36} \)
Simplify the fraction:
\( P(A \cup B) = \frac{5}{9} \)
In simple words: We listed all possible outcomes when rolling two dice. Then, we found how many times we get an even number on the first die and how many times the numbers add up to 8. We also found how many times both of these happen together. Using a formula, we calculated the chance of either one of these events happening.

🎯 Exam Tip: When dealing with two dice, always remember that \( n(S) = 36 \). Be careful to list all outcomes for each event and their intersection accurately to avoid calculation errors. Simplify fractions to their lowest terms for the final answer.

Question 7. From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen.
Answer: The total number of cards in a pack is 52.
So, the size of the sample space is \( n(S) = 52 \).

Let A be the event of getting a red king.
There are two red kings in a pack of cards (King of Hearts and King of Diamonds).
So, \( n(A) = 2 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{2}{52} \).

Let B be the event of getting a black queen.
There are two black queens in a pack of cards (Queen of Spades and Queen of Clubs).
So, \( n(B) = 2 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{2}{52} \).

The events A (getting a red king) and B (getting a black queen) are mutually exclusive because a card cannot be both a red king and a black queen at the same time. This means \( P(A \cap B) = 0 \).

The probability of getting either a red king or a black queen is given by the union of these events:
\( P(A \cup B) = P(A) + P(B) \)
\( P(A \cup B) = \frac{2}{52} + \frac{2}{52} \)
\( P(A \cup B) = \frac{4}{52} \)
Simplify the fraction:
\( P(A \cup B) = \frac{1}{13} \)
In simple words: We found the total number of cards. Then, we counted how many red kings there are and how many black queens there are. Since a card cannot be both, we just added their chances together and simplified the answer.

🎯 Exam Tip: For card problems, always remember the standard deck composition (52 cards, 4 suits, 2 colors, 13 ranks). Clearly identify if events are mutually exclusive (cannot happen at the same time) or not, as this affects the union formula.

Question 8. A box contains cards numbered 3, 5, 7, 9,..., 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Answer: First, list all the numbers in the sample space (S):
\( S = \{3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37\} \)
The total number of cards is \( n(S) = 18 \).

Let A be the event of getting a multiple of 7.
The multiples of 7 in the sample space are:
\( A = \{7, 21, 35\} \)
The number of outcomes in event A is \( n(A) = 3 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{3}{18} \).

Let B be the event of getting a prime number.
Prime numbers in the sample space are numbers greater than 1 that have only two factors: 1 and themselves.
\( B = \{3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37\} \)
The number of outcomes in event B is \( n(B) = 11 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{11}{18} \).

Now, find the intersection of A and B, which means numbers that are both multiples of 7 AND prime.
\( A \cap B = \{7\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 1 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{18} \).

Finally, find the probability of getting either a multiple of 7 OR a prime number, using the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = \frac{3}{18} + \frac{11}{18} - \frac{1}{18} \)
\( P(A \cup B) = \frac{3 + 11 - 1}{18} \)
\( P(A \cup B) = \frac{13}{18} \)
In simple words: First, we listed all the cards. Then, we found cards that are multiples of 7 and cards that are prime numbers. We also found any cards that are both. Using a special formula, we added the chances for each event and subtracted the chance of both happening to find the final probability.

🎯 Exam Tip: Be careful when identifying prime numbers and multiples within a given set. Double-check your lists for accuracy, especially for the intersection, as a small error can lead to a wrong final probability. Always list all numbers in the sample space clearly first.

Question 9. Three unbiased coins are tossed once. Find the probability of getting at most 2 tails or at least 2 heads.
Answer: When three unbiased coins are tossed once, the sample space (S) consists of \( 2^3 = 8 \) possible outcomes:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
So, \( n(S) = 8 \).

Let A be the event of getting at most 2 tails (meaning 0, 1, or 2 tails).
\( A = \{HHH, HHT, HTH, HTT, THH, THT, TTH\} \)
The number of outcomes in event A is \( n(A) = 7 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{7}{8} \).

Let B be the event of getting at least 2 heads (meaning 2 or 3 heads).
\( B = \{HHH, HHT, HTH, THH\} \)
The number of outcomes in event B is \( n(B) = 4 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{4}{8} \).

Now, find the intersection of A and B, which means getting at most 2 tails AND at least 2 heads.
\( A \cap B = \{HHH, HHT, HTH, THH\} \)
The number of outcomes in \( A \cap B \) is \( n(A \cap B) = 4 \).
The probability of \( A \cap B \) is \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{4}{8} \).

Finally, find the probability of getting at most 2 tails OR at least 2 heads, using the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = \frac{7}{8} + \frac{4}{8} - \frac{4}{8} \)
\( P(A \cup B) = \frac{7 + 4 - 4}{8} \)
\( P(A \cup B) = \frac{7}{8} \)
In simple words: We listed all possible results when tossing three coins. Then we counted how many ways we could get 0, 1, or 2 tails, and how many ways we could get 2 or 3 heads. We also found the results that fit both descriptions. Using a formula, we calculated the chance of either of these things happening.

🎯 Exam Tip: Be careful with phrases like "at most" and "at least." "At most 2 tails" means 0, 1, or 2 tails. "At least 2 heads" means 2 or 3 heads. Listing out the sample space clearly helps in correctly identifying the outcomes for each event and their intersection.

Question 10. The probability that a person will get an electrification contract is \( \frac{3}{5} \) and the probability that he will not get plumbing contract is \( \frac{5}{8} \). The probability of getting at least one contract is \( \frac{5}{7} \). What is the probability that he will get both?
Answer: Let E be the event that the person gets an electrification contract.
Let P be the event that the person gets a plumbing contract.

We are given:
\( P(E) = \frac{3}{5} \)
\( P(\text{not P}) = P(\overline{P}) = \frac{5}{8} \)
Probability of getting at least one contract is \( P(E \cup P) = \frac{5}{7} \).

First, find the probability of getting a plumbing contract, \( P(P) \):
\( P(P) = 1 - P(\overline{P}) \)
\( P(P) = 1 - \frac{5}{8} = \frac{8-5}{8} = \frac{3}{8} \)

We need to find the probability that he will get both contracts, which is \( P(E \cap P) \).
Use the union formula for two events:
\( P(E \cup P) = P(E) + P(P) - P(E \cap P) \)
Substitute the known values into the formula:
\( \frac{5}{7} = \frac{3}{5} + \frac{3}{8} - P(E \cap P) \)
Rearrange the equation to solve for \( P(E \cap P) \):
\( P(E \cap P) = \frac{3}{5} + \frac{3}{8} - \frac{5}{7} \)
Find a common denominator for 5, 8, and 7, which is \( 5 \times 8 \times 7 = 280 \).
\( P(E \cap P) = \frac{3 \times 56}{280} + \frac{3 \times 35}{280} - \frac{5 \times 40}{280} \)
\( P(E \cap P) = \frac{168}{280} + \frac{105}{280} - \frac{200}{280} \)
\( P(E \cap P) = \frac{168 + 105 - 200}{280} \)
\( P(E \cap P) = \frac{273 - 200}{280} \)
\( P(E \cap P) = \frac{73}{280} \)
In simple words: We were given the chance of getting an electrical contract, not getting a plumbing contract, and getting at least one contract. First, we figured out the chance of getting a plumbing contract. Then, using a formula that connects all these chances, we found the probability of getting both types of contracts.

🎯 Exam Tip: Pay close attention to "not get plumbing contract" and convert it to \( P(P) \) using \( 1 - P(\overline{P}) \). The phrase "at least one contract" always refers to the union of events. Be meticulous with fraction addition and subtraction using common denominators.

Question 11. In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Answer: Total number of people in the town, \( n(S) = 8000 \).

Let A be the event that a chosen individual is female.
Number of females, \( n(A) = 3000 \).
Probability of choosing a female, \( P(A) = \frac{n(A)}{n(S)} = \frac{3000}{8000} \).

Let B be the event that a chosen individual is over 50 years.
Number of people over 50 years, \( n(B) = 1300 \).
Probability of choosing someone over 50 years, \( P(B) = \frac{n(B)}{n(S)} = \frac{1300}{8000} \).

We are given that 30% of the females are over 50 years.
This means the number of females who are also over 50 years is \( n(A \cap B) \).
\( n(A \cap B) = 30\% \text{ of } 3000 = \frac{30}{100} \times 3000 = 900 \).
Probability of choosing a female who is also over 50 years, \( P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{900}{8000} \).

We need to find the probability that a chosen individual is either a female OR over 50 years, which is \( P(A \cup B) \).
Using the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = \frac{3000}{8000} + \frac{1300}{8000} - \frac{900}{8000} \)
\( P(A \cup B) = \frac{3000 + 1300 - 900}{8000} \)
\( P(A \cup B) = \frac{4300 - 900}{8000} \)
\( P(A \cup B) = \frac{3400}{8000} \)
Simplify the fraction:
\( P(A \cup B) = \frac{34}{80} = \frac{17}{40} \)
In simple words: We calculated the chance of picking a female and the chance of picking someone over 50 years. We also found the number of people who are both female and over 50. Using a formula, we added the individual chances and subtracted the chance of both, to find the probability of picking someone who is either female or over 50.

🎯 Exam Tip: This problem involves conditional information ("30% of the females are over 50"). Make sure to calculate the number in the intersection (females AND over 50) correctly before applying the union formula. Always simplify fractions for the final answer.

Question 12. A coin is tossed thrice. Find the probability of getting exactly two heads or at least one tail or two consecutive heads.
Answer: When a coin is tossed thrice, the sample space (S) consists of \( 2^3 = 8 \) possible outcomes:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
So, \( n(S) = 8 \).

Let A be the event of getting exactly two heads.
\( A = \{HHT, HTH, THH\} \)
\( n(A) = 3 \)
\( P(A) = \frac{n(A)}{n(S)} = \frac{3}{8} \)

Let B be the event of getting at least one tail.
This means 1, 2, or 3 tails. This is all outcomes except HHH.
\( B = \{HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
\( n(B) = 7 \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{7}{8} \)

Let C be the event of getting two consecutive heads.
\( C = \{HHH, HHT, THH\} \)
\( n(C) = 3 \)
\( P(C) = \frac{n(C)}{n(S)} = \frac{3}{8} \)

Now, find the intersections:
\( A \cap B \) (exactly two heads AND at least one tail):
\( A \cap B = \{HHT, HTH, THH\} \)
\( n(A \cap B) = 3 \)
\( P(A \cap B) = \frac{3}{8} \)

\( B \cap C \) (at least one tail AND two consecutive heads):
\( B \cap C = \{HHT, THH\} \)
\( n(B \cap C) = 2 \)
\( P(B \cap C) = \frac{2}{8} \)

\( A \cap C \) (exactly two heads AND two consecutive heads):
\( A \cap C = \{HHT, THH\} \)
\( n(A \cap C) = 2 \)
\( P(A \cap C) = \frac{2}{8} \)

\( A \cap B \cap C \) (exactly two heads AND at least one tail AND two consecutive heads):
\( A \cap B \cap C = \{HHT, THH\} \)
\( n(A \cap B \cap C) = 2 \)
\( P(A \cap B \cap C) = \frac{2}{8} \)

Finally, use the Principle of Inclusion-Exclusion for three events to find \( P(A \cup B \cup C) \):
\( P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \)
\( P(A \cup B \cup C) = \frac{3}{8} + \frac{7}{8} + \frac{3}{8} - \frac{3}{8} - \frac{2}{8} - \frac{2}{8} + \frac{2}{8} \)
\( P(A \cup B \cup C) = \frac{3 + 7 + 3 - 3 - 2 - 2 + 2}{8} \)
\( P(A \cup B \cup C) = \frac{13 - 7 + 2}{8} \)
\( P(A \cup B \cup C) = \frac{8}{8} \)
\( P(A \cup B \cup C) = 1 \)
In simple words: We listed all outcomes for three coin tosses. Then we identified outcomes for three separate conditions: exactly two heads, at least one tail, and two heads in a row. We also found which outcomes fit two or all three conditions. By using a special formula for combining probabilities of three events, we found that it is certain (probability 1) that at least one of these conditions will happen.

🎯 Exam Tip: For problems involving three events, remember the full Principle of Inclusion-Exclusion formula. Systematically list all elements for A, B, C, then their pairwise intersections, and finally the three-way intersection. This helps avoid missing outcomes and ensures correct calculations.

Question 13. If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if \( P(A \cap B) = \frac{1}{6} \), \( P(B \cap C) = \frac{1}{4} \), \( P(A \cap C) = \frac{1}{8} \), \( P(A \cup B \cup C) = \frac{9}{10} \) and \( P(A \cap B \cap C) = \frac{1}{15} \), then find P(A), P(B) and P(C)?
Answer: Let \( P(A) = x \).
From the given conditions:
\( P(B) = 2 \times P(A) \implies P(B) = 2x \)
\( P(C) = 3 \times P(A) \implies P(C) = 3x \)

We are given the following probabilities:
\( P(A \cap B) = \frac{1}{6} \)
\( P(B \cap C) = \frac{1}{4} \)
\( P(A \cap C) = \frac{1}{8} \)
\( P(A \cup B \cup C) = \frac{9}{10} \)
\( P(A \cap B \cap C) = \frac{1}{15} \)

Now, use the Principle of Inclusion-Exclusion for three events:
\( P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \)
Substitute all the known values and expressions in terms of x:
\( \frac{9}{10} = x + 2x + 3x - \frac{1}{6} - \frac{1}{4} - \frac{1}{8} + \frac{1}{15} \)
Combine the terms with x:
\( \frac{9}{10} = 6x - \frac{1}{6} - \frac{1}{4} - \frac{1}{8} + \frac{1}{15} \)
Move all the constant terms to the left side:
\( 6x = \frac{9}{10} + \frac{1}{6} + \frac{1}{4} + \frac{1}{8} - \frac{1}{15} \)
Find the least common multiple (LCM) of the denominators (10, 6, 4, 8, 15), which is 120.
\( 6x = \frac{9 \times 12}{120} + \frac{1 \times 20}{120} + \frac{1 \times 30}{120} + \frac{1 \times 15}{120} - \frac{1 \times 8}{120} \)
\( 6x = \frac{108}{120} + \frac{20}{120} + \frac{30}{120} + \frac{15}{120} - \frac{8}{120} \)
\( 6x = \frac{108 + 20 + 30 + 15 - 8}{120} \)
\( 6x = \frac{173 - 8}{120} \)
\( 6x = \frac{165}{120} \)
Simplify the fraction \( \frac{165}{120} \) by dividing by 15:
\( 6x = \frac{11}{8} \)
Now, solve for x:
\( x = \frac{11}{8 \times 6} \)
\( x = \frac{11}{48} \)

Now find \( P(A) \), \( P(B) \), and \( P(C) \):
\( P(A) = x = \frac{11}{48} \)
\( P(B) = 2x = 2 \times \frac{11}{48} = \frac{22}{48} = \frac{11}{24} \)
\( P(C) = 3x = 3 \times \frac{11}{48} = \frac{33}{48} = \frac{11}{16} \)
In simple words: We used a given relationship to write the probabilities of B and C in terms of A. Then, we used a big formula that combines the probabilities of three events and their overlaps. By putting all the numbers and expressions into this formula, we solved for the basic probability of A, and then found the probabilities of B and C.

🎯 Exam Tip: When probabilities are related (e.g., \( P(B) = 2P(A) \)), express them all in terms of a single variable. This makes solving the Principle of Inclusion-Exclusion equation much simpler. Always find the LCM for fractions to perform calculations accurately.

Question 14. In a class of 35, students are numbered from 1 to 35. The ratio of boys and girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Answer: Total number of students in the class, \( n(S) = 35 \).
The sample space (S) is the set of roll numbers from 1 to 35: \( S = \{1, 2, 3, ..., 35\} \).

The ratio of boys to girls is 4:3. So, the total ratio parts are \( 4 + 3 = 7 \).
Number of boys \( = \frac{4}{7} \times 35 = 20 \). The boys' roll numbers are \( \{1, 2, ..., 20\} \).
Number of girls \( = \frac{3}{7} \times 35 = 15 \). The girls' roll numbers are \( \{21, 22, ..., 35\} \).

Let A be the event of getting a boy with a prime roll number.
Prime numbers among boys' roll numbers (1-20) are \( \{2, 3, 5, 7, 11, 13, 17, 19\} \).
\( n(A) = 8 \)
\( P(A) = \frac{n(A)}{n(S)} = \frac{8}{35} \).

Let B be the event of getting a girl with a composite roll number.
Girls' roll numbers are \( \{21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35\} \).
Composite numbers among these are numbers greater than 1 that are not prime.
\( B = \{21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35\} \)
Note: 23, 29, 31 are prime and are excluded.
\( n(B) = 12 \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{12}{35} \).

Let C be the event of getting an even roll number.
Even numbers among 1-35 are \( \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34\} \).
\( n(C) = 17 \)
\( P(C) = \frac{n(C)}{n(S)} = \frac{17}{35} \).

Now, find the intersections:
\( A \cap B \): A boy with a prime roll number AND a girl with a composite roll number. This is impossible as a student cannot be both a boy and a girl.
\( A \cap B = \{\} \)
\( n(A \cap B) = 0 \implies P(A \cap B) = 0 \).

\( B \cap C \) (a girl with a composite roll number AND an even roll number):
Girls' roll numbers are from 21-35. Composite and even among these are:
\( B \cap C = \{22, 24, 26, 28, 30, 32, 34\} \)
\( n(B \cap C) = 7 \)
\( P(B \cap C) = \frac{7}{35} \).

\( A \cap C \) (a boy with a prime roll number AND an even roll number):
Boys' roll numbers are from 1-20. Prime and even among these:
The only even prime number is 2.
\( A \cap C = \{2\} \)
\( n(A \cap C) = 1 \)
\( P(A \cap C) = \frac{1}{35} \).

\( A \cap B \cap C \) (a boy with a prime roll number AND a girl with a composite roll number AND an even roll number): This is also impossible as \( A \cap B \) is empty.
\( A \cap B \cap C = \{\} \)
\( n(A \cap B \cap C) = 0 \implies P(A \cap B \cap C) = 0 \).

Finally, use the Principle of Inclusion-Exclusion for three events to find \( P(A \cup B \cup C) \):
\( P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C) \)
\( P(A \cup B \cup C) = \frac{8}{35} + \frac{12}{35} + \frac{17}{35} - 0 - \frac{7}{35} - \frac{1}{35} + 0 \)
\( P(A \cup B \cup C) = \frac{8 + 12 + 17 - 7 - 1}{35} \)
\( P(A \cup B \cup C) = \frac{37 - 8}{35} \)
\( P(A \cup B \cup C) = \frac{29}{35} \)
In simple words: We first divided the class into boys and girls by their roll numbers. Then, we found the chance of picking a boy with a prime roll number, a girl with a composite roll number, and any student with an even roll number. We also found the overlaps between these groups. Using a formula for combining probabilities, we calculated the total chance of picking a student who meets any of these conditions.

🎯 Exam Tip: Clearly define the ranges for boys and girls and carefully list prime and composite numbers within those ranges. Remember that "boy" and "girl" events are mutually exclusive. The "Principle of Inclusion-Exclusion" formula is crucial for combining three events. Be sure to calculate all intersections correctly.