Samacheer Kalvi Class 11 Maths Solutions Chapter 12 Introduction to Probability Theory Exercise 12.1

Question 1. An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible.
(i) \( P(A) = 0.15 \), \( P(B) = 0.30 \), \( P(C) = 0.43 \), \( P(D) = 0.12 \)
Answer: For probabilities to be permissible, two conditions must be met: first, each probability must be between 0 and 1 (inclusive), and second, the sum of all probabilities must be exactly 1.
(i) Given probabilities are \( P(A) = 0.15 \), \( P(B) = 0.30 \), \( P(C) = 0.43 \), \( P(D) = 0.12 \).
All these probabilities are between 0 and 1, which satisfies the first condition.
Now, let's sum them up:
\( P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 \)
\( \implies P(A) + P(B) + P(C) + P(D) = 1.00 \)
Since the sum is exactly 1, the second condition is also satisfied. Therefore, this assignment of probability is permissible.
(ii) \( P(A) = 0.22 \), \( P(B) = 0.38 \), \( P(C) = 0.16 \), \( P(D) = 0.34 \)
Answer:
(ii) Given probabilities are \( P(A) = 0.22 \), \( P(B) = 0.38 \), \( P(C) = 0.16 \), \( P(D) = 0.34 \).
All these probabilities are greater than or equal to 0, satisfying one part of the first condition. Also, they are all less than or equal to 1.
Let's find their sum:
\( P(A) + P(B) + P(C) + P(D) = 0.22 + 0.38 + 0.16 + 0.34 \)
\( \implies P(A) + P(B) + P(C) + P(D) = 1.10 \)
Since the sum of the probabilities is 1.10, which is greater than 1, this assignment is not permissible. Probabilities must sum up to exactly 1.
(iii) \( P(A) = \frac{2}{5} \), \( P(B) = \frac{3}{5} \), \( P(C) = -\frac{1}{5} \), \( P(D) = \frac{1}{5} \)
Answer:
(iii) Given probabilities are \( P(A) = \frac{2}{5} \), \( P(B) = \frac{3}{5} \), \( P(C) = -\frac{1}{5} \), \( P(D) = \frac{1}{5} \).
We know that the probability of any event must be between 0 and 1, inclusive.
Here, \( P(C) = -\frac{1}{5} \).
Since \( P(C) \) is a negative value, it violates the basic rule that a probability cannot be less than 0. Therefore, this assignment of probability is not permissible.
In simple words: For probabilities to be correct, each probability must be a number from 0 to 1, and all of them added together must make exactly 1. If any probability is negative, or if the total sum is more or less than 1, then the probabilities are not valid.

🎯 Exam Tip: Remember the two fundamental rules of probability: individual probabilities must be between 0 and 1, and their sum for all possible outcomes must equal 1. Always check both conditions.

Question 2. If two coins are tossed simultaneously, then find the probability of getting
(i) one head and one tail
Answer: When two coins are tossed at the same time, we need to list all the possible results. This set of all possible results is called the sample space.
The sample space \( S \) for tossing two coins is:
\( S = \{HH, HT, TH, TT\} \)
The total number of outcomes in the sample space, \( n(S) \), is 4.
(i) Let \( A \) be the event of getting one head and one tail.
The outcomes for event \( A \) are:
\( A = \{HT, TH\} \)
The number of outcomes for event \( A \), \( n(A) \), is 2.
The probability of event \( A \) is calculated as:
\( P(A) = \frac{n(A)}{n(S)} = \frac{2}{4} = \frac{1}{2} \)
(ii) atmost two tails
Answer:
(ii) Let \( B \) be the event of getting at most two tails. "At most two tails" means we can have zero tails, one tail, or two tails.
Outcomes with zero tails: \( HH \)
Outcomes with one tail: \( HT, TH \)
Outcomes with two tails: \( TT \)
So, the outcomes for event \( B \) are:
\( B = \{HH, HT, TH, TT\} \)
The number of outcomes for event \( B \), \( n(B) \), is 4.
The probability of event \( B \) is:
\( P(B) = \frac{n(B)}{n(S)} = \frac{4}{4} = 1 \)
In simple words: When you flip two coins, there are four possible outcomes. To find the chance of something happening, count how many ways that specific thing can happen and divide by the total number of all possible outcomes. For example, getting one head and one tail can happen in two ways out of four total ways. Getting "at most two tails" means getting zero, one, or two tails, which covers all possible outcomes, so the probability is 1.

🎯 Exam Tip: Always list the full sample space first to clearly see all possible outcomes. This helps avoid missing any specific events.

Question 3. Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that
(i) one is a mango and the other is an apple
Answer: First, let's find the total number of fruits. There are 5 mangoes and 4 apples, so the total number of fruits is \( 5 + 4 = 9 \).
We need to find the total number of ways to choose 2 fruits from these 9 fruits. This is our sample space, \( n(S) \).
\( n(S) = \text{C}(9, 2) = \frac{9 \times 8}{2 \times 1} = 36 \)
(i) Let \( A \) be the event that one fruit is a mango and the other is an apple.
To get one mango, we choose 1 from 5 mangoes: \( \text{C}(5, 1) = 5 \) ways.
To get one apple, we choose 1 from 4 apples: \( \text{C}(4, 1) = 4 \) ways.
The number of ways to get one mango and one apple, \( n(A) \), is:
\( n(A) = \text{C}(5, 1) \times \text{C}(4, 1) = 5 \times 4 = 20 \)
The probability of event \( A \) is:
\( P(A) = \frac{n(A)}{n(S)} = \frac{20}{36} = \frac{5}{9} \)
(ii) both are of the same variety.
Answer:
(ii) Let \( B \) be the event that both fruits are of the same variety. This means either both are mangoes OR both are apples.
Number of ways to choose 2 mangoes from 5: \( \text{C}(5, 2) = \frac{5 \times 4}{2 \times 1} = 10 \)
Number of ways to choose 2 apples from 4: \( \text{C}(4, 2) = \frac{4 \times 3}{2 \times 1} = 6 \)
Since choosing two mangoes and choosing two apples are mutually exclusive events (they cannot happen at the same time), we add their probabilities.
The total number of ways to choose two fruits of the same variety, \( n(B) \), is:
\( n(B) = 10 + 6 = 16 \)
The probability of event \( B \) is:
\( P(B) = \frac{n(B)}{n(S)} = \frac{16}{36} = \frac{4}{9} \)
In simple words: When picking items from a group, first figure out all possible ways to pick the items (this is the total sample space). Then, count how many of those ways match what you are looking for. Divide the second number by the first to get the probability. If there are different ways to get what you want (like two mangoes or two apples), add those ways together before dividing.

🎯 Exam Tip: For problems involving combinations (choosing items without order), always use the \( nCr \) formula. Remember to add probabilities for "OR" events if they are mutually exclusive.

Question 4. What is the chance that
(i) Non - leap year
Answer:
(i) A non-leap year has 365 days.
To find the number of weeks and remaining days, divide 365 by 7:
\( 365 \div 7 = 52 \) weeks and 1 day remaining.
This means there are 52 full Sundays in a non-leap year. We need to find the probability of the *remaining one day* being a Sunday.
The remaining one day can be any of the seven days of the week: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}.
So, the sample space for this remaining day, \( n(S) \), is 7.
Let \( A \) be the event that this remaining day is a Sunday.
There is only one Sunday in the list of possible single days. So, \( n(A) = 1 \).
The probability of getting 53 Sundays in a non-leap year is:
\( P(A) = \frac{n(A)}{n(S)} = \frac{1}{7} \)
(ii) Leap year should have fifty-three Sundays?
Answer:
(ii) A leap year has 366 days.
To find the number of weeks and remaining days, divide 366 by 7:
\( 366 \div 7 = 52 \) weeks and 2 days remaining.
This means there are 52 full Sundays in a leap year. We need to find the probability that *at least one of the remaining two days* is a Sunday to make 53 Sundays in total.
The remaining two days can be any of these combinations: {Saturday-Sunday, Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday}.
So, the sample space for these two remaining days, \( n(S) \), is 7 (listing the first day of each pair).
Let \( A \) be the event that these two remaining days include a Sunday.
The combinations that include Sunday are: {Saturday-Sunday, Sunday-Monday}.
So, the number of outcomes for event \( A \), \( n(A) \), is 2.
The probability of getting 53 Sundays in a leap year is:
\( P(A) = \frac{n(A)}{n(S)} = \frac{2}{7} \)
In simple words: To find the chance of having 53 Sundays, first see how many extra days are left after 52 full weeks. For a normal year, there's 1 extra day; for a leap year, there are 2 extra days. Then, count how many of these extra day arrangements include a Sunday. Divide that by the total possible extra day arrangements.

🎯 Exam Tip: Remember that a non-leap year has 365 days (52 weeks + 1 day), and a leap year has 366 days (52 weeks + 2 days). This is crucial for determining the sample space of the 'extra' days.

Question 5. Eight coins are tossed once, find the probability of getting
(i) exactly two tails
Answer: When eight coins are tossed simultaneously, each coin can land in 2 ways (Head or Tail).
The total number of possible outcomes, \( n(S) \), is \( 2^8 \).
\( n(S) = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 256 \)
(i) Let \( A \) be the event of getting exactly two tails.
To find the number of ways to get exactly two tails out of 8 tosses, we use combinations:
\( n(A) = \text{C}(8, 2) = \frac{8 \times 7}{2 \times 1} = 28 \)
The probability of getting exactly two tails is:
\( P(A) = \frac{n(A)}{n(S)} = \frac{28}{256} \)
We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4:
\( P(A) = \frac{28 \div 4}{256 \div 4} = \frac{7}{64} \)
(ii) at least two tails
Answer:
(ii) Let \( B \) be the event of getting at least two tails. "At least two tails" means getting 2, 3, 4, 5, 6, 7, or 8 tails.
It's often easier to calculate the complementary event: "not at least two tails," which means getting less than two tails (i.e., zero tails or one tail).
Number of ways to get zero tails: \( \text{C}(8, 0) = 1 \) (all heads)
Number of ways to get one tail: \( \text{C}(8, 1) = 8 \)
Number of ways to get less than two tails = \( \text{C}(8, 0) + \text{C}(8, 1) = 1 + 8 = 9 \)
The number of ways to get at least two tails, \( n(B) \), is the total outcomes minus the outcomes with less than two tails:
\( n(B) = n(S) - (\text{C}(8, 0) + \text{C}(8, 1)) = 256 - 9 = 247 \)
The probability of getting at least two tails is:
\( P(B) = \frac{n(B)}{n(S)} = \frac{247}{256} \)
(iii) at most two tails
Answer:
(iii) Let \( C \) be the event of getting at most two tails. "At most two tails" means getting zero tails, one tail, or two tails.
Number of ways to get zero tails: \( \text{C}(8, 0) = 1 \)
Number of ways to get one tail: \( \text{C}(8, 1) = 8 \)
Number of ways to get two tails: \( \text{C}(8, 2) = \frac{8 \times 7}{2 \times 1} = 28 \)
The total number of ways to get at most two tails, \( n(C) \), is:
\( n(C) = \text{C}(8, 0) + \text{C}(8, 1) + \text{C}(8, 2) = 1 + 8 + 28 = 37 \)
The probability of getting at most two tails is:
\( P(C) = \frac{n(C)}{n(S)} = \frac{37}{256} \)
In simple words: When flipping many coins, first find all possible results (like \( 2^8 \) for 8 coins). Then, use combinations to count how many of these results match your specific condition (like exactly two tails, or at least two tails, or at most two tails). Divide your count by the total to get the chance. Remember that "at least" means that number or more, and "at most" means that number or less.

🎯 Exam Tip: For "at least" type probability questions, it's often simpler to calculate the probability of the complementary event (the opposite) and subtract it from 1. For "at most," directly sum the probabilities of the specified number of events and all fewer events.

Question 6. An integer is chosen at random from the first 100 positive integers. What is the probability that the integer chosen is a prime or multiple of 8 ?
Answer: We are choosing an integer from the first 100 positive integers.
The sample space \( S \) is the set of integers from 1 to 100:
\( S = \{1, 2, 3, \ldots, 100\} \)
The total number of possible outcomes, \( n(S) \), is 100.
Let \( A \) be the event that the chosen integer is a prime number.
The prime numbers between 1 and 100 are:
\( A = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\} \)
The number of prime numbers, \( n(A) \), is 25.
Let \( B \) be the event that the chosen integer is a multiple of 8.
The multiples of 8 between 1 and 100 are:
\( B = \{8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\} \)
The number of multiples of 8, \( n(B) \), is 12.
We need to find the probability that the integer is a prime OR a multiple of 8, which is \( P(A \cup B) \).
First, check if events \( A \) and \( B \) have any common elements (i.e., if any prime number is also a multiple of 8).
Prime numbers are generally not multiples of 8 (except 2, but 2 is not a multiple of 8). So, there are no common elements between set A and set B. This means \( A \cap B = \Phi \) (empty set).
Since \( A \) and \( B \) are mutually exclusive events, the probability of their union is the sum of their individual probabilities:
\( P(A \cup B) = P(A) + P(B) \)
\( P(A) = \frac{n(A)}{n(S)} = \frac{25}{100} \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{12}{100} \)
\( P(A \cup B) = \frac{25}{100} + \frac{12}{100} = \frac{37}{100} \)
In simple words: To find the chance of picking a prime number or a multiple of 8 from 1 to 100, first list all the primes and then all the multiples of 8. Count how many of each there are. Since no prime number is also a multiple of 8, you can just add up the counts for primes and multiples of 8, and then divide by the total numbers (100).

🎯 Exam Tip: When finding the probability of "A or B," always first check if events A and B are mutually exclusive. If they are (no overlap), simply add their probabilities. If they are not, you must subtract the probability of their intersection.

Question 7. A bag contains 7 red 4 black balls, 3 balls are drawn at random. Find the probability that
(i) all are red
Answer: We have a bag with 7 red balls and 4 black balls.
Total number of balls in the bag is \( 7 + 4 = 11 \).
We are drawing 3 balls at random from these 11 balls. The total number of ways to choose 3 balls from 11 is our sample space, \( n(S) \).
\( n(S) = \text{C}(11, 3) = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 11 \times 5 \times 3 = 165 \)
(i) Let \( A \) be the event that all 3 balls drawn are red.
To choose 3 red balls, we must choose them from the 7 red balls available.
\( n(A) = \text{C}(7, 3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35 \)
The probability of event \( A \) is:
\( P(A) = \frac{n(A)}{n(S)} = \frac{35}{165} \)
We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 5:
\( P(A) = \frac{35 \div 5}{165 \div 5} = \frac{7}{33} \)
(ii) one red and 2 black.
Answer:
(ii) Let \( B \) be the event that one ball is red and two balls are black.
To choose 1 red ball, we select 1 from 7 red balls: \( \text{C}(7, 1) = 7 \) ways.
To choose 2 black balls, we select 2 from 4 black balls: \( \text{C}(4, 2) = \frac{4 \times 3}{2 \times 1} = 6 \) ways.
The number of ways to choose one red and two black balls, \( n(B) \), is:
\( n(B) = \text{C}(7, 1) \times \text{C}(4, 2) = 7 \times 6 = 42 \)
The probability of event \( B \) is:
\( P(B) = \frac{n(B)}{n(S)} = \frac{42}{165} \)
We can simplify this fraction by dividing both numerator and denominator by their greatest common divisor, which is 3:
\( P(B) = \frac{42 \div 3}{165 \div 3} = \frac{14}{55} \)
In simple words: When picking a few items from a larger mixed group, first find out how many different ways you can pick *any* items (this is the total possible results). Then, for each specific condition, calculate how many ways that condition can happen. Divide the specific ways by the total ways to get the probability. Always simplify your fractions if possible.

🎯 Exam Tip: For problems with multiple selection criteria (e.g., "1 red AND 2 black"), remember to multiply the combinations for each part. Always simplify your final probability fraction.

Question 8. A single card is drawn from a pack of 52 cards. What is the probability that
(i) the card is an ace or a king
Answer: A standard pack of playing cards has 52 cards.
When a single card is drawn, the total number of possible outcomes, \( n(S) \), is 52.
(i) Let \( A \) be the event that the card drawn is an ace.
There are 4 aces in a pack of 52 cards.
\( n(A) = \text{C}(4, 1) = 4 \)
Let \( B \) be the event that the card drawn is a king.
There are 4 kings in a pack of 52 cards.
\( n(B) = \text{C}(4, 1) = 4 \)
Since a card cannot be both an ace and a king at the same time, events \( A \) and \( B \) are mutually exclusive.
The probability of getting an ace or a king is \( P(A \cup B) = P(A) + P(B) \).
\( P(A) = \frac{n(A)}{n(S)} = \frac{4}{52} \)
\( P(B) = \frac{n(B)}{n(S)} = \frac{4}{52} \)
\( P(A \cup B) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} \)
Simplify the fraction by dividing both by 4:
\( P(A \cup B) = \frac{8 \div 4}{52 \div 4} = \frac{2}{13} \)
(ii) the card will be 6 or smaller
Answer:
(ii) Let \( A \) be the event that the card drawn is a 6.
There are 4 cards of rank 6 (one for each suit).
\( n(A) = \text{C}(4, 1) = 4 \)
Let \( B \) be the event that the card drawn is smaller than 6. This means the card can be an Ace (which is often considered 1), 2, 3, 4, or 5.
Each of these ranks (Ace, 2, 3, 4, 5) has 4 cards. So, the total number of cards smaller than 6 is \( 5 \times 4 = 20 \).
\( n(B) = \text{C}(20, 1) = 20 \)
Since a card cannot be both a 6 and smaller than 6 at the same time, events \( A \) and \( B \) are mutually exclusive.
The probability of getting a card that is 6 or smaller is \( P(A \cup B) = P(A) + P(B) \).
\( P(A) = \frac{4}{52} \)
\( P(B) = \frac{20}{52} \)
\( P(A \cup B) = \frac{4}{52} + \frac{20}{52} = \frac{24}{52} \)
Simplify the fraction by dividing both by 4:
\( P(A \cup B) = \frac{24 \div 4}{52 \div 4} = \frac{6}{13} \)
(iii) the card is either a queen or 9?
Answer:
(iii) Let \( A \) be the event that the card drawn is a Queen.
There are 4 Queens in a pack of 52 cards.
\( n(A) = \text{C}(4, 1) = 4 \)
Let \( B \) be the event that the card drawn is a 9.
There are 4 nines in a pack of 52 cards.
\( n(B) = \text{C}(4, 1) = 4 \)
Since a card cannot be both a Queen and a 9 at the same time, events \( A \) and \( B \) are mutually exclusive.
The probability of getting a Queen or a 9 is \( P(A \cup B) = P(A) + P(B) \).
\( P(A) = \frac{4}{52} \)
\( P(B) = \frac{4}{52} \)
\( P(A \cup B) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} \)
Simplify the fraction by dividing both by 4:
\( P(A \cup B) = \frac{8 \div 4}{52 \div 4} = \frac{2}{13} \)
In simple words: When drawing one card from a deck, the total number of possibilities is 52. To find the chance of getting one type of card OR another (like an Ace or a King), count how many of each type there are and add those counts together. Then, divide this sum by 52. Make sure the two types of cards can't be the same card.

🎯 Exam Tip: For card problems, quickly recall the number of cards in a suit (13), number of suits (4), and the count of specific cards (e.g., 4 Aces, 4 Kings). Always simplify fractions to their lowest terms.

Question 9. A cricket club has 16 members, of whom only 5 can bowl. What is the probability that in a team of 11 members at least 3 bowlers are selected?
Answer: Total members in the cricket club = 16.
Number of members who can bowl (bowlers) = 5.
Number of members who cannot bowl (batters) = \( 16 - 5 = 11 \).
A team of 11 members is to be selected.
The total number of ways to select 11 members from 16 is our sample space, \( n(S) \).
\( n(S) = \text{C}(16, 11) = \text{C}(16, 16-11) = \text{C}(16, 5) \)
\( n(S) = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} = 16 \times 3 \times 7 \times 13 = 4368 \)
We need to find the probability that at least 3 bowlers are selected. "At least 3 bowlers" means the team can have 3 bowlers, 4 bowlers, or 5 bowlers (since there are only 5 bowlers available in total).
This involves three mutually exclusive cases:
**Case 1: 3 bowlers and 8 batters**
Number of ways = \( \text{C}(5, 3) \times \text{C}(11, 8) \)
\( \text{C}(5, 3) = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \)
\( \text{C}(11, 8) = \text{C}(11, 3) = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 11 \times 5 \times 3 = 165 \)
Ways for Case 1 = \( 10 \times 165 = 1650 \)
**Case 2: 4 bowlers and 7 batters**
Number of ways = \( \text{C}(5, 4) \times \text{C}(11, 7) \)
\( \text{C}(5, 4) = \text{C}(5, 1) = 5 \)
\( \text{C}(11, 7) = \text{C}(11, 4) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 11 \times 10 \times 3 = 330 \)
Ways for Case 2 = \( 5 \times 330 = 1650 \)
**Case 3: 5 bowlers and 6 batters**
Number of ways = \( \text{C}(5, 5) \times \text{C}(11, 6) \)
\( \text{C}(5, 5) = 1 \)
\( \text{C}(11, 6) = \text{C}(11, 5) = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 11 \times 2 \times 3 \times 7 = 462 \)
Ways for Case 3 = \( 1 \times 462 = 462 \)
Total number of ways to select at least 3 bowlers, \( n(E) \), is the sum of ways from these three cases:
\( n(E) = 1650 + 1650 + 462 = 3762 \)
The probability is:
\( P(E) = \frac{n(E)}{n(S)} = \frac{3762}{4368} \)
To simplify, we can divide by common factors. Both are divisible by 6:
\( \frac{3762 \div 6}{4368 \div 6} = \frac{627}{728} \)
In simple words: To pick a team with "at least 3 bowlers," you need to find all the ways you can pick 3, 4, or 5 bowlers along with enough other players to make a team of 11. Calculate the possibilities for each case separately using combinations, then add them up. Divide this total by the grand total ways to pick any 11 players from 16 to get the final chance.

🎯 Exam Tip: When dealing with "at least" or "at most" scenarios in combinations, remember to consider all possible cases that satisfy the condition and sum their individual combinations. Double-check your combinations calculations, especially for larger numbers.

Question 10. (i) The odds that event A occurs is 5 to 7, find P(A).
Answer:
(i) The odds that event A occurs are given as 5 to 7. This means the ratio of the probability of A happening, \( P(A) \), to the probability of A not happening, \( P(A') \), is 5:7.
So, \( \frac{P(A)}{P(A')} = \frac{5}{7} \)
We know that \( P(A') = 1 - P(A) \). Substitute this into the equation:
\( \frac{P(A)}{1 - P(A)} = \frac{5}{7} \)
Now, cross-multiply to solve for \( P(A) \):
\( 7 \times P(A) = 5 \times (1 - P(A)) \)
\( 7 P(A) = 5 - 5 P(A) \)
Add \( 5 P(A) \) to both sides:
\( 7 P(A) + 5 P(A) = 5 \)
\( 12 P(A) = 5 \)
Divide by 12:
\( P(A) = \frac{5}{12} \)
(ii) Suppose p (B) = \( \frac{2}{5} \). Express the odds that the event B occurs.
Answer:
(ii) We are given the probability of event B as \( P(B) = \frac{2}{5} \).
To express the odds that event B occurs, we need the ratio of \( P(B) \) to \( P(B') \).
First, find the probability that event B does not occur, \( P(B') \):
\( P(B') = 1 - P(B) = 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5} \)
Now, express the odds for event B, which is \( P(B) : P(B') \):
Odds for B = \( \frac{P(B)}{P(B')} = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{3} \)
So, the odds that event B occurs are 2 to 3.
In simple words: "Odds 5 to 7" means for every 5 times something happens, it doesn't happen 7 times. To find the probability from odds, add the two numbers (5+7=12), then the probability of it happening is the first number divided by the total (5/12). If you have the probability and want to find the odds, first find the chance of it *not* happening, then write the ratio of the chance it happens to the chance it doesn't.

🎯 Exam Tip: Remember the clear distinction between odds and probability: probability is favorable outcomes divided by total outcomes, while odds are favorable outcomes compared to unfavorable outcomes. \( P(E) = \frac{\text{odds in favor}}{(\text{odds in favor} + \text{odds against})} \).

Chapter 1

Question 1. An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible.
(i) P (A) = 0.15, P (B) = 0.30, P (C) = 0.43, P (D) = 0.12
(ii) P (A) = 0.22, P (B) = 0.38, P (C) = 0.16, P (D) = 0.34
(iii) P(A) = \( \frac{2}{5} \), P(B) = \( \frac{3}{5} \), P(C) = \( -\frac{1}{5} \), P(D) = \( \frac{1}{5} \)
Answer:
(i) For probabilities to be permissible, two conditions must be met: each probability must be between 0 and 1 (inclusive), and the sum of all probabilities must be 1. Let's check these conditions for the given assignment.
We have: \( P(A) = 0.15 \), \( P(B) = 0.30 \), \( P(C) = 0.43 \), \( P(D) = 0.12 \).
Each of these probabilities is between 0 and 1.
Now, let's sum them up:
\( P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 = 1 \).
Since the sum is 1, and all individual probabilities are valid, this assignment is permissible. This means that these probabilities correctly describe a possible outcome for the experiment.
In simple words: All probabilities are positive, and they add up to exactly 1. So, this set of probabilities is valid.

(ii) For this assignment, we have: \( P(A) = 0.22 \), \( P(B) = 0.38 \), \( P(C) = 0.16 \), \( P(D) = 0.34 \).
All individual probabilities are between 0 and 1, which is a good start.
Now, let's find their sum:
\( P(S) = P(A) + P(B) + P(C) + P(D) = 0.22 + 0.38 + 0.16 + 0.34 = 1.1 \).
Since the sum of the probabilities is \( 1.1 \), which is greater than 1, this assignment is not permissible. The total probability for all possible outcomes must always be 1. A sum greater than 1 suggests an error or an impossible scenario.
In simple words: The probabilities add up to more than 1, so this set of probabilities is not valid.

(iii) For this assignment, we have: \( P(A) = \frac{2}{5} \), \( P(B) = \frac{3}{5} \), \( P(C) = -\frac{1}{5} \), \( P(D) = \frac{1}{5} \).
A fundamental rule of probability is that the probability of any event must be between 0 and 1 (inclusive).
Here, \( P(C) = -\frac{1}{5} \). This is a negative value.
Probabilities cannot be negative. Therefore, this assignment is not permissible. A probability of -1/5 is not possible in real-world events.
In simple words: One of the probabilities is a negative number, which is not allowed. So, this set of probabilities is not valid.

🎯 Exam Tip: Remember the two key rules for permissible probabilities: each probability \( P(E) \) must satisfy \( 0 \le P(E) \le 1 \), and the sum of probabilities for all possible outcomes must be exactly 1.

Question 2. If two coins are tossed simultaneously, then find the probability of getting
(i) one head and one tail
(ii) atmost two tails
Answer:
When two coins are tossed at the same time, the possible outcomes form the sample space \( S \).
The sample space is \( S = \{HH, HT, TH, TT\} \).
The total number of outcomes, \( n(S) \), is 4.

(i) Let A be the event of getting one head and one tail. The outcomes for this event are \( A = \{HT, TH\} \).
The number of outcomes for event A, \( n(A) \), is 2.
The probability of event A, \( P(A) \), is given by the formula \( P(A) = \frac{n(A)}{n(S)} \).
\( P(A) = \frac{2}{4} = \frac{1}{2} \).
This means there's a 50% chance of getting one head and one tail. This is a common probability outcome for coin tosses.
In simple words: There are 4 possible results when you toss two coins. Two of these results give one head and one tail. So, the chance is 2 out of 4, which simplifies to 1/2.

(ii) Let B be the event of getting at most two tails. "At most two tails" means 0 tails, 1 tail, or 2 tails.
From our sample space \( S = \{HH, HT, TH, TT\} \):
- HH has 0 tails.
- HT has 1 tail.
- TH has 1 tail.
- TT has 2 tails.
So, all possible outcomes satisfy the condition of having at most two tails. Therefore, the event B includes all outcomes: \( B = \{HH, HT, TH, TT\} \).
The number of outcomes for event B, \( n(B) \), is 4.
The probability of event B, \( P(B) \), is:
\( P(B) = \frac{n(B)}{n(S)} = \frac{4}{4} = 1 \).
A probability of 1 means the event is certain to happen. This is expected since it's impossible to get more than two tails when tossing only two coins.
In simple words: "At most two tails" means you can have zero, one, or two tails. All the ways two coins can land fit this. So, the chance is 4 out of 4, which is 1 (certain).

🎯 Exam Tip: When calculating probabilities, always start by listing the complete sample space, then identify the specific outcomes for the event, and finally apply the formula \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

Question 3. Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that
(i) one is a mango and the other is an apple
(ii) both are of the same variety.
Answer:
Total number of fruits in the box is \( 5 \text{ (mangoes)} + 4 \text{ (apples)} = 9 \) fruits.
We are choosing 2 fruits at random from these 9 fruits. The total number of ways to do this is given by combinations, \( n(S) = \text{9C2} \).
\( n(S) = \frac{9 \times 8}{1 \times 2} = 9 \times 4 = 36 \).
So, there are 36 possible ways to choose 2 fruits from the box. This represents all the different pairs of fruits we could pick.

(i) Let A be the event that one fruit is a mango and the other is an apple.
To choose 1 mango from 5 mangoes, the number of ways is \( \text{5C1} = 5 \).
To choose 1 apple from 4 apples, the number of ways is \( \text{4C1} = 4 \).
To get one mango AND one apple, we multiply these possibilities:
\( n(A) = \text{5C1} \times \text{4C1} = 5 \times 4 = 20 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{20}{36} \).
Simplifying the fraction, we divide both numerator and denominator by 4:
\( P(A) = \frac{20 \div 4}{36 \div 4} = \frac{5}{9} \).
This means there's a good chance (more than 50%) that you'll pick one of each type of fruit. Probability involves understanding all possible outcomes and the specific outcomes that meet the criteria.
In simple words: There are 36 ways to pick any two fruits. There are 20 ways to pick one mango and one apple. So, the chance is 20 out of 36, or 5/9.

(ii) Let B be the event that both fruits are of the same variety.
This means either both are mangoes OR both are apples.
Number of ways to choose 2 mangoes from 5 mangoes: \( \text{5C2} = \frac{5 \times 4}{1 \times 2} = 10 \).
Number of ways to choose 2 apples from 4 apples: \( \text{4C2} = \frac{4 \times 3}{1 \times 2} = 6 \).
Since these are mutually exclusive events (you can't pick two mangoes AND two apples at the same time), we add their possibilities:
\( n(B) = \text{5C2} + \text{4C2} = 10 + 6 = 16 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{16}{36} \).
Simplifying the fraction, we divide both numerator and denominator by 4:
\( P(B) = \frac{16 \div 4}{36 \div 4} = \frac{4}{9} \).
This implies a slightly lower chance of picking two fruits of the same type compared to picking one of each. Understanding how combinations work is key to solving such problems.
In simple words: There are 10 ways to pick two mangoes and 6 ways to pick two apples. So, there are 16 ways to pick two fruits of the same kind. The chance is 16 out of 36, or 4/9.

🎯 Exam Tip: When dealing with "and" conditions for combinations, you multiply the number of ways. For "or" conditions (mutually exclusive events), you add the number of ways. Always simplify your final probability fractions.

Question 4. What is the chance that
(i) Non - leap year
(ii) Leap year should have fifty-three Sundays?
Answer:
(i) For a non-leap year, there are 365 days. To find the number of weeks, we divide 365 by 7:
\( 365 \text{ days} = \frac{365}{7} \text{ weeks} = 52 \text{ weeks and 1 day} \).
A non-leap year always has 52 full weeks, which means it has 52 Sundays. The remaining 1 day can be any one of the seven days of the week: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday.
So, the sample space for this extra day, \( n(S) \), is 7 (each day is equally likely).
For the year to have 53 Sundays, this extra day must be a Sunday. Let A be the event that the extra day is a Sunday. The number of outcomes for event A, \( n(A) \), is 1.
The probability of having 53 Sundays in a non-leap year is \( P(A) = \frac{n(A)}{n(S)} = \frac{1}{7} \).
This shows there's a 1 in 7 chance for the extra day to be a Sunday. This concept is fundamental to understanding calendar probabilities.
In simple words: A regular year has 52 full weeks plus one extra day. For it to have 53 Sundays, that extra day must be a Sunday. There are 7 choices for that extra day, so the chance is 1 out of 7.

(ii) For a leap year, there are 366 days.
\( 366 \text{ days} = \frac{366}{7} \text{ weeks} = 52 \text{ weeks and 2 days} \).
A leap year also has 52 full weeks, meaning 52 Sundays. The remaining 2 days can be any combination of two consecutive days of the week. The possible pairs for these 2 extra days are:
\( S = \{\text{Saturday-Sunday, Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday}\} \).
So, the total number of possible outcomes for these two days, \( n(S) \), is 7.
For the leap year to have 53 Sundays, one of these two extra days must be a Sunday. The events where Sunday is included are: {Saturday-Sunday, Sunday-Monday}.
Let A be the event that one of these extra two days is a Sunday. The number of outcomes for event A, \( n(A) \), is 2.
The probability of having 53 Sundays in a leap year is \( P(A) = \frac{n(A)}{n(S)} = \frac{2}{7} \).
This implies a higher chance of having 53 Sundays in a leap year compared to a non-leap year because there are two extra days, increasing the likelihood of one being a Sunday. This difference highlights the impact of an extra day in a leap year.
In simple words: A leap year has 52 full weeks plus two extra days. For it to have 53 Sundays, one of these two extra days must be a Sunday. There are 7 possible pairs for these two days, and 2 of those pairs include a Sunday. So, the chance is 2 out of 7.

🎯 Exam Tip: When dealing with days and weeks, remember that a non-leap year has 1 extra day and a leap year has 2 extra days after 52 full weeks. This small difference significantly impacts the probability calculations for an extra weekday.

Question 5. Eight coins are tossed once, find the probability of getting
(i) exactly two tails
(ii) at least two tails
(iii) at most two tails
Answer:
When eight coins are tossed simultaneously, each coin can land in 2 ways (Head or Tail). So, for 8 coins, the total number of possible outcomes in the sample space \( S \) is \( n(S) = 2^8 = 256 \). This is a large number of combinations.

(i) Let A be the event of getting exactly two tails.
This is a combination problem: we need to choose 2 positions out of 8 for the tails, and the rest will be heads. The number of ways to do this is \( \text{8C2} \).
\( n(A) = \text{8C2} = \frac{8 \times 7}{1 \times 2} = 4 \times 7 = 28 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{28}{256} \).
Simplifying the fraction by dividing both by 4:
\( P(A) = \frac{28 \div 4}{256 \div 4} = \frac{7}{64} \).
This shows the specific chance of getting exactly two tails among eight tosses. Combinations are crucial for such probability calculations.
In simple words: Out of 256 total results, there are 28 ways to get exactly two tails. So, the chance is 28 out of 256, which simplifies to 7/64.

(ii) Let B be the event of getting at least two tails.
"At least two tails" means 2 tails, or 3 tails, or ... up to 8 tails. This can be calculated directly by summing combinations (\( \text{8C2} + \text{8C3} + \dots + \text{8C8} \)), but it's easier to use the complement rule.
The complement of "at least two tails" is "less than two tails", which means 0 tails or 1 tail.
Number of ways to get 0 tails (all heads): \( \text{8C0} = 1 \). (This is one specific outcome: HHHHHHHH)
Number of ways to get 1 tail: \( \text{8C1} = 8 \). (This means one T and seven H's, like HTHHHHHH, etc.)
So, the number of outcomes with less than two tails is \( 1 + 8 = 9 \).
The number of outcomes for event B (at least two tails) is the total outcomes minus the outcomes with less than two tails:
\( n(B) = n(S) - (\text{Number of outcomes with 0 tails + Number of outcomes with 1 tail}) \)
\( n(B) = 256 - (1 + 8) = 256 - 9 = 247 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{247}{256} \).
This high probability indicates that getting two or more tails is very common when tossing eight coins. The complement rule simplifies calculations greatly in such cases.
In simple words: "At least two tails" means 2 tails or more. It's easier to find the opposite (0 tails or 1 tail) and subtract from the total. There are 9 ways to get 0 or 1 tail. So, 256 - 9 = 247 ways to get at least two tails. The chance is 247 out of 256.

(iii) Let C be the event of getting at most two tails.
"At most two tails" means 0 tails, 1 tail, or 2 tails.
Number of ways to get 0 tails: \( \text{8C0} = 1 \).
Number of ways to get 1 tail: \( \text{8C1} = 8 \).
Number of ways to get 2 tails: \( \text{8C2} = \frac{8 \times 7}{1 \times 2} = 28 \).
The total number of outcomes for event C is the sum of these:
\( n(C) = \text{8C0} + \text{8C1} + \text{8C2} = 1 + 8 + 28 = 37 \).
The probability of event C is \( P(C) = \frac{n(C)}{n(S)} = \frac{37}{256} \).
This shows the specific likelihood of getting up to two tails. Understanding what "at most" means is key here.
In simple words: "At most two tails" means 0 tails, 1 tail, or 2 tails. There's 1 way for 0 tails, 8 ways for 1 tail, and 28 ways for 2 tails. Adding them gives 37 ways. So, the chance is 37 out of 256.

🎯 Exam Tip: Be precise with the terms "exactly," "at least," and "at most." "Exactly X" means only X. "At least X" means X or more. "At most X" means X or less. For "at least" problems, consider using the complement rule if it simplifies calculations.

Question 6. An integer is chosen at random from the first 100 positive integers. What is the probability that the integer chosen is a prime or multiple of 8 ?
Answer:
The sample space \( S \) consists of the first 100 positive integers:
\( S = \{1, 2, 3, \dots, 100\} \).
The total number of outcomes, \( n(S) \), is 100.

Let A be the event of choosing a prime number from these integers.
The prime numbers between 1 and 100 are:
\( A = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\} \).
Counting these, we find that the number of prime numbers, \( n(A) \), is 25.
The probability of choosing a prime number is \( P(A) = \frac{n(A)}{n(S)} = \frac{25}{100} = \frac{1}{4} \).

Let B be the event of choosing an integer that is a multiple of 8 from these integers.
The multiples of 8 between 1 and 100 are:
\( B = \{8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96\} \).
Counting these, we find that the number of multiples of 8, \( n(B) \), is 12.
The probability of choosing a multiple of 8 is \( P(B) = \frac{n(B)}{n(S)} = \frac{12}{100} \).

We need to find the probability that the chosen integer is a prime OR a multiple of 8, which is \( P(A \text{ or } B) = P(A \cup B) \).
First, we check if events A and B are mutually exclusive (i.e., if there's any overlap). Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. Multiples of 8 (except 8 itself) are composite numbers. 8 is not a prime number. Since the only even prime number is 2, and 2 is not a multiple of 8, there is no number that is both prime and a multiple of 8. Thus, \( A \cap B = \emptyset \) (empty set), meaning the events are mutually exclusive.
For mutually exclusive events, \( P(A \cup B) = P(A) + P(B) \).
\( P(A \cup B) = \frac{25}{100} + \frac{12}{100} \).
\( P(A \cup B) = \frac{25 + 12}{100} = \frac{37}{100} \).
This means there's a 37% chance of picking a number that is either prime or a multiple of 8. It's important to identify whether events overlap or not. The concept of mutual exclusivity is key here.
In simple words: There are 25 prime numbers and 12 multiples of 8 up to 100. No number is both prime and a multiple of 8. So, we just add the number of primes and multiples of 8 (25 + 12 = 37) and divide by the total numbers (100). The chance is 37 out of 100.

🎯 Exam Tip: When calculating probabilities for "A or B," always first determine if events A and B are mutually exclusive. If they are, simply add their individual probabilities. If they are not, you must subtract the probability of their intersection.

Question 7. A bag contains 7 red 4 black balls, 3 balls are drawn at random. Find the probability that
(i) all are red
(ii) one red and 2 black.
Answer:
Total number of balls in the bag is \( 7 \text{ red} + 4 \text{ black} = 11 \) balls.
We are drawing 3 balls at random from these 11 balls. The total number of ways to do this is \( n(S) = \text{11C3} \).
\( n(S) = \frac{11 \times 10 \times 9}{1 \times 2 \times 3} = 11 \times 5 \times 3 = 165 \).
So, there are 165 different combinations of 3 balls that can be drawn from the bag. This forms the basis for our probability calculations.

(i) Let A be the event that all 3 balls drawn are red.
To draw 3 red balls from 7 red balls, the number of ways is \( \text{7C3} \).
\( n(A) = \text{7C3} = \frac{7 \times 6 \times 5}{1 \times 2 \times 3} = 7 \times 5 = 35 \).
The probability of event A is \( P(A) = \frac{n(A)}{n(S)} = \frac{35}{165} \).
Simplifying the fraction by dividing both numerator and denominator by 5:
\( P(A) = \frac{35 \div 5}{165 \div 5} = \frac{7}{33} \).
This means there's a certain chance of drawing only red balls, showing the impact of having more red balls in the bag. Using combinations is essential for solving such problems.
In simple words: There are 165 ways to pick any 3 balls. There are 35 ways to pick 3 red balls. So, the chance is 35 out of 165, which simplifies to 7/33.

(ii) Let B be the event that one ball is red and 2 balls are black.
To draw 1 red ball from 7 red balls, the number of ways is \( \text{7C1} = 7 \).
To draw 2 black balls from 4 black balls, the number of ways is \( \text{4C2} = \frac{4 \times 3}{1 \times 2} = 6 \).
To get one red AND two black balls, we multiply these possibilities:
\( n(B) = \text{7C1} \times \text{4C2} = 7 \times 6 = 42 \).
The probability of event B is \( P(B) = \frac{n(B)}{n(S)} = \frac{42}{165} \).
Simplifying the fraction by dividing both numerator and denominator by 3:
\( P(B) = \frac{42 \div 3}{165 \div 3} = \frac{14}{55} \).
This shows the specific probability of drawing a mixed set of balls. Understanding how to combine different selection possibilities is crucial in probability.
In simple words: There are 7 ways to pick 1 red ball and 6 ways to pick 2 black balls. So, there are 7 x 6 = 42 ways to pick one red and two black balls. The chance is 42 out of 165, which simplifies to 14/55.

🎯 Exam Tip: Always calculate the total number of possible outcomes (sample space) first. Then, identify the number of favorable outcomes for each specific event using combinations, and finally, divide the favorable outcomes by the total outcomes to find the probability.

Question 8. A single card is drawn from a pack of 52 cards. What is the probability that
(i) the card is an ace or a king
(ii) the card will be 6 or smaller
(iii) the card is either a queen or 9?
Answer:
When a single card is drawn from a standard deck of 52 cards, the total number of possible outcomes, \( n(S) \), is 52. Each card has an equal chance of being drawn.

(i) Let A be the event of getting an ace. There are 4 aces in a deck, so \( n(A) = 4 \).
Let B be the event of getting a king. There are 4 kings in a deck, so \( n(B) = 4 \).
We need to find the probability that the card is an ace OR a king. Events A and B are mutually exclusive because a card cannot be both an ace and a king at the same time. Therefore, \( P(A \cup B) = P(A) + P(B) \).
\( P(A) = \frac{4}{52} \)
\( P(B) = \frac{4}{52} \)
\( P(A \cup B) = \frac{4}{52} + \frac{4}{52} = \frac{4+4}{52} = \frac{8}{52} \).
Simplifying the fraction by dividing both by 4:
\( P(A \cup B) = \frac{8 \div 4}{52 \div 4} = \frac{2}{13} \).
This shows there's a 2 in 13 chance of drawing either an ace or a king. Understanding mutually exclusive events simplifies such probability calculations.
In simple words: There are 4 aces and 4 kings in a deck of 52 cards. You can't draw both at once. So, there are 4 + 4 = 8 favorable cards. The chance is 8 out of 52, which is 2/13.

(ii) Let A be the event of getting a card with the number 6. There are 4 sixes in a deck, so \( n(A) = 4 \).
Let B be the event of getting a card with a number smaller than 6. These are cards 2, 3, 4, 5. For each number, there are 4 suits, so \( 4 \times 4 = 16 \) cards.
So, \( n(B) = 16 \).
We need the probability that the card is 6 OR smaller than 6. These events are also mutually exclusive. The cards are 2, 3, 4, 5, 6.
\( P(A \cup B) = P(A) + P(B) = \frac{4}{52} + \frac{16}{52} \).
\( P(A \cup B) = \frac{4+16}{52} = \frac{20}{52} \).
Simplifying the fraction by dividing both by 4:
\( P(A \cup B) = \frac{20 \div 4}{52 \div 4} = \frac{5}{13} \).
This indicates a 5 in 13 chance of drawing a card that is a 6 or a lower number. Knowing how to identify and count cards in different categories is key here.
In simple words: There are 4 cards with the number 6. There are 16 cards (2, 3, 4, 5) that are smaller than 6. These are 4+16 = 20 favorable cards. The chance is 20 out of 52, which simplifies to 5/13.

(iii) Let A be the event of getting a queen. There are 4 queens in a deck, so \( n(A) = 4 \).
Let B be the event of getting a 9. There are 4 nines in a deck, so \( n(B) = 4 \).
We need the probability that the card is either a queen OR a 9. These events are mutually exclusive.
\( P(A \cup B) = P(A) + P(B) = \frac{4}{52} + \frac{4}{52} \).
\( P(A \cup B) = \frac{4+4}{52} = \frac{8}{52} \).
Simplifying the fraction by dividing both by 4:
\( P(A \cup B) = \frac{8 \div 4}{52 \div 4} = \frac{2}{13} \).
This calculation is similar to the first part, reinforcing the concept of mutually exclusive events in card probability. The probability is the same as drawing an ace or a king.
In simple words: There are 4 queens and 4 nines in a deck of 52 cards. These are 4+4 = 8 favorable cards. The chance is 8 out of 52, which simplifies to 2/13.

🎯 Exam Tip: When calculating probabilities with cards, always remember the composition of a standard 52-card deck: 4 suits (clubs, diamonds, hearts, spades) and 13 ranks (A, 2-10, J, Q, K) per suit. Clearly define your events and check for mutual exclusivity.

Question 9. A cricket club has 16 members, of whom only 5 can bowl. What is the probability that in a team of 11 members at least 3 bowlers are selected?
Answer:
Total number of members in the cricket club is 16.
Number of bowlers = 5.
Number of non-bowlers (batters/fielders) = \( 16 - 5 = 11 \).
We need to select a team of 11 members from the 16 available members. The total number of ways to do this is \( n(S) = \text{16C11} \).
\( \text{16C11} = \text{16C(16-11)} = \text{16C5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4 \times 5} \).
\( \text{16C5} = 16 \times 7 \times 13 \times \frac{12}{2 \times 3 \times 4 \times 5} \) (simplifying 15 with 3, 5; 16 with 4, 2; 14 with 2)
\( \text{16C5} = 16 \times 1 \times 7 \times 13 \times \frac{1}{1 \times 1 \times 1 \times 5} = 4368 \).
So, \( n(S) = 4368 \). There are 4368 ways to form an 11-member team.

We want the probability that at least 3 bowlers are selected. "At least 3 bowlers" means the team can have 3 bowlers, or 4 bowlers, or 5 bowlers (since there are only 5 bowlers available in total).
This involves three mutually exclusive cases:
**Case 1: Exactly 3 bowlers and 8 non-bowlers.**
Number of ways to choose 3 bowlers from 5: \( \text{5C3} = \frac{5 \times 4 \times 3}{1 \times 2 \times 3} = 10 \).
Number of ways to choose 8 non-bowlers from 11: \( \text{11C8} = \text{11C(11-8)} = \text{11C3} = \frac{11 \times 10 \times 9}{1 \times 2 \times 3} = 11 \times 5 \times 3 = 165 \).
Number of ways for Case 1: \( \text{5C3} \times \text{11C8} = 10 \times 165 = 1650 \).

**Case 2: Exactly 4 bowlers and 7 non-bowlers.**
Number of ways to choose 4 bowlers from 5: \( \text{5C4} = \text{5C1} = 5 \).
Number of ways to choose 7 non-bowlers from 11: \( \text{11C7} = \text{11C(11-7)} = \text{11C4} = \frac{11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4} = 11 \times 10 \times 3 = 330 \).
Number of ways for Case 2: \( \text{5C4} \times \text{11C7} = 5 \times 330 = 1650 \).

**Case 3: Exactly 5 bowlers and 6 non-bowlers.**
Number of ways to choose 5 bowlers from 5: \( \text{5C5} = 1 \).
Number of ways to choose 6 non-bowlers from 11: \( \text{11C6} = \text{11C(11-6)} = \text{11C5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4 \times 5} = 11 \times 2 \times 3 \times 7 = 462 \).
Number of ways for Case 3: \( \text{5C5} \times \text{11C6} = 1 \times 462 = 462 \).

The total number of favorable outcomes for "at least 3 bowlers" is the sum of these cases:
\( n(\text{at least 3 bowlers}) = 1650 + 1650 + 462 = 3762 \).

The probability of selecting at least 3 bowlers is:
\( P(\text{at least 3 bowlers}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3762}{4368} \).
To simplify the fraction, we can divide by common factors. Both are divisible by 6:
\( \frac{3762 \div 6}{4368 \div 6} = \frac{627}{728} \).
This demonstrates how to handle "at least" problems by breaking them into mutually exclusive sub-cases and summing their probabilities. Combinations are a key tool in this process.
In simple words: First, find all ways to pick 11 players from 16 (4368 ways). Then, count ways to pick 3, 4, or 5 bowlers and the rest non-bowlers. Add these ways (1650+1650+462 = 3762). Divide this by the total ways: 3762/4368, which simplifies to 627/728.

🎯 Exam Tip: For "at least" or "at most" problems involving combinations, clearly list all possible scenarios that satisfy the condition. Remember that \( \text{nCr} = \text{nC(n-r)} \), which can simplify calculations for large 'r' values.

Question 10.
(i) The odds that event A occurs is 5 to 7, find P(A).
(ii) Suppose P(B) = \( \frac{2}{5} \). Express the odds that the event B occurs.
Answer:
(i) When the odds that an event A occurs are given as \( m \) to \( n \), it means that for every \( m \) times event A happens, there are \( n \) times event A does not happen.
The probability of event A, \( P(A) \), is given by the formula: \( P(A) = \frac{m}{m+n} \).
Given odds for event A are 5 to 7.
Here, \( m = 5 \) and \( n = 7 \).
So, \( P(A) = \frac{5}{5+7} = \frac{5}{12} \).
This probability means that event A is expected to occur 5 out of every 12 times. Understanding the relationship between odds and probability is crucial.
In simple words: Odds 5 to 7 mean event A happens 5 times and does not happen 7 times. So, the total number of chances is 5 + 7 = 12. The probability of A is 5 out of 12.

(ii) Given the probability of event B is \( P(B) = \frac{2}{5} \).
We need to express the odds that event B occurs. The odds in favor of an event are given by the ratio \( P(B) : P(B') \), where \( P(B') \) is the probability that event B does not occur.
First, find \( P(B') \):
\( P(B') = 1 - P(B) \).
\( P(B') = 1 - \frac{2}{5} \).
To subtract, find a common denominator: \( 1 = \frac{5}{5} \).
\( P(B') = \frac{5}{5} - \frac{2}{5} = \frac{5-2}{5} = \frac{3}{5} \).
Now, the odds that event B occurs are \( P(B) : P(B') \).
Odds = \( \frac{2}{5} : \frac{3}{5} \).
To express this as a ratio of integers, we can multiply both sides by 5:
Odds = \( 2 : 3 \).
This means for every 2 times event B occurs, it does not occur 3 times. This conversion is a common task in probability.
In simple words: If the chance of B happening is 2/5, then the chance of B not happening is 1 - 2/5 = 3/5. The odds of B happening are 2/5 to 3/5, which simplifies to 2 to 3.

🎯 Exam Tip: Remember the clear distinction and conversion formulas: if odds are \( m:n \), then \( P(event) = \frac{m}{m+n} \). If \( P(event) = p \), then \( P(not\ event) = 1-p \), and odds are \( p : (1-p) \). Always simplify ratios to their smallest whole number form.