Questions and Answers Section

Question 1

Three coins are tossed once. Find the probability of getting (i) exactly two heads, (ii) at least two heads.

Solution: 1. Define the Sample Space (S):

The sample space is the set of all possible outcomes. When tossing three coins, the possibilities are:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.

The total number of outcomes is $n(S)=8$.

(i) Probability of getting exactly two heads:

• Identify Favorable Outcomes: The event E1 = 'getting exactly two heads' corresponds to the outcomes {HHT, HTH, THH}.
• Count Favorable Outcomes: The number of favorable outcomes is $n(E1) = 3$.
• Calculate Probability: $P(E1) = \frac{n(E1)}{n(S)} = \frac{3}{8}$.

(ii) Probability of getting at least two heads:

• Identify Favorable Outcomes: The event E2 = 'getting at least two heads' means getting either two heads OR three heads. The outcomes are {HHT, HTH, THH, HHH}.
• Count Favorable Outcomes: The number of favorable outcomes is $n(E2) = 4$.
• Calculate Probability: $P(E2) = \frac{n(E2)}{n(S)} = \frac{4}{8} = \frac{1}{2}$.

Answer: (i) 3/8, (ii) 1/2.

Question 2

A card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card will be (i) a diamond, (ii) not a diamond, (iii) a black card?

Solution:

The total number of possible outcomes (the sample space) is the total number of cards, so $n(S) = 52$.

(i) A diamond:

• There are 13 diamond cards in a deck.
• The number of favorable outcomes is $n(\text{Diamond}) = 13$.
• The probability is $P(\text{Diamond}) = \frac{13}{52} = \frac{1}{4}$.

(ii) Not a diamond:

• This is the complement of the event 'drawing a diamond'. We can use the complement rule: $P(A') = 1 - P(A)$.
• $P(\text{not a diamond}) = 1 - P(\text{Diamond}) = 1 - \frac{1}{4} = \frac{3}{4}$.
• Alternatively, there are $52 - 13 = 39$ cards that are not diamonds. The probability is $\frac{39}{52} = \frac{3}{4}$.

(iii) A black card:

• A deck has two black suits: 13 spades and 13 clubs.
• The total number of black cards is $13 + 13 = 26$.
• The number of favorable outcomes is $n(\text{Black}) = 26$.
• The probability is $P(\text{Black}) = \frac{26}{52} = \frac{1}{2}$.

Answer: (i) 1/4, (ii) 3/4, (iii) 1/2.

Question 3

Given $P(A) = 3/5$ and $P(B) = 1/5$. Find $P(A \text{ or } B)$, if A and B are mutually exclusive events.

Solution:

1. Understand the Terminology:

   • The event 'A or B' is the union of the two events, denoted as $A \cup B$.
   • Mutually exclusive events are events that cannot happen at the same time. This means their intersection is empty, and $P(A \cap B) = 0$.

2. Use the Addition Rule for Mutually Exclusive Events: The general addition rule is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Since the events are mutually exclusive, $P(A \cap B) = 0$, and the rule simplifies to: $P(A \cup B) = P(A) + P(B)$.

3. Calculate the Probability: $P(A \cup B) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$.

Answer: 4/5.

Question 4

A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that the letter is (i) a vowel, (ii) a consonant.

Solution: 1. Analyze the Sample Space: First, count the total number of letters in the word 'ASSASSINATION'. Total letters, $n(S) = 13$.

(i) Probability of choosing a vowel:

• Identify and Count Vowels: The vowels in the word are A, I, O.
  • 'A' appears 3 times.
  • 'I' appears 2 times.
  • 'O' appears 1 time.
• Total Favorable Outcomes: Number of vowels, $n(\text{Vowel}) = 3+2+1=6$.
• Calculate Probability: $P(\text{Vowel}) = \frac{n(\text{Vowel})}{n(S)} = \frac{6}{13}$.

(ii) Probability of choosing a consonant:

• Identify and Count Consonants: The consonants are S, N, T.

  • 'S' appears 4 times.
  • 'N' appears 2 times.
  • 'T' appears 1 time.

• Total Favorable Outcomes: Number of consonants, $n(\text{Consonant}) = 4+2+1=7$.

• Calculate Probability: $P(\text{Consonant}) = \frac{n(\text{Consonant})}{n(S)} = \frac{7}{13}$.

• Alternative Method (using complement): A letter is either a vowel or a consonant. So, $P(\text{Consonant}) = 1 - P(\text{Vowel}) = 1 - 6/13 = 7/13$.

Answer: (i) 6/13, (ii) 7/13.

Question 5

Two dice are thrown simultaneously. Find the probability of getting an even number as the sum.

Solution: 1. Determine the Sample Space: When two dice are thrown, each die has 6 possible outcomes. The total number of outcomes in the sample space is $n(S) = 6 \times 6 = 36$.

2. Identify the Favorable Events: An even sum can be obtained in two ways:

• Case 1: (Even on 1st die) + (Even on 2nd die)
  • Even numbers on a die are {2, 4, 6} (3 possibilities).
  • Number of ways = $3 \times 3 = 9$.
• Case 2: (Odd on 1st die) + (Odd on 2nd die)
  • Odd numbers on a die are {1, 3, 5} (3 possibilities).
  • Number of ways = $3 \times 3 = 9$.

3. Calculate Total Favorable Outcomes: The total number of ways to get an even sum is the sum of the ways from both cases. $n(E) = 9 + 9 = 18$.

4. Calculate the Probability: $P(\text{Even Sum}) = \frac{n(E)}{n(S)} = \frac{18}{36} = \frac{1}{2}$.

Answer: 1/2.

Question 6

If $P(A) = 0.54$, $P(B) = 0.69$ and $P(A \cap B) = 0.35$, find $P(A' \cap B')$.

Solution: 1. Understand the Goal:

We need to find $P(A' \cap B')$, which represents the probability that neither A nor B occurs.

2. Apply De Morgan's Law:

De Morgan's Law states that $A' \cap B' = (A \cup B)'$. This means the event 'neither A nor B occurs' is the complement of the event 'at least one of A or B occurs'.

So, we need to find $P((A \cup B)')$.

3. Use the Complement Rule: $$P((A \cup B)') = 1 - P(A \cup B)$$

4. Find $P(A \cup B)$ using the Addition Rule: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ $$P(A \cup B) = 0.54 + 0.69 - 0.35 = 1.23 - 0.35 = 0.88$$

5. Calculate the Final Probability: $P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.88 = 0.12$.

Answer: 0.12.

Question 7

A box contains 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Solution: Method 1: Direct Calculation

1. Total Outcomes: The total number of balls is $n(S) = 8+7+6 = 21$.
2. Favorable Outcomes: The event 'neither red nor green' is logically the same as the event 'the ball is blue'. The number of blue balls is $n(\text{Blue}) = 7$.
3. Calculate Probability: $P(\text{neither red nor green}) = P(\text{Blue}) = \frac{7}{21} = \frac{1}{3}$.

Method 2: Using Complements

1. Define Events: Let R be the event 'drawing a red ball' and G be 'drawing a green ball'. We want to find $P(R' \cap G')$.
2. Apply De Morgan's Law: $P(R' \cap G') = P((R \cup G)')$.
3. Use the Complement Rule: $P((R \cup G)') = 1 - P(R \cup G)$.
4. Find $P(R \cup G)$: Drawing a red ball and drawing a green ball are mutually exclusive events. $P(R) = 8/21$, $P(G) = 6/21$. $P(R \cup G) = P(R) + P(G) = \frac{8}{21} + \frac{6}{21} = \frac{14}{21} = \frac{2}{3}$.
5. Calculate Final Probability: $P(R' \cap G') = 1 - \frac{2}{3} = \frac{1}{3}$.

Answer: 1/3.

Question 8

Three students A, B and C are given a problem to solve. The probabilities of them solving it are 1/3, 1/4 and 1/5 respectively. What is the probability that the problem is solved?

Solution: 1. Identify the Strategy: The problem is solved if 'at least one' of them solves it. Calculating this directly would involve multiple cases (only A solves, only B solves, A and B solve, etc.). It is much easier to calculate the probability of the complementary event and subtract from 1.

• Event E = 'the problem is solved' (at least one solves it).
• Complement E' = 'the problem is not solved'.

2. Calculate the Probability of the Complement (E'): The problem is not solved only if A fails AND B fails AND C fails. Assuming the events are independent:

• $P(A' ) = 1 - P(A) = 1 - 1/3 = 2/3$.
• $P(B' ) = 1 - P(B) = 1 - 1/4 = 3/4$.
• $P(C' ) = 1 - P(C) = 1 - 1/5 = 4/5$.

Since the events are independent, the probability of all three failing is the product of their individual probabilities:

$P(E') = P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C')$

$P(E') = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{24}{60} = \frac{2}{5}$.

3. Calculate the Final Probability: The probability that the problem is solved is $P(E) = 1 - P(E')$.

$P(E) = 1 - \frac{2}{5} = \frac{3}{5}$.

Answer: 3/5.

Question 9

A leap year is selected at random. What is the probability that it will contain 53 Sundays?

Solution:

1. Analyze the Number of Days: A leap year has 366 days.

2. Find the Number of Complete Weeks: We divide the total days by 7 to see how many full weeks there are. $366 \div 7 = 52$ with a remainder of 2. This means a leap year consists of 52 full weeks and 2 extra days.

3. Determine the Condition for 53 Sundays: The 52 full weeks guarantee 52 Sundays. To have 53 Sundays in total, one of the two extra days must be a Sunday.

4. List the Sample Space for the Extra Days: The two extra days must be consecutive. The possible pairs are: $S = \{(\text{Sun, Mon}), (\text{Mon, Tue}), (\text{Tue, Wed}), (\text{Wed, Thu}), (\text{Thu, Fri}), (\text{Fri, Sat}), (\text{Sat, Sun})\}$. There are $n(S) = 7$ possible pairs.

5. Identify Favorable Outcomes: The pairs that contain a Sunday are: E = { (Sun, Mon), (Sat, Sun) }. The number of favorable outcomes is $n(E) = 2$.

6. Calculate the Probability: $P(53 \text{ Sundays}) = \frac{n(E)}{n(S)} = \frac{2}{7}$.

Answer: 2/7.

Question 10

From a group of 5 men and 4 women, a committee of 3 members is to be formed. Find the probability that the committee has 1 man and 2 women.

Solution: This is a problem of selection where the order does not matter, so we will use combinations.

1. Calculate the Total Number of Possible Committees (Sample Space):

• Total people = 5 men + 4 women = 9.
• We are selecting a committee of 3.
• Total number of ways, $n(S) = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$.

2. Calculate the Number of Favorable Outcomes:

• The desired committee has '1 man AND 2 women'.
• Number of ways to choose 1 man from 5: $\binom{5}{1} = 5$.
• Number of ways to choose 2 women from 4: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
• The total number of favorable ways is the product of these two: $$n(E) = \binom{5}{1} \times \binom{4}{2} = 5 \times 6 = 30$$

3. Calculate the Probability:

$P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{n(E)}{n(S)} = \frac{30}{84}$.

Simplify the fraction: $\frac{30}{84} = \frac{5 \times 6}{14 \times 6} = \frac{5}{14}$.

Answer: 5/14.

Question 11

In a single throw of three dice, find the probability of getting a total of 17.

Solution: 1. Determine the Sample Space: Each of the three dice has 6 possible outcomes. The total number of outcomes in the sample space is $n(S) = 6 \times 6 \times 6 = 216$.

2. Identify the Favorable Outcomes: We need to find the combinations of numbers on the three dice that sum to 17. The maximum sum is 18 (6,6,6), so 17 is possible. Let the outcomes on the three dice be $(d_1, d_2, d_3)$. The only way to get a sum of 17 is with the numbers 6, 6, and 5. The possible arrangements (permutations) of these numbers are: E = { (6, 6, 5), (6, 5, 6), (5, 6, 6) }.

3. Count Favorable Outcomes: There are $n(E) = 3$ favorable outcomes.

4. Calculate the Probability: $P(\text{Sum is 17}) = \frac{n(E)}{n(S)} = \frac{3}{216} = \frac{1}{72}$.

Answer: 1/72.

Question 12

If E and F are events such that $P(E) = 1/4$, $P(F) = 1/2$ and $P(E \text{ and } F) = 1/8$, find $P(E' \text{ and } F')$.

Solution: 1. Understand the Goal: We need to find $P(E' \cap F')$, which is the probability that neither E nor F occurs.

2. Apply De Morgan's Law: De Morgan's Law states that $E' \cap F' = (E \cup F)'$. This means we need to find the probability of the complement of the union of E and F.

3. Use the Complement Rule: $P((E \cup F)') = 1 - P(E \cup F)$.

4. Find $P(E \cup F)$ using the Addition Rule: $$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$ $$P(E \cup F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$$

To add these fractions, find a common denominator (8): $P(E \cup F) = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$.

5. Calculate the Final Probability: $P(E' \cap F') = 1 - P(E \cup F) = 1 - \frac{5}{8} = \frac{3}{8}$.

Answer: 3/8.

Question 13

What is the probability that a number selected from the numbers {1, 2, 3, …, 25} is a prime number, given that it is greater than 10?

Solution: This is a conditional probability problem. The phrase "given that it is greater than 10" reduces our sample space.

1. Define the Original Sample Space: $S = \{1, 2, ..., 25\}$ with $n(S) = 25$.

2. Define the Reduced Sample Space: The condition is that the number is greater than 10. Our new sample space, let's call it $S'$, includes only these numbers:

$S' = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25\}$.

The total number of outcomes in this reduced space is $n(S') = 15$.

3. Identify Favorable Outcomes within the Reduced Sample Space:

We need to find the prime numbers within the set $S'$.

The prime numbers are: {11, 13, 17, 19, 23}.

4. Count Favorable Outcomes:

There are 5 favorable outcomes.

5. Calculate the Probability:

The required probability is the ratio of favorable outcomes to the total outcomes within the reduced sample space.

Probability = $\frac{5}{15} = \frac{1}{3}$.

Answer: 1/3.

Question 14

A die is loaded in such a way that an even number is twice as likely to occur as an odd number. Find the probability of getting a number less than 4.

Solution: 1. Set up the Probabilities Algebraically:

Let the probability of getting any specific odd number {1, 3, 5} be $x$. So, $P(1)=P(3)=P(5)=x$.

We are told an even number is twice as likely. So, the probability of getting any specific even number {2, 4, 6} is $2x$. So, $P(2)=P(4)=P(6)=2x$.

2. Use the Axiom of Total Probability:

The sum of the probabilities of all possible outcomes in the sample space must be 1.

$P(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1$.

$x + 2x + x + 2x + x + 2x = 1$.

$9x = 1 \implies x = 1/9$.

3. Determine the Probabilities of Each Outcome:

• Probability of any odd number = $x = 1/9$.
• Probability of any even number = $2x = 2/9$.

4. Calculate the Probability of the Desired Event:

The event E = 'getting a number less than 4' corresponds to the outcomes {1, 2, 3}.

Since these are mutually exclusive outcomes, we can add their probabilities:

$P(E) = P(1) + P(2) + P(3)$.

$P(E) = (1/9) + (2/9) + (1/9) = 4/9$.

Answer: 4/9.

Question 15

Find the probability of getting a doublet in a single throw of two dice.

Solution: 1. Determine the Sample Space:

When two fair dice are rolled, the total number of possible outcomes is $n(S) = 6 \times 6 = 36$.

2. Identify Favorable Outcomes:

A 'doublet' means getting the same number on both dice.

The set of favorable outcomes, E, is: E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

3. Count Favorable Outcomes:

There are $n(E) = 6$ favorable outcomes.

4. Calculate the Probability:

$P(\text{Doublet}) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

Answer: 1/6.

Question 16

From a set of 100 cards numbered 1 to 100, one card is drawn at random. Find the probability that the number on the card is divisible by 6 or 8.

Solution: 1. Define Events and the Goal:

• Sample Space $S = \{1, 2, ..., 100\}$, so $n(S) = 100$.
• Event A = 'the number is divisible by 6'.
• Event B = 'the number is divisible by 8'.
• We want to find the probability of 'A or B', which is $P(A \cup B)$.

2. Use the General Addition Rule: Since a number can be divisible by both 6 and 8 (e.g., 24), the events are not mutually exclusive. We must use: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

3. Calculate the Probabilities of Each Part:

• P(A): The number of multiples of 6 up to 100 is $n(A) = \lfloor \frac{100}{6} \rfloor = 16$. So, $P(A) = 16/100$.

• P(B): The number of multiples of 8 up to 100 is $n(B) = \lfloor \frac{100}{8} \rfloor = 12$. So, $P(B) = 12/100$.

• P(A ∩ B): This is the event that the number is divisible by both 6 and 8.

  This means the number must be divisible by the Least Common Multiple (LCM) of 6 and 8.

  LCM(6,8) = 24.

  The number of multiples of 24 up to 100 is $n(A \cap B) = \lfloor \frac{100}{24} \rfloor = 4$.

  So, $P(A \cap B) = 4/100$.

4. Solve:

$P(A \cup B) = \frac{16}{100} + \frac{12}{100} - \frac{4}{100} = \frac{28 - 4}{100} = \frac{24}{100} = \frac{6}{25}$.

Answer: 6/25.

Question 17

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

Solution: 1. Translate the Information into Symbolic Form:

• Let E = 'passing English' and H = 'passing Hindi'.
• $P(E \cap H) = 0.5$ (passing in both).
• $P(E' \cap H') = 0.1$ (passing in neither).
• $P(E) = 0.75$.
• We need to find $P(H)$.

2. Use De Morgan's Law and the Complement Rule:

We know that $$P(E' \cap H') = P((E \cup H)')$$ Using the complement rule, $P((E \cup H)') = 1 - P(E \cup H)$.

So, $0.1 = 1 - P(E \cup H)$, which implies $P(E \cup H) = 1 - 0.1 = 0.9$.

3. Use the General Addition Rule:

Now we use the formula $$P(E \cup H) = P(E) + P(H) - P(E \cap H)$$ to find the missing value, $P(H)$.

$0.9 = 0.75 + P(H) - 0.5$.

$0.9 = 0.25 + P(H)$.

$P(H) = 0.9 - 0.25 = 0.65$.

Answer: The probability of passing the Hindi examination is 0.65.

Question 18

What is the probability of getting 53 Fridays in a non-leap year?

Solution:

1. Analyze the Number of Days: A non-leap year (an ordinary year) has 365 days.

2. Find the Number of Complete Weeks: $365 \div 7 = 52$ with a remainder of 1. This means a non-leap year consists of 52 full weeks and 1 extra day.

3. Determine the Condition for 53 Fridays: The 52 full weeks guarantee 52 Fridays. The 53rd Friday depends entirely on what this one extra day is.

4. List the Sample Space and Favorable Outcome:

   The sample space for this single extra day is {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}. There are 7 possible outcomes.

   The single favorable outcome for having a 53rd Friday is that the extra day is a 'Friday'.

5. Calculate the Probability: Probability = $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{1}{7}$.

Answer: 1/7.

Question 19

If four-digit numbers are formed using the digits 1, 3, 5, 7, 9 without repetition, what is the probability that the number is divisible by 5?

Solution: This is a problem involving permutations.

1. Calculate the Total Number of Possible Numbers (Sample Space): We are forming 4-digit numbers from 5 available digits without repetition.

• The first digit can be chosen in 5 ways.
• The second digit can be chosen in 4 remaining ways.
• The third digit in 3 ways.
• The fourth digit in 2 ways. Using the multiplication principle, the total number of outcomes is $n(S) = 5 \times 4 \times 3 \times 2 = 120$. (This is also $^5P_4$).

2. Calculate the Number of Favorable Outcomes: The event E is 'the number is divisible by 5'. For this to be true, the last digit (the units place) must be 5.

• Last digit: Must be 5 (1 choice).
• First digit: Can be chosen from the 4 remaining digits.
• Second digit: Can be chosen from the 3 remaining digits.
• Third digit: Can be chosen from the 2 remaining digits. The number of favorable outcomes is $n(E) = 4 \times 3 \times 2 \times 1 = 24$.

3. Calculate the Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{24}{120} = \frac{1}{5}$.

Answer: 1/5.

Question 20

Given two mutually exclusive events A and B such that $P(A) = 0.45$ and $P(B) = 0.35$. Find $P(A' \cap B')$.

Solution: 1. Identify the Goal:

We need to find $P(A' \cap B')$, which is the probability that neither A nor B occurs.

2. Apply De Morgan's Law:

$P(A' \cap B') = P((A \cup B)')$.

3. Use the Complement Rule:

$P((A \cup B)') = 1 - P(A \cup B)$.

4. Calculate $P(A \cup B)$:

We are given that the events A and B are mutually exclusive. Therefore, we use the simplified addition rule:

$P(A \cup B) = P(A) + P(B)$.

$P(A \cup B) = 0.45 + 0.35 = 0.80$.

5. Calculate the Final Probability:

$P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.80 = 0.20$.

Answer: 0.20.

Question 21

Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that you both enter the same section?

Solution: The event 'you both enter the same section' can be broken down into two mutually exclusive cases:

• Case 1: You both enter the section of 40.
• Case 2: You both enter the section of 60.

The total probability will be the sum of the probabilities of these two cases.

Case 1: Both in the section of 40 (let's call it Section A)

• The probability that you are in Section A is $\frac{40}{100}$.
• Given that you are in Section A, there are now 39 spots left in the section and 99 students remaining.
• The probability that your friend is also in Section A is $\frac{39}{99}$.
• $P(\text{Both in A}) = \frac{40}{100} \times \frac{39}{99}$.

Case 2: Both in the section of 60 (let's call it Section B)

• The probability that you are in Section B is $\frac{60}{100}$.
• Given that you are in it, there are 59 spots left and 99 students remaining.
• The probability that your friend is also in Section B is $\frac{59}{99}$.
• $P(\text{Both in B}) = \frac{60}{100} \times \frac{59}{99}$.

Total Probability:

$P(\text{Same Section}) = P(\text{Both in A}) + P(\text{Both in B})$

$P(\text{Same Section}) = \frac{40 \times 39}{100 \times 99} + \frac{60 \times 59}{100 \times 99} = \frac{1560 + 3540}{9900} = \frac{5100}{9900} = \frac{51}{99} = \frac{17}{33}$.

Answer: 17/33.

Question 22

Two cards are drawn from a pack of 52 cards. Find the probability that both cards are aces.

Solution:

We are selecting 2 items from a larger set, so we use combinations.

1. Calculate the Total Number of Outcomes (Sample Space):

Total ways to draw any 2 cards from 52 is:

$n(S) = \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$.

2. Calculate the Number of Favorable Outcomes:

The event E is 'drawing 2 aces'. There are 4 aces in the deck. The number of ways to draw 2 aces from the 4 available is:

$n(E) = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.

3. Calculate the Probability:

$P(E) = \frac{n(E)}{n(S)} = \frac{6}{1326} = \frac{1}{221}$.

Answer: 1/221.

Question 23

A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability of getting at least one tail.

Solution: 1. Determine the Probabilities for a Single Toss:

• Let $P(T) = x$. Then $P(H) = 3x$.
• The sum of probabilities for all outcomes must be 1: $P(H) + P(T) = 1$.
• $3x + x = 1 \implies 4x=1 \implies x=1/4$.
• So, for a single toss: $P(T)=1/4$ and $P(H)=3/4$.

2. Use the Complement Strategy: It's easier to find the probability of the complement event (E') and subtract from 1.

• Event E = 'at least one tail'.
• Complement E' = 'no tails', which means 'both tosses are heads' (HH).

3. Calculate the Probability of the Complement:

Since the tosses are independent, we can multiply their probabilities:

$P(E') = P(HH) = P(H) \times P(H) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$.

4. Calculate the Final Probability:

$P(E) = 1 - P(E') = 1 - \frac{9}{16} = \frac{7}{16}$.

Answer: 7/16.

Question 24

What is the probability that an ordinary year has 53 Mondays?

Solution:

1. Analyze the Number of Days: An ordinary (non-leap) year has 365 days.

2. Find the Number of Complete Weeks: $365 \div 7 = 52$ with a remainder of 1. This means a non-leap year consists of 52 full weeks and 1 extra day.

3. Determine the Condition for 53 Mondays: The 52 full weeks provide 52 Mondays. The 53rd Monday depends entirely on what this one extra day is.

4. List the Sample Space and Favorable Outcome: The sample space for the extra day is {Sun, Mon, Tue, Wed, Thu, Fri, Sat}. Total outcomes = 7. The single favorable outcome is 'Monday'. Favorable outcomes = 1.

5. Calculate the Probability: Probability = $\frac{1}{7}$.

Answer: 1/7.

Question 25

If $P(A) = 0.5$, $P(B) = 0.3$ and the events are independent, find $P(A \text{ but not } B)$.

Solution:

1. Translate the Goal:

   The event 'A but not B' is written as $A \cap B'$.

2. Use the Property of Independent Events:

   A key property of independent events is that if A and B are independent, then their complements (A' and B') are also independent of each other and of the original events. Specifically, A and B' are independent.

3. Apply the Multiplication Rule for Independent Events:

   For independent events, the probability of their intersection is the product of their probabilities.

   $P(A \cap B') = P(A) \times P(B')$.

4. Calculate $P(B')$ and Solve: First, find the probability of the complement of B:

   $P(B') = 1 - P(B) = 1 - 0.3 = 0.7$.

   Now, calculate the final probability:

   $P(A \cap B') = 0.5 \times 0.7 = 0.35$.

Answer: 0.35.

Question 26

Two dice are rolled. Find the probability that the sum is a prime number.

Solution:

1. Sample Space: Total outcomes = $6 \times 6 = 36$.

2. Identify Favorable Sums: The possible sums range from 2 to 12. The prime numbers in this range are {2, 3, 5, 7, 11}.

3. Count Ways for Each Favorable Sum:

   • Sum = 2: (1,1) -> 1 way
   • Sum = 3: (1,2), (2,1) -> 2 ways
   • Sum = 5: (1,4), (4,1), (2,3), (3,2) -> 4 ways
   • Sum = 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) -> 6 ways
   • Sum = 11: (5,6), (6,5) -> 2 ways

4. Calculate Total Favorable Outcomes: $n(E) = 1+2+4+6+2 = 15$.

5. Calculate Probability: $P(\text{Prime Sum}) = \frac{15}{36} = \frac{5}{12}$.

Answer: 5/12.

Question 27

A committee of two persons is selected from two men and two women. What is the probability that the committee will have no man?

Solution: 1. Calculate the Total Number of Outcomes (Sample Space):

• Total people = 4.
• Total ways to select a committee of 2 from 4 people is $n(S) = \binom{4}{2} = \frac{4 \times 3}{2} = 6$.

2. Calculate the Number of Favorable Outcomes:

• The event E = 'the committee has no man'. This is equivalent to 'the committee has 2 women'.
• The number of ways to select 2 women from the 2 available women is $n(E) = \binom{2}{2} = 1$.

3. Calculate the Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{1}{6}$.

Answer: 1/6.

Question 28

For any two events A and B, show that $P(A \cup B) \le P(A) + P(B)$.

Solution: 1. Start with the General Addition Rule:

We know from the axioms of probability that for any two events A and B:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

2. Use the First Axiom of Probability (Non-negativity):

The first axiom states that the probability of any event must be greater than or equal to zero. This applies to the intersection event as well:

$P(A \cap B) \ge 0$.

3. Combine the Statements:

Since $P(A \cap B)$ is a non-negative quantity, subtracting it from $P(A) + P(B)$ will either leave the sum unchanged (if $P(A \cap B)=0$) or decrease it.

Therefore, it must be true that:

$P(A) + P(B) - P(A \cap B) \le P(A) + P(B)$.

4. Conclusion:

By substituting the left side with its equivalent from the addition rule, we get:

$P(A \cup B) \le P(A) + P(B)$.

This important result is known as Boole's Inequality or the union bound.

Answer: Proven.

Question 29

A bag contains 5 green and 7 red balls. Two balls are drawn. Find the probability that one is green and the other is red.

Solution: 1. Calculate the Total Number of Outcomes:

• Total balls = 5 + 7 = 12.

• We are drawing 2 balls from 12. The total number of possible pairs is:

  $n(S) = \binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66$.

2. Calculate the Number of Favorable Outcomes:

• The event E is 'drawing 1 green ball AND 1 red ball'.
• Number of ways to choose 1 green ball from 5: $\binom{5}{1} = 5$.
• Number of ways to choose 1 red ball from 7: $\binom{7}{1} = 7$.
• The total number of favorable outcomes is the product of these two: $n(E) = 5 \times 7 = 35$.

3. Calculate the Probability: $P(E) = \frac{n(E)}{n(S)} = \frac{35}{66}$.

Answer: 35/66.

Question 30

What is the probability of throwing a number greater than 2 or an even number with a single fair die?

Solution: 1. Define Events:

• Sample Space, S = {1, 2, 3, 4, 5, 6}.
• Let A = 'number > 2' = {3, 4, 5, 6}. So, $P(A) = 4/6$.
• Let B = 'even number' = {2, 4, 6}. So, $P(B) = 3/6$.

2. Check for Mutual Exclusivity:

The events are not mutually exclusive because the numbers 4 and 6 are in both sets. We must use the general addition rule.

3. Find the Intersection:

The intersection is $A \cap B = \{4, 6\}$.

The probability of the intersection is $P(A \cap B) = 2/6$.

4. Apply the General Addition Rule:

We need to find $P(A \cup B)$.

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{4}{6} + \frac{3}{6} - \frac{2}{6} = \frac{5}{6}$.

Answer: 5/6.