Samacheer Kalvi Class 11 Maths Solutions Chapter 12 Introduction to Probability Theory Exercise 12.3

Question 1. Can two events be mutually exclusive and independent simultaneously?
Answer: No, two events usually cannot be both mutually exclusive and independent at the same time, unless one of the events has zero probability. Mutually exclusive events mean they cannot happen together, so their intersection probability \( P(A \cap B) = 0 \). Independent events mean the probability of both happening is the product of their individual probabilities, \( P(A \cap B) = P(A)P(B) \). For both conditions to be true, \( P(A)P(B) \) must be equal to 0. This implies that either \( P(A) = 0 \) or \( P(B) = 0 \), meaning one of the events is impossible. Therefore, for non-trivial events (events that can actually happen), they cannot be both mutually exclusive and independent.
In simple words: If two things cannot happen at the same time, they are mutually exclusive. If one thing happening does not affect the other, they are independent. It's rare for both to be true unless one of the things can never happen.

🎯 Exam Tip: Remember the definitions: mutually exclusive means no overlap (\( P(A \cap B) = 0 \)), and independent means the probability of both is a product (\( P(A \cap B) = P(A)P(B) \)). Test these conditions to see if they can hold true together.

Question 2. If A and B are two events such that \( P(A \cup B) = 0.7 \), \( P(A \cap B) = 0.2 \), and \( P(B) = 0.5 \), then show that A and B are independent.
Answer: To show that events A and B are independent, we need to prove that \( P(A \cap B) = P(A) \cdot P(B) \). We are given \( P(A \cup B) = 0.7 \), \( P(A \cap B) = 0.2 \), and \( P(B) = 0.5 \). Let's first find \( P(A) \) using the formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given values into the formula:
\( 0.7 = P(A) + 0.5 - 0.2 \)
Simplify the right side of the equation:
\( 0.7 = P(A) + 0.3 \)
Now, subtract 0.3 from both sides to find \( P(A) \):
\( P(A) = 0.7 - 0.3 \)
\( P(A) = 0.4 \)
Next, we calculate the product of \( P(A) \) and \( P(B) \):
\( P(A) \cdot P(B) = 0.4 \times 0.5 \)
\( P(A) \cdot P(B) = 0.20 \)
We see that the calculated value \( P(A) \cdot P(B) = 0.20 \) is equal to the given \( P(A \cap B) = 0.2 \). This confirms the condition for independence.
Therefore, A and B are independent events.
In simple words: To check if two events are independent, we see if the chance of both happening is the same as multiplying their individual chances. First, we found the chance of event A. Then, we multiplied the chances of A and B. Since this matched the given chance of A and B both happening, they are independent.

🎯 Exam Tip: When proving independence, always calculate \( P(A) \cdot P(B) \) and compare it directly to \( P(A \cap B) \). Don't just assume they are equal and work backwards.

Question 3. If A and B are two independent events such that \( P(A \cup B) = 0.6 \), \( P(A) = 0.2 \), find \( P(B) \).
Answer: We are given that A and B are independent events. For independent events, the probability of their intersection is the product of their individual probabilities: \( P(A \cap B) = P(A) \cdot P(B) \).
We also know the formula for the probability of the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Since A and B are independent, we can substitute \( P(A) \cdot P(B) \) for \( P(A \cap B) \) in the union formula:
\( P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B) \)
Now, substitute the given values \( P(A \cup B) = 0.6 \) and \( P(A) = 0.2 \):
\( 0.6 = 0.2 + P(B) - 0.2 \cdot P(B) \)
We can factor out \( P(B) \) from the terms on the right side:
\( 0.6 = 0.2 + P(B) (1 - 0.2) \)
\( 0.6 = 0.2 + P(B) (0.8) \)
Subtract 0.2 from both sides of the equation:
\( 0.6 - 0.2 = P(B) (0.8) \)
\( 0.4 = P(B) (0.8) \)
Finally, divide by 0.8 to solve for \( P(B) \):
\( P(B) = \frac{0.4}{0.8} \)
\( P(B) = \frac{4}{8} \)
\( P(B) = \frac{1}{2} \)
\( P(B) = 0.5 \)
The probability of event B happening is 0.5.
In simple words: We know A and B are independent, so the chance of both A and B happening is just the chance of A times the chance of B. We used a special formula for independent events involving their union and the chance of A. By putting in the numbers we already knew, we could work out the chance of B.

🎯 Exam Tip: For independent events, directly use the simplified union formula \( P(A \cup B) = P(A) + P(B) - P(A)P(B) \) to quickly solve for an unknown probability.

Question 4. If \( P(A) = 0.5 \), \( P(B) = 0.8 \) and \( P(B/A) = 0.8 \), find \( P(A/B) \) and \( P(A \cup B) \).
Answer: We are given \( P(A) = 0.5 \), \( P(B) = 0.8 \), and \( P(B/A) = 0.8 \). We need to find \( P(A/B) \) and \( P(A \cup B) \).
First, let's find \( P(A \cap B) \) using the conditional probability formula for \( P(B/A) \):
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
Substitute the known values:
\( 0.8 = \frac{P(A \cap B)}{0.5} \)
Multiply both sides by 0.5 to find \( P(A \cap B) \):
\( P(A \cap B) = 0.8 \times 0.5 \)
\( P(A \cap B) = 0.40 \)
Now, we can find \( P(A/B) \) using its conditional probability formula:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Substitute the calculated \( P(A \cap B) \) and the given \( P(B) \):
\( P(A/B) = \frac{0.40}{0.8} \)
\( P(A/B) = \frac{4}{8} \)
\( P(A/B) = \frac{1}{2} \)
\( P(A/B) = 0.5 \)
Next, let's find \( P(A \cup B) \) using the general formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given \( P(A) \) and \( P(B) \), and the calculated \( P(A \cap B) \):
\( P(A \cup B) = 0.5 + 0.8 - 0.40 \)
\( P(A \cup B) = 1.3 - 0.40 \)
\( P(A \cup B) = 0.90 \)
So, \( P(A/B) = 0.5 \) and \( P(A \cup B) = 0.90 \). It's interesting to note here that \( P(B/A) = P(B) \), which means A and B are independent events.
In simple words: We used special formulas to find the chance of A happening given B, and the chance of A or B happening. First, we found the chance of A and B both happening. Then, we used that value with the given numbers to solve for the two things the question asked for.

🎯 Exam Tip: Always calculate \( P(A \cap B) \) first when given conditional probabilities like \( P(B/A) \), as this value is often a key intermediate step for finding other probabilities.

Question 5. If for two events A and B, \( P(A) = \frac{3}{4} \), \( P(B) = \frac{2}{5} \) and \( A \cup B = S \) (sample space), find the conditional probability \( P(A/B) \).
Answer: We are given \( P(A) = \frac{3}{4} \), \( P(B) = \frac{2}{5} \), and \( A \cup B = S \). When the union of two events equals the entire sample space \( S \), it means that the probability of their union is 1: \( P(A \cup B) = P(S) = 1 \). We need to find \( P(A/B) \).
The formula for conditional probability \( P(A/B) \) is:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
First, we need to find \( P(A \cap B) \). We can use the formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the known values:
\( 1 = \frac{3}{4} + \frac{2}{5} - P(A \cap B) \)
To add the fractions, find a common denominator, which is 20:
\( 1 = \frac{3 \times 5}{4 \times 5} + \frac{2 \times 4}{5 \times 4} - P(A \cap B) \)
\( 1 = \frac{15}{20} + \frac{8}{20} - P(A \cap B) \)
\( 1 = \frac{15 + 8}{20} - P(A \cap B) \)
\( 1 = \frac{23}{20} - P(A \cap B) \)
Now, solve for \( P(A \cap B) \):
\( P(A \cap B) = \frac{23}{20} - 1 \)
\( P(A \cap B) = \frac{23}{20} - \frac{20}{20} \)
\( P(A \cap B) = \frac{3}{20} \)
Now that we have \( P(A \cap B) \), we can find \( P(A/B) \):
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Substitute \( P(A \cap B) = \frac{3}{20} \) and \( P(B) = \frac{2}{5} \):
\( P(A/B) = \frac{\frac{3}{20}}{\frac{2}{5}} \)
To divide fractions, multiply by the reciprocal of the denominator:
\( P(A/B) = \frac{3}{20} \times \frac{5}{2} \)
\( P(A/B) = \frac{3 \times 5}{20 \times 2} \)
\( P(A/B) = \frac{15}{40} \)
Simplify the fraction:
\( P(A/B) = \frac{3}{8} \)
The conditional probability \( P(A/B) \) is \( \frac{3}{8} \).
In simple words: The problem asks for the chance of A happening if we know B has already happened. We first used the fact that A or B must happen (since their union is the whole sample space) to find the chance of A and B both happening. Then, we used this result with the chance of B to find the answer.

🎯 Exam Tip: Recognize that if \( A \cup B = S \) (the sample space), then \( P(A \cup B) = 1 \). This is a common simplification to look for in probability problems.

Question 6. A problem in mathematics is given to three students whose chances of solving it are \( \frac{1}{3} \), \( \frac{1}{4} \) and \( \frac{1}{5} \).
(i) What is the probability that the problem is solved?
Answer: Let \( A_1 \), \( A_2 \), and \( A_3 \) be the events that the first, second, and third students solve the problem, respectively. We are given their probabilities:
\( P(A_1) = \frac{1}{3} \)
\( P(A_2) = \frac{1}{4} \)
\( P(A_3) = \frac{1}{5} \)
The students solve the problem independently. The problem is solved if at least one of them solves it. This is \( P(A_1 \cup A_2 \cup A_3) \).
It's easier to calculate the probability that *none* of them solve the problem and subtract that from 1. Let \( \overline{A_1}, \overline{A_2}, \overline{A_3} \) be the events that the students *do not* solve the problem.
\( P(\overline{A_1}) = 1 - P(A_1) = 1 - \frac{1}{3} = \frac{2}{3} \)
\( P(\overline{A_2}) = 1 - P(A_2) = 1 - \frac{1}{4} = \frac{3}{4} \)
\( P(\overline{A_3}) = 1 - P(A_3) = 1 - \frac{1}{5} = \frac{4}{5} \)
Since \( A_1, A_2, A_3 \) are independent events, their complements \( \overline{A_1}, \overline{A_2}, \overline{A_3} \) are also independent.
The probability that none of them solve the problem is:
\( P(\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}) = P(\overline{A_1}) \cdot P(\overline{A_2}) \cdot P(\overline{A_3}) \)
\( P(\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}) = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \)
\( P(\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}) = \frac{2 \times 3 \times 4}{3 \times 4 \times 5} \)
\( P(\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}) = \frac{24}{60} \)
\( P(\overline{A_1} \cap \overline{A_2} \cap \overline{A_3}) = \frac{2}{5} \)
The probability that the problem is solved is:
\( P(\text{problem is solved}) = 1 - P(\text{none solve the problem}) \)
\( P(\text{problem is solved}) = 1 - \frac{2}{5} \)
\( P(\text{problem is solved}) = \frac{5}{5} - \frac{2}{5} \)
\( P(\text{problem is solved}) = \frac{3}{5} \)
This means there is a 60% chance the math problem will be successfully solved.
In simple words: To find the chance that the problem is solved, we first figure out the chance that *no one* solves it. We multiply the individual chances of each student *not* solving it. Then, we subtract this "none solve" chance from 1 to get the chance that at least one student solves it.

🎯 Exam Tip: For "at least one" probability questions, it's generally easier and more accurate to calculate 1 minus the probability of "none" of the events occurring, especially when events are independent.

Question 6. A problem in mathematics is given to three students whose chances of solving it are \( \frac{1}{3} \), \( \frac{1}{4} \) and \( \frac{1}{5} \).
(ii) What is the probability that exactly one of them will solve it?
Answer: Let \( A_1, A_2, A_3 \) be the events that the first, second, and third students solve the problem, and \( \overline{A_1}, \overline{A_2}, \overline{A_3} \) be the events that they do not solve it. We have:
\( P(A_1) = \frac{1}{3} \), \( P(\overline{A_1}) = \frac{2}{3} \)
\( P(A_2) = \frac{1}{4} \), \( P(\overline{A_2}) = \frac{3}{4} \)
\( P(A_3) = \frac{1}{5} \), \( P(\overline{A_3}) = \frac{4}{5} \)
For exactly one student to solve the problem, one of three scenarios must occur:
1. Student 1 solves, and Students 2 and 3 do not: \( P(A_1 \cap \overline{A_2} \cap \overline{A_3}) \)
2. Student 2 solves, and Students 1 and 3 do not: \( P(\overline{A_1} \cap A_2 \cap \overline{A_3}) \)
3. Student 3 solves, and Students 1 and 2 do not: \( P(\overline{A_1} \cap \overline{A_2} \cap A_3) \)
Since the events are independent, we can multiply their probabilities for each scenario:
\( P(A_1 \cap \overline{A_2} \cap \overline{A_3}) = P(A_1) \cdot P(\overline{A_2}) \cdot P(\overline{A_3}) = \frac{1}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{12}{60} = \frac{1}{5} \)
\( P(\overline{A_1} \cap A_2 \cap \overline{A_3}) = P(\overline{A_1}) \cdot P(A_2) \cdot P(\overline{A_3}) = \frac{2}{3} \times \frac{1}{4} \times \frac{4}{5} = \frac{8}{60} = \frac{2}{15} \)
\( P(\overline{A_1} \cap \overline{A_2} \cap A_3) = P(\overline{A_1}) \cdot P(\overline{A_2}) \cdot P(A_3) = \frac{2}{3} \times \frac{3}{4} \times \frac{1}{5} = \frac{6}{60} = \frac{1}{10} \)
The probability that exactly one of them solves the problem is the sum of these three mutually exclusive scenarios:
\( P(\text{exactly one solves}) = \frac{1}{5} + \frac{2}{15} + \frac{1}{10} \)
To add these fractions, find a common denominator, which is 30:
\( P(\text{exactly one solves}) = \frac{1 \times 6}{5 \times 6} + \frac{2 \times 2}{15 \times 2} + \frac{1 \times 3}{10 \times 3} \)
\( P(\text{exactly one solves}) = \frac{6}{30} + \frac{4}{30} + \frac{3}{30} \)
\( P(\text{exactly one solves}) = \frac{6 + 4 + 3}{30} \)
\( P(\text{exactly one solves}) = \frac{13}{30} \)
The probability that exactly one student solves the problem is \( \frac{13}{30} \). This calculation involves considering all possible ways one person could succeed while others fail.
In simple words: To find the chance that only one student solves the problem, we look at three different situations: student 1 solves (and others don't), student 2 solves (and others don't), or student 3 solves (and others don't). We calculate the chance for each situation and then add them up.

🎯 Exam Tip: For "exactly one" or "exactly N" problems, break it down into mutually exclusive cases (e.g., A works, B and C don't; B works, A and C don't, etc.) and sum their individual probabilities.

Question 7. The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15.
(i) If the oil had to be changed, what is the probability that a new oil filter is needed?
Answer: Let A be the event that the oil needs changing, and B be the event that a new oil filter is needed. We are given:
\( P(A) = 0.30 \) (probability of needing an oil change)
\( P(B) = 0.40 \) (probability of needing a new oil filter)
\( P(A \cap B) = 0.15 \) (probability of needing both oil change and new filter)
We need to find the probability that a new oil filter is needed *given* that the oil had to be changed. This is \( P(B/A) \).
The formula for conditional probability \( P(B/A) \) is:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
Substitute the given values:
\( P(B/A) = \frac{0.15}{0.30} \)
\( P(B/A) = \frac{15}{30} \)
\( P(B/A) = \frac{1}{2} \)
\( P(B/A) = 0.5 \)
So, if the oil is changed, there is a 50% chance that a new oil filter is also needed. This shows a clear relationship between the two maintenance tasks.
In simple words: We want to know the chance of needing a new oil filter if we already know the oil is being changed. We use a special formula that divides the chance of both things happening by the chance of the oil being changed.

🎯 Exam Tip: Carefully identify which event is the "given" condition and which event's probability you are trying to find. This determines the denominator in the conditional probability formula.

Question 7. The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15.
(ii) If a new oil filter is needed, what is the probability that the oil has to be changed?
Answer: Let A be the event that the oil needs changing, and B be the event that a new oil filter is needed. We have:
\( P(A) = 0.30 \)
\( P(B) = 0.40 \)
\( P(A \cap B) = 0.15 \)
We need to find the probability that the oil has to be changed *given* that a new oil filter is needed. This is \( P(A/B) \).
The formula for conditional probability \( P(A/B) \) is:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Substitute the given values:
\( P(A/B) = \frac{0.15}{0.40} \)
\( P(A/B) = \frac{15}{40} \)
\( P(A/B) = \frac{3}{8} \)
\( P(A/B) = 0.375 \)
Thus, if a new oil filter is needed, there is a 37.5% chance that the oil also needs to be changed. This is a common situation where knowing one event has occurred changes the likelihood of another.
In simple words: Here, we want to know the chance of needing an oil change if we already know a new oil filter is needed. Again, we use the conditional probability formula, dividing the chance of both happening by the chance of needing the oil filter.

🎯 Exam Tip: Be careful not to confuse \( P(A/B) \) with \( P(B/A) \). They are generally not the same, as the "given" condition changes the sample space for the probability calculation.

Question 8. One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that
(i) both are white
Answer: Let's define the contents of each bag:
Bag 1: 5 white balls, 3 black balls. Total balls = \( 5 + 3 = 8 \).
Bag 2: 4 white balls, 6 black balls. Total balls = \( 4 + 6 = 10 \).
We are drawing one ball from each bag. The draws are independent events.
For part (i), we want the probability that both balls drawn are white.
Probability of drawing a white ball from Bag 1: \( P(W_1) = \frac{\text{Number of white balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{5}{8} \)
Probability of drawing a white ball from Bag 2: \( P(W_2) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{4}{10} \)
Since the draws are independent, the probability that both are white is the product of their individual probabilities:
\( P(\text{both are white}) = P(W_1) \times P(W_2) \)
\( P(\text{both are white}) = \frac{5}{8} \times \frac{4}{10} \)
\( P(\text{both are white}) = \frac{20}{80} \)
\( P(\text{both are white}) = \frac{1}{4} \)
The probability of drawing two white balls is \( \frac{1}{4} \). This is like rolling two dice and wanting specific numbers on each.
In simple words: We want to find the chance of picking a white ball from the first bag AND a white ball from the second bag. Since these choices don't affect each other, we simply multiply the chance of picking a white ball from each bag.

🎯 Exam Tip: When calculating probabilities for multiple independent events occurring together, always multiply their individual probabilities.

Question 8. One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that
(ii) both are black
Answer: As before, we have:
Bag 1: 5 white, 3 black. Total = 8.
Bag 2: 4 white, 6 black. Total = 10.
For part (ii), we want the probability that both balls drawn are black.
Probability of drawing a black ball from Bag 1: \( P(B_1) = \frac{\text{Number of black balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{3}{8} \)
Probability of drawing a black ball from Bag 2: \( P(B_2) = \frac{\text{Number of black balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{6}{10} \)
Since the draws are independent, the probability that both are black is the product of their individual probabilities:
\( P(\text{both are black}) = P(B_1) \times P(B_2) \)
\( P(\text{both are black}) = \frac{3}{8} \times \frac{6}{10} \)
\( P(\text{both are black}) = \frac{18}{80} \)
\( P(\text{both are black}) = \frac{9}{40} \)
The probability of drawing two black balls is \( \frac{9}{40} \). This calculation is similar to finding the likelihood of two separate coin flips both landing on tails.
In simple words: To find the chance of picking a black ball from the first bag AND a black ball from the second bag, we multiply the chance of picking a black ball from the first bag by the chance of picking a black ball from the second bag.

🎯 Exam Tip: Always clearly list the number of favorable outcomes and total outcomes for each event before calculating probabilities to avoid errors.

Question 8. One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that
(iii) one white and one black
Answer: We have:
Bag 1: 5 white, 3 black. Total = 8.
Bag 2: 4 white, 6 black. Total = 10.
For part (iii), we want the probability of drawing one white ball and one black ball. There are two ways this can happen:
Case 1: White from Bag 1 AND Black from Bag 2
Case 2: Black from Bag 1 AND White from Bag 2
These two cases are mutually exclusive, so we can add their probabilities.
Probability of Case 1: \( P(W_1 \cap B_2) = P(W_1) \times P(B_2) \)
\( P(W_1) = \frac{5}{8} \)
\( P(B_2) = \frac{6}{10} \)
\( P(W_1 \cap B_2) = \frac{5}{8} \times \frac{6}{10} = \frac{30}{80} = \frac{3}{8} \)
Probability of Case 2: \( P(B_1 \cap W_2) = P(B_1) \times P(W_2) \)
\( P(B_1) = \frac{3}{8} \)
\( P(W_2) = \frac{4}{10} \)
\( P(B_1 \cap W_2) = \frac{3}{8} \times \frac{4}{10} = \frac{12}{80} = \frac{3}{20} \)
The total probability of drawing one white and one black ball is the sum of the probabilities of these two cases:
\( P(\text{one white and one black}) = P(W_1 \cap B_2) + P(B_1 \cap W_2) \)
\( P(\text{one white and one black}) = \frac{3}{8} + \frac{3}{20} \)
To add these fractions, find a common denominator, which is 40:
\( P(\text{one white and one black}) = \frac{3 \times 5}{8 \times 5} + \frac{3 \times 2}{20 \times 2} \)
\( P(\text{one white and one black}) = \frac{15}{40} + \frac{6}{40} \)
\( P(\text{one white and one black}) = \frac{15 + 6}{40} \)
\( P(\text{one white and one black}) = \frac{21}{40} \)
The probability of drawing one white and one black ball is \( \frac{21}{40} \). This scenario requires considering all possible sequences of desired outcomes.
In simple words: We want to pick one white ball and one black ball. This can happen in two ways: white from the first bag and black from the second, OR black from the first and white from the second. We calculate the chance for each way and then add them together.

🎯 Exam Tip: For "one of each" type problems with two sources, remember there are usually two distinct scenarios that lead to the desired outcome, and their probabilities must be added.

Question 9. Two-thirds of students in a class are boys and the rest girls. It is known that the probability of a girl getting the first grade is 0.85 and that of boys is 0.70. Find the probability that a student chosen at random will get first-grade marks.
Answer: Let B be the event of choosing a boy, and G be the event of choosing a girl. We are given:
\( P(B) = \frac{2}{3} \)
Since the rest are girls, \( P(G) = 1 - P(B) = 1 - \frac{2}{3} = \frac{1}{3} \).
Let \( F_B \) be the event that a boy gets the first grade, and \( F_G \) be the event that a girl gets the first grade. We are given:
\( P(F_B) = 0.70 \)
\( P(F_G) = 0.85 \)
We want to find the total probability that a randomly chosen student gets first-grade marks. This is an application of the Law of Total Probability.
\( P(\text{first grade}) = P(\text{first grade and boy}) + P(\text{first grade and girl}) \)
Since getting a first grade is independent of whether a student is a boy or a girl:
\( P(\text{first grade}) = P(B) \cdot P(F_B) + P(G) \cdot P(F_G) \)
Substitute the given probabilities:
\( P(\text{first grade}) = \frac{2}{3} \times 0.70 + \frac{1}{3} \times 0.85 \)
\( P(\text{first grade}) = \frac{1.40}{3} + \frac{0.85}{3} \)
\( P(\text{first grade}) = \frac{1.40 + 0.85}{3} \)
\( P(\text{first grade}) = \frac{2.25}{3} \)
\( P(\text{first grade}) = 0.75 \)
The probability that a student chosen at random will get first-grade marks is 0.75 or 75%. This weighted average reflects the proportions of boys and girls and their respective success rates.
In simple words: We want the overall chance of a student getting a first grade. We know the chance of picking a boy or a girl, and their separate chances of getting a first grade. So, we combine these chances by adding the probability of a boy getting a first grade to the probability of a girl getting a first grade.

🎯 Exam Tip: The Law of Total Probability is key for problems where an event can occur through several mutually exclusive scenarios, each with its own probability.

Question 10. Given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \). Find \( P(B) \) if
(i) A and B are mutually exclusive
Answer: We are given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \).
(i) If A and B are mutually exclusive, then they cannot happen at the same time, which means \( P(A \cap B) = 0 \).
The formula for the union of two events simplifies in this case:
\( P(A \cup B) = P(A) + P(B) \)
Substitute the given values into this simplified formula:
\( 0.7 = 0.4 + P(B) \)
Subtract 0.4 from both sides to find \( P(B) \):
\( P(B) = 0.7 - 0.4 \)
\( P(B) = 0.3 \)
So, if A and B are mutually exclusive, the probability of B is 0.3. This is a direct application of the addition rule for mutually exclusive events.
In simple words: When two things cannot happen together, the chance of either of them happening is just the sum of their individual chances. We used this simple rule to find the chance of B.

🎯 Exam Tip: Always remember that for mutually exclusive events, \( P(A \cap B) = 0 \), which greatly simplifies the union formula to \( P(A \cup B) = P(A) + P(B) \).

Question 10. Given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \). Find \( P(B) \) if
(ii) A and B are independent events
Answer: We are given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \).
(ii) If A and B are independent events, then the probability of their intersection is the product of their individual probabilities: \( P(A \cap B) = P(A) \cdot P(B) \).
The general formula for the union of two events is:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute \( P(A) \cdot P(B) \) for \( P(A \cap B) \) because the events are independent:
\( P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B) \)
Now, substitute the given values \( P(A \cup B) = 0.7 \) and \( P(A) = 0.4 \):
\( 0.7 = 0.4 + P(B) - 0.4 \cdot P(B) \)
Subtract 0.4 from both sides:
\( 0.7 - 0.4 = P(B) - 0.4 \cdot P(B) \)
\( 0.3 = P(B) (1 - 0.4) \)
\( 0.3 = P(B) (0.6) \)
Divide by 0.6 to solve for \( P(B) \):
\( P(B) = \frac{0.3}{0.6} \)
\( P(B) = \frac{3}{6} \)
\( P(B) = 0.5 \)
So, if A and B are independent, the probability of B is 0.5. This shows how assuming independence changes the result significantly from the mutually exclusive case.
In simple words: If two things are independent, we use a different way to link their chances. We put the known chances into the formula for independent events and then solve it to find the chance of B.

🎯 Exam Tip: For independent events, remember the key relationship \( P(A \cap B) = P(A)P(B) \), which you can substitute into the general union formula to find unknowns.

Question 10. Given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \). Find \( P(B) \) if
(iii) \( P(A/B) = 0.4 \)
Answer: We are given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \).
(iii) We are given \( P(A/B) = 0.4 \).
The formula for conditional probability \( P(A/B) \) is:
\( P(A/B) = \frac{P(A \cap B)}{P(B)} \)
Substitute the given \( P(A/B) \):
\( 0.4 = \frac{P(A \cap B)}{P(B)} \)
This implies \( P(A \cap B) = 0.4 \cdot P(B) \). Let's call this Equation (1).
Now use the general formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given \( P(A \cup B) = 0.7 \) and \( P(A) = 0.4 \):
\( 0.7 = 0.4 + P(B) - P(A \cap B) \)
Rearrange this to solve for \( P(A \cap B) \):
\( P(A \cap B) = 0.4 + P(B) - 0.7 \)
\( P(A \cap B) = P(B) - 0.3 \). Let's call this Equation (2).
Now, we equate the two expressions for \( P(A \cap B) \) from Equation (1) and Equation (2):
\( 0.4 \cdot P(B) = P(B) - 0.3 \)
Move all terms with \( P(B) \) to one side:
\( 0.3 = P(B) - 0.4 \cdot P(B) \)
\( 0.3 = P(B) (1 - 0.4) \)
\( 0.3 = P(B) (0.6) \)
Divide by 0.6 to solve for \( P(B) \):
\( P(B) = \frac{0.3}{0.6} \)
\( P(B) = \frac{3}{6} \)
\( P(B) = 0.5 \)
So, if \( P(A/B) = 0.4 \), the probability of B is 0.5. This result also means that \( P(A) = P(A/B) \), which indicates A and B are independent events.
In simple words: We used the given conditional chance of A happening if B has happened. This helped us find a link between the chance of A and B both happening and the chance of B. Then, we used the main union formula and put in all the information to find the chance of B.

🎯 Exam Tip: When dealing with conditional probabilities, it's often useful to express \( P(A \cap B) \) in terms of \( P(B) \) (or \( P(A) \)) using the conditional formula, and then substitute this into the union formula.

Question 10. Given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \). Find \( P(B) \) if
(iv) \( P(B/A) = 0.5 \)
Answer: We are given \( P(A) = 0.4 \) and \( P(A \cup B) = 0.7 \).
(iv) We are given \( P(B/A) = 0.5 \).
The formula for conditional probability \( P(B/A) \) is:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
Substitute the given \( P(B/A) \) and \( P(A) \):
\( 0.5 = \frac{P(A \cap B)}{0.4} \)
Multiply both sides by 0.4 to find \( P(A \cap B) \):
\( P(A \cap B) = 0.5 \times 0.4 \)
\( P(A \cap B) = 0.2 \)
Now, we use the general formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given \( P(A \cup B) = 0.7 \), \( P(A) = 0.4 \), and the calculated \( P(A \cap B) = 0.2 \):
\( 0.7 = 0.4 + P(B) - 0.2 \)
Simplify the right side of the equation:
\( 0.7 = P(B) + (0.4 - 0.2) \)
\( 0.7 = P(B) + 0.2 \)
Subtract 0.2 from both sides to find \( P(B) \):
\( P(B) = 0.7 - 0.2 \)
\( P(B) = 0.5 \)
So, if \( P(B/A) = 0.5 \), the probability of B is 0.5. It is important to carefully use the correct conditional probability formula to find the intersection.
In simple words: We used the given chance of B happening if A has happened. This helped us directly find the chance of both A and B happening. Then, we used the main formula for "A or B" and put in all the known numbers to find the chance of B.

🎯 Exam Tip: Understand that \( P(B/A) \) tells you the probability of B given A, so if you know \( P(A) \), you can easily find \( P(A \cap B) \) by multiplying them.

Question 11. What is the probability that a randomly chosen year contains 53 Sundays?
(i) it contains 53 Sundays
Answer: To find the probability that a randomly chosen year contains 53 Sundays, we need to consider two types of years: non-leap years and leap years.
A non-leap year has 365 days. \( 365 = 52 \times 7 + 1 \). This means a non-leap year has 52 full weeks and 1 extra day. For it to have 53 Sundays, this extra day must be a Sunday. The extra day can be any of the 7 days (Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday), each with equal probability. So, the probability that this extra day is a Sunday is \( \frac{1}{7} \).
A leap year has 366 days. \( 366 = 52 \times 7 + 2 \). This means a leap year has 52 full weeks and 2 extra days. For it to have 53 Sundays, at least one of these two extra days must be a Sunday. The pair of extra days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), or (Saturday, Sunday). There are 7 possible pairs, and 2 of them include a Sunday (Sunday, Monday) and (Saturday, Sunday). So, the probability that at least one of these two extra days is a Sunday is \( \frac{2}{7} \).
The probability of a randomly chosen year being a leap year is \( \frac{1}{4} \) (approximately, as there's a more complex rule every 100/400 years, but for general purposes, \( \frac{1}{4} \) is used).
The probability of a randomly chosen year being a non-leap year is \( 1 - \frac{1}{4} = \frac{3}{4} \).
Now, we combine these probabilities using the Law of Total Probability:
\( P(\text{53 Sundays}) = P(\text{53 Sundays | non-leap year}) \cdot P(\text{non-leap year}) + P(\text{53 Sundays | leap year}) \cdot P(\text{leap year}) \)
\( P(\text{53 Sundays}) = \frac{1}{7} \times \frac{3}{4} + \frac{2}{7} \times \frac{1}{4} \)
\( P(\text{53 Sundays}) = \frac{3}{28} + \frac{2}{28} \)
\( P(\text{53 Sundays}) = \frac{3+2}{28} \)
\( P(\text{53 Sundays}) = \frac{5}{28} \)
The probability that a randomly chosen year contains 53 Sundays is \( \frac{5}{28} \). This calculation accounts for the different structures of years and their likelihoods.
In simple words: A year can be a normal year or a leap year. A normal year has one extra day, and a leap year has two. We find the chance of that extra day (or days) being a Sunday for both types of years. Then, we combine these chances, considering how often normal years and leap years happen.

🎯 Exam Tip: For calendar-related probability questions, always divide the total days by 7 to find full weeks and remaining days. Then, consider how those remaining days can form the desired outcome for both non-leap and leap years.

Question 11. What is the probability that a randomly chosen year contains 53 Sundays?
(ii) it is a leap year which contains 53 Sundays.
Answer: We want to find the probability that a randomly chosen year is a leap year *and* contains 53 Sundays. This is an intersection of two events.
A leap year has 366 days. This means it has 52 full weeks and 2 extra days. For a leap year to have 53 Sundays, at least one of these two extra days must be a Sunday. The possible pairs for these two extra days are:
(Sunday, Monday)
(Monday, Tuesday)
(Tuesday, Wednesday)
(Wednesday, Thursday)
(Thursday, Friday)
(Friday, Saturday)
(Saturday, Sunday)
There are 7 equally likely combinations for the two extra days. Out of these 7 combinations, 2 of them contain a Sunday (Sunday, Monday and Saturday, Sunday).
So, the probability that a leap year contains 53 Sundays is \( \frac{2}{7} \).
The probability of a randomly chosen year being a leap year is \( \frac{1}{4} \). This probability arises from the fact that roughly one in every four years is a leap year.
To find the probability that a year is a leap year *and* contains 53 Sundays, we multiply these two probabilities (assuming these are independent in the context of year selection):
\( P(\text{leap year and 53 Sundays}) = P(\text{leap year}) \times P(\text{53 Sundays | leap year}) \)
\( P(\text{leap year and 53 Sundays}) = \frac{1}{4} \times \frac{2}{7} \)
\( P(\text{leap year and 53 Sundays}) = \frac{2}{28} \)
\( P(\text{leap year and 53 Sundays}) = \frac{1}{14} \)
The probability that a randomly chosen year is a leap year and contains 53 Sundays is \( \frac{1}{14} \).
In simple words: We want to know the chance that a year is a leap year AND has 53 Sundays. We find the chance of a year being a leap year, and then the chance that a leap year has 53 Sundays. We multiply these two chances together.

🎯 Exam Tip: For compound probabilities (Event A AND Event B), always multiply the probability of A by the conditional probability of B given A (or just P(B) if A and B are independent).

Question 12. The probability of hitting a target by person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously. What is the probability that the target is damaged by exactly 2 hits?
Answer: Let X, Y, and Z be the events that persons X, Y, and Z hit the target, respectively. We are given their probabilities:
\( P(X) = \frac{3}{4} \)
\( P(Y) = \frac{4}{5} \)
\( P(Z) = \frac{2}{3} \)
Let \( \overline{X}, \overline{Y}, \overline{Z} \) be the events that they miss the target:
\( P(\overline{X}) = 1 - P(X) = 1 - \frac{3}{4} = \frac{1}{4} \)
\( P(\overline{Y}) = 1 - P(Y) = 1 - \frac{4}{5} = \frac{1}{5} \)
\( P(\overline{Z}) = 1 - P(Z) = 1 - \frac{2}{3} = \frac{1}{3} \)
We want to find the probability that the target is damaged by *exactly 2 hits*. This can happen in three mutually exclusive ways:
1. X and Y hit, Z misses: \( P(X \cap Y \cap \overline{Z}) \)
2. X and Z hit, Y misses: \( P(X \cap \overline{Y} \cap Z) \)
3. Y and Z hit, X misses: \( P(\overline{X} \cap Y \cap Z) \)
Since their shots are independent, we can multiply their individual probabilities for each case:
**Case 1: X hits, Y hits, Z misses**
\( P(X \cap Y \cap \overline{Z}) = P(X) \cdot P(Y) \cdot P(\overline{Z}) = \frac{3}{4} \times \frac{4}{5} \times \frac{1}{3} = \frac{12}{60} = \frac{1}{5} \)
**Case 2: X hits, Y misses, Z hits**
\( P(X \cap \overline{Y} \cap Z) = P(X) \cdot P(\overline{Y}) \cdot P(Z) = \frac{3}{4} \times \frac{1}{5} \times \frac{2}{3} = \frac{6}{60} = \frac{1}{10} \)
**Case 3: X misses, Y hits, Z hits**
\( P(\overline{X} \cap Y \cap Z) = P(\overline{X}) \cdot P(Y) \cdot P(Z) = \frac{1}{4} \times \frac{4}{5} \times \frac{2}{3} = \frac{8}{60} = \frac{2}{15} \)
The total probability that the target is damaged by exactly 2 hits is the sum of these three probabilities:
\( P(\text{exactly 2 hits}) = \frac{1}{5} + \frac{1}{10} + \frac{2}{15} \)
To add these fractions, find a common denominator, which is 60:
\( P(\text{exactly 2 hits}) = \frac{1 \times 12}{5 \times 12} + \frac{1 \times 6}{10 \times 6} + \frac{2 \times 4}{15 \times 4} \)
\( P(\text{exactly 2 hits}) = \frac{12}{60} + \frac{6}{60} + \frac{8}{60} \)
\( P(\text{exactly 2 hits}) = \frac{12 + 6 + 8}{60} \)
\( P(\text{exactly 2 hits}) = \frac{26}{60} \)
Simplify the fraction:
\( P(\text{exactly 2 hits}) = \frac{13}{30} \)
The probability that the target is damaged by exactly 2 hits is \( \frac{13}{30} \). This type of problem requires careful consideration of all successful combinations.
In simple words: We want to find the chance that exactly two people hit the target. This can happen in three ways: X and Y hit (Z misses), X and Z hit (Y misses), or Y and Z hit (X misses). We calculate the chance for each of these three situations and then add them up.

🎯 Exam Tip: For "exactly N" success scenarios, list all combinations of N successes and the remaining failures. Calculate the probability for each combination (multiplying individual probabilities) and then sum these results, as the combinations are mutually exclusive.