RBSE Solutions Class 12 Maths Chapter 16 Probability and Probability Distribution Exercise 16.2

Question 1. If A and B be two events such that \( P(A) = \frac { 1 }{ 4 } \); \( P(B) = \frac { 1 }{ 2 } \) and \( P(A \cap B) = \frac { 1 }{ 8 } \), then find \( P(\overline { A } \cap \overline {B}) \).
Answer:Given that \( P(A) = \frac { 1 }{ 4 } \), \( P(B) = \frac { 1 }{ 2 } \), and \( P(A \cap B) = \frac { 1 }{ 8 } \).
First, we find the probability of the union of events A and B using the formula:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substitute the given values:
\( P(A \cup B) = \frac { 1 }{ 4 } + \frac { 1 }{ 2 } - \frac { 1 }{ 8 } \)
To add and subtract these fractions, we find a common denominator, which is 8.
\( P(A \cup B) = \frac { 2 }{ 8 } + \frac { 4 }{ 8 } - \frac { 1 }{ 8 } \)
\( P(A \cup B) = \frac { 2 + 4 - 1 }{ 8 } \)
\( P(A \cup B) = \frac { 5 }{ 8 } \)
Next, we need to find \( P(\overline { A } \cap \overline {B}) \). By De Morgan's Law, this is equal to \( P(\overline {A \cup B}) \).
The probability of the complement of an event is 1 minus the probability of the event.
\( P(\overline { A } \cap \overline {B}) = P(\overline {A \cup B}) \)
\( P(\overline { A } \cap \overline {B}) = 1 - P(A \cup B) \)
Substitute the value of \( P(A \cup B) \):
\( P(\overline { A } \cap \overline {B}) = 1 - \frac { 5 }{ 8 } \)
\( P(\overline { A } \cap \overline {B}) = \frac { 8 - 5 }{ 8 } \)
\( P(\overline { A } \cap \overline {B}) = \frac { 3 }{ 8 } \)
In simple words: We first find the chance of either A or B happening. Then, we use that result to find the chance that neither A nor B happens. This is like finding the opposite of A or B happening.

🎯 Exam Tip: Remember De Morgan's Laws: \( P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) \) and \( P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) \). They are very useful for simplifying probability problems involving complements.

Question 2. If P(A) = 0.4, P(B) = p and P(A ∪ B) = 0.6 and A and B are independent events, then find the value of p.
Answer:Given:
\( P(A) = 0.4 \)
\( P(B) = p \)
\( P(A \cup B) = 0.6 \)
A and B are independent events.
For independent events, the probability of their intersection is the product of their individual probabilities:
\( P(A \cap B) = P(A) \cdot P(B) \)
\( P(A \cap B) = 0.4 \cdot p \)
We also know the formula for the probability of the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Now, substitute the given values and the expression for \( P(A \cap B) \):
\( 0.6 = 0.4 + p - (0.4 \cdot p) \)
\( 0.6 = 0.4 + p - 0.4p \)
Subtract 0.4 from both sides:
\( 0.6 - 0.4 = p - 0.4p \)
\( 0.2 = (1 - 0.4)p \)
\( 0.2 = 0.6p \)
To find p, divide 0.2 by 0.6:
\( p = \frac { 0.2 }{ 0.6 } \)
\( p = \frac { 2 }{ 6 } \)
\( p = \frac { 1 }{ 3 } \)
So, the value of p is \( \frac{1}{3} \). This calculation helps determine an unknown probability when events are independent.
In simple words: We are given some chances for events A and B, and that they don't affect each other. We use the rule for independent events and the union formula to find the missing chance for event B.

🎯 Exam Tip: When events are independent, the key is to use \( P(A \cap B) = P(A) \cdot P(B) \). If they are mutually exclusive, then \( P(A \cap B) = 0 \).

Question 3. If A and B are independent events, and P(A) = 0.3, P(B) = 0.4, then find:
(i)P(A ∩ B)
(ii)P(A ∪ B)
(iii) P(\( \frac { A }{ B } \))
(iv) P(\( \frac { B }{ A } \))
Answer:Given that A and B are independent events.
\( P(A) = 0.3 \)
\( P(B) = 0.4 \)

(i) **Probability of A and B (Intersection):**
Since A and B are independent, \( P(A \cap B) = P(A) \cdot P(B) \).
\( P(A \cap B) = 0.3 \cdot 0.4 \)
\( P(A \cap B) = 0.12 \)

(ii) **Probability of A or B (Union):**
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = 0.3 + 0.4 - 0.12 \)
\( P(A \cup B) = 0.7 - 0.12 \)
\( P(A \cup B) = 0.58 \)

(iii) **Conditional Probability of A given B:**
For independent events, \( P(A|B) = P(A) \).
So, \( P(\frac { A }{ B }) = P(A) \)
\( P(\frac { A }{ B }) = 0.3 \)
Alternatively, using the general formula \( P(A|B) = \frac{P(A \cap B)}{P(B)} \):
\( P(\frac { A }{ B }) = \frac { 0.12 }{ 0.4 } \)
\( P(\frac { A }{ B }) = \frac { 12 }{ 40 } \)
\( P(\frac { A }{ B }) = \frac { 3 }{ 10 } \)
\( P(\frac { A }{ B }) = 0.3 \)

(iv) **Conditional Probability of B given A:**
For independent events, \( P(B|A) = P(B) \).
So, \( P(\frac { B }{ A }) = P(B) \)
\( P(\frac { B }{ A }) = 0.4 \)
Alternatively, using the general formula \( P(B|A) = \frac{P(A \cap B)}{P(A)} \):
\( P(\frac { B }{ A }) = \frac { 0.12 }{ 0.3 } \)
\( P(\frac { B }{ A }) = \frac { 12 }{ 30 } \)
\( P(\frac { B }{ A }) = \frac { 2 }{ 5 } \)
\( P(\frac { B }{ A }) = 0.4 \)
These results demonstrate how independence simplifies probability calculations.
In simple words: When two events are independent, the chance of both happening is just their chances multiplied. The chance of one happening if we already know the other happened is just the chance of the first one alone.

🎯 Exam Tip: Remember that for independent events A and B, \( P(A|B) = P(A) \) and \( P(B|A) = P(B) \). This is a common shortcut for conditional probability questions involving independence.

Question 4. If A and B are independent events., where P(A) = 0.3, P(B) = 0.6, then find:
(i)P(A ∩ B)
(ii) P(A \( \cup \overline {B} \))
(iii) P(A \( \cup \) B)
(iv) P(\( \overline {A} \cap \overline {B} \))
Answer:Given that A and B are independent events.
\( P(A) = 0.3 \)
\( P(B) = 0.6 \)
Since A and B are independent, their complements \( \overline{A} \) and \( \overline{B} \) are also independent.
\( P(\overline{A}) = 1 - P(A) = 1 - 0.3 = 0.7 \)
\( P(\overline{B}) = 1 - P(B) = 1 - 0.6 = 0.4 \)

(i) **Probability of A and B (Intersection):**
For independent events, \( P(A \cap B) = P(A) \cdot P(B) \).
\( P(A \cap B) = 0.3 \cdot 0.6 \)
\( P(A \cap B) = 0.18 \)

(ii) **Probability of A or not B:**
\( P(A \cup \overline {B}) = P(A) + P(\overline {B}) - P(A \cap \overline {B}) \)
Since A and \( \overline{B} \) are independent, \( P(A \cap \overline {B}) = P(A) \cdot P(\overline {B}) \).
\( P(A \cap \overline {B}) = 0.3 \cdot 0.4 = 0.12 \)
Now, substitute these values into the union formula:
\( P(A \cup \overline {B}) = 0.3 + 0.4 - 0.12 \)
\( P(A \cup \overline {B}) = 0.7 - 0.12 \)
\( P(A \cup \overline {B}) = 0.58 \)

(iii) **Probability of A or B (Union):**
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = 0.3 + 0.6 - 0.18 \)
\( P(A \cup B) = 0.90 - 0.18 \)
\( P(A \cup B) = 0.72 \)

(iv) **Probability of not A and not B:**
Using De Morgan's Law, \( P(\overline {A} \cap \overline {B}) = P(\overline {A \cup B}) \).
This is also \( 1 - P(A \cup B) \).
\( P(\overline {A} \cap \overline {B}) = 1 - 0.72 \)
\( P(\overline {A} \cap \overline {B}) = 0.28 \)
Alternatively, since \( \overline{A} \) and \( \overline{B} \) are independent:
\( P(\overline {A} \cap \overline {B}) = P(\overline {A}) \cdot P(\overline {B}) \)
\( P(\overline {A} \cap \overline {B}) = 0.7 \cdot 0.4 \)
\( P(\overline {A} \cap \overline {B}) = 0.28 \)
This confirms the properties of independent events and their complements.
In simple words: We find the chances of different combinations of A and B happening, remembering that they don't affect each other. This means we can multiply their individual chances to find the chance of both happening.

🎯 Exam Tip: When A and B are independent, remember that \( A \) and \( \overline{B} \) are also independent, as are \( \overline{A} \) and \( B \), and \( \overline{A} \) and \( \overline{B} \). This simplifies many intersection calculations.

Question 5. A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one without replacement, find the probability of getting all white balls.
Answer:Given:
Number of white balls = 5
Number of red balls = 7
Number of black balls = 8
Total number of balls in the bag = \( 5 + 7 + 8 = 20 \)
We need to find the probability of drawing four white balls one by one without replacement.

1. **Probability of getting the first white ball:**
There are 5 white balls out of 20 total balls.
\( P(\text{1st white}) = \frac { 5 }{ 20 } = \frac { 1 }{ 4 } \)
After drawing one white ball, the total number of balls remaining is 19, and the number of white balls remaining is 4.

2. **Probability of getting the second white ball:**
There are 4 white balls left out of 19 total balls.
\( P(\text{2nd white}) = \frac { 4 }{ 19 } \)
After drawing the second white ball, there are 18 total balls remaining, and 3 white balls remaining.

3. **Probability of getting the third white ball:**
There are 3 white balls left out of 18 total balls.
\( P(\text{3rd white}) = \frac { 3 }{ 18 } = \frac { 1 }{ 6 } \)
After drawing the third white ball, there are 17 total balls remaining, and 2 white balls remaining.

4. **Probability of getting the fourth white ball:**
There are 2 white balls left out of 17 total balls.
\( P(\text{4th white}) = \frac { 2 }{ 17 } \)

The probability of getting all four white balls is the product of these individual probabilities, because the draws are dependent events (without replacement).
\( P(\text{all white balls}) = P(\text{1st white}) \cdot P(\text{2nd white}) \cdot P(\text{3rd white}) \cdot P(\text{4th white}) \)
\( P(\text{all white balls}) = \frac { 5 }{ 20 } \cdot \frac { 4 }{ 19 } \cdot \frac { 3 }{ 18 } \cdot \frac { 2 }{ 17 } \)
\( P(\text{all white balls}) = \frac { 1 }{ 4 } \cdot \frac { 4 }{ 19 } \cdot \frac { 1 }{ 6 } \cdot \frac { 2 }{ 17 } \)
\( P(\text{all white balls}) = \frac { 1 \cdot 4 \cdot 1 \cdot 2 }{ 4 \cdot 19 \cdot 6 \cdot 17 } \)
\( P(\text{all white balls}) = \frac { 8 }{ 7752 } \)
\( P(\text{all white balls}) = \frac { 1 }{ 969 } \)
This type of problem illustrates how probability changes with each successive draw when items are not replaced.
In simple words: We take out balls one by one and don't put them back. For each turn, the chance of picking a white ball changes because there are fewer balls left. We multiply these chances together to find the overall chance of picking only white balls four times in a row.

🎯 Exam Tip: In "without replacement" problems, always adjust the total number of items and the number of favorable items after each draw. This changes the probability for subsequent draws.

Question 6. A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer:When a fair die is tossed, the possible outcomes are {1, 2, 3, 4, 5, 6}. The total number of outcomes is 6.
Odd numbers are {1, 3, 5}. The number of odd outcomes is 3.
Even numbers are {2, 4, 6}. The number of even outcomes is 3.
The probability of getting an odd number in a single toss is \( P(\text{odd}) = \frac { 3 }{ 6 } = \frac { 1 }{ 2 } \).
The probability of getting an even number in a single toss is \( P(\text{even}) = \frac { 3 }{ 6 } = \frac { 1 }{ 2 } \).

The die is tossed thrice. We want to find the probability of getting an odd number at least once.
It's often easier to calculate the probability of the complementary event, which is "not getting an odd number at least once". This means "getting an even number every time".

Probability of getting an even number in the first toss = \( \frac { 1 }{ 2 } \).
Probability of getting an even number in the second toss = \( \frac { 1 }{ 2 } \).
Probability of getting an even number in the third toss = \( \frac { 1 }{ 2 } \).

Since each toss is an independent event, the probability of getting an even number in all three tosses is the product of individual probabilities:
\( P(\text{all even}) = P(\text{even}) \cdot P(\text{even}) \cdot P(\text{even}) \)
\( P(\text{all even}) = \frac { 1 }{ 2 } \cdot \frac { 1 }{ 2 } \cdot \frac { 1 }{ 2 } \)
\( P(\text{all even}) = \frac { 1 }{ 8 } \)

Now, the probability of getting an odd number at least once is 1 minus the probability of getting all even numbers:
\( P(\text{at least one odd}) = 1 - P(\text{all even}) \)
\( P(\text{at least one odd}) = 1 - \frac { 1 }{ 8 } \)
\( P(\text{at least one odd}) = \frac { 8 - 1 }{ 8 } \)
\( P(\text{at least one odd}) = \frac { 7 }{ 8 } \)
This method simplifies calculations by considering the opposite scenario.
In simple words: We want to find the chance of rolling an odd number at least one time in three tries. It's easier to find the chance of NOT rolling any odd numbers (meaning all rolls are even) and then subtract that from 1.

🎯 Exam Tip: For "at least once" probability problems, always consider using the complement rule: \( P(\text{at least one}) = 1 - P(\text{none}) \). This often simplifies complex calculations, especially with multiple trials.

Question 7. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Answer:Given a standard pack of 52 playing cards.
Total number of cards = 52.
Number of black cards in a pack = 26 (13 spades and 13 clubs).
Number of red cards in a pack = 26 (13 hearts and 13 diamonds).
We are drawing two cards without replacement.

1. **Probability of drawing the first black card:**
There are 26 black cards out of 52 total cards.
\( P(\text{1st black card}) = \frac { 26 }{ 52 } = \frac { 1 }{ 2 } \)
After drawing one black card, there are 51 cards left in the pack, and 25 black cards remaining.

2. **Probability of drawing the second black card (given the first was black):**
There are 25 black cards left out of 51 total cards.
\( P(\text{2nd black card | 1st black card}) = \frac { 25 }{ 51 } \)

The probability that both cards drawn are black is the product of these two probabilities, as the events are dependent (without replacement).
\( P(\text{both black}) = P(\text{1st black}) \cdot P(\text{2nd black | 1st black}) \)
\( P(\text{both black}) = \frac { 26 }{ 52 } \cdot \frac { 25 }{ 51 } \)
\( P(\text{both black}) = \frac { 1 }{ 2 } \cdot \frac { 25 }{ 51 } \)
\( P(\text{both black}) = \frac { 25 }{ 102 } \)
This calculation shows how the probability changes after each card is removed.
In simple words: We want to pick two black cards without putting the first one back. We find the chance of picking a black card first. Then, because one black card is gone, the chance changes for the second pick. We multiply these two chances to get the final answer.

🎯 Exam Tip: For "without replacement" problems, always remember that the total number of items and the number of specific items decrease after each successful draw. This impacts the probabilities for subsequent draws.

Question 8. Two coins are tossed. What is the probability of coming up two heads if it is known that at least one head comes up?
Answer:Let S be the sample space for tossing two coins. The possible outcomes are:
\( S = \{HH, HT, TH, TT\} \)
The total number of possible outcomes is 4.

Let A be the event that "two heads come up".
\( A = \{HH\} \)
The number of outcomes in A is 1.

Let B be the event that "at least one head comes up".
\( B = \{HH, HT, TH\} \)
The number of outcomes in B is 3.

We need to find the probability of event A occurring, given that event B has already occurred. This is a conditional probability, written as \( P(A|B) \).
The formula for conditional probability is: \( P(A|B) = \frac { P(A \cap B) }{ P(B) } \)

First, find the intersection of A and B, \( A \cap B \). These are the outcomes that are in both A and B.
\( A \cap B = \{HH\} \)
The number of outcomes in \( A \cap B \) is 1.

Now, calculate the probabilities:
\( P(A \cap B) = \frac { \text{Number of outcomes in } (A \cap B) }{ \text{Total number of outcomes in S} } = \frac { 1 }{ 4 } \)
\( P(B) = \frac { \text{Number of outcomes in B} }{ \text{Total number of outcomes in S} } = \frac { 3 }{ 4 } \)

Finally, substitute these probabilities into the conditional probability formula:
\( P(A|B) = \frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } \)
\( P(A|B) = \frac { 1 }{ 3 } \)
This problem demonstrates how knowing some information can change the likelihood of an event.
In simple words: We want to know the chance of getting two heads, but only when we already know that at least one head appeared. We look at all the ways to get at least one head, and then see how many of those ways also have two heads.

🎯 Exam Tip: For conditional probability problems, clearly define the two events, determine their intersection, and then apply the formula \( P(A|B) = \frac { P(A \cap B) }{ P(B) } \). Remember that \( P(B) \) cannot be zero.

Question 9. Hindi and English newspapers. A student is selected at random :
(i) find the probability that he reads neither Hindi nor English newspapers.
(ii) If he reads Hindi newspaper, find the probability that he reads English newspaper.
(iii) If he reads English newspaper. Find the probability that he reads Hindi newspaper.
Answer:Let H be the event that a student reads Hindi newspaper.
Let E be the event that a student reads English newspaper.

From the context of similar problems, typically data is provided as percentages or fractions. Assuming 60% read Hindi, 40% read English, and 20% read both (these numbers align with the solution's calculations):
\( P(H) = \frac { 60 }{ 100 } = 0.6 \)
\( P(E) = \frac { 40 }{ 100 } = 0.4 \)
\( P(H \cap E) = \frac { 20 }{ 100 } = 0.2 \)

(i) **Probability that he reads neither Hindi nor English newspapers:**
First, find the probability that he reads at least one newspaper, which is \( P(H \cup E) \).
\( P(H \cup E) = P(H) + P(E) - P(H \cap E) \)
\( P(H \cup E) = 0.6 + 0.4 - 0.2 \)
\( P(H \cup E) = 1.0 - 0.2 \)
\( P(H \cup E) = 0.8 \)
The probability that he reads neither Hindi nor English newspapers is the complement of reading at least one.
\( P(\overline{H} \cap \overline{E}) = P(\overline{H \cup E}) = 1 - P(H \cup E) \)
\( P(\overline{H} \cap \overline{E}) = 1 - 0.8 \)
\( P(\overline{H} \cap \overline{E}) = 0.2 \)
This means 20% of students read neither newspaper. This helps understand the segment of students not engaging with either.

(ii) **If he reads Hindi newspaper, find the probability that he reads English newspaper:**
This is conditional probability: \( P(E|H) \).
\( P(E|H) = \frac { P(E \cap H) }{ P(H) } \)
\( P(E|H) = \frac { 0.2 }{ 0.6 } \)
\( P(E|H) = \frac { 2 }{ 6 } = \frac { 1 }{ 3 } \)

(iii) **If he reads English newspaper, find the probability that he reads Hindi newspaper:**
This is conditional probability: \( P(H|E) \).
\( P(H|E) = \frac { P(H \cap E) }{ P(E) } \)
\( P(H|E) = \frac { 0.2 }{ 0.4 } \)
\( P(H|E) = \frac { 2 }{ 4 } = \frac { 1 }{ 2 } \)
In simple words: We look at students and their newspaper reading habits. First, we find the chance that a student reads no newspaper at all. Then, we find the chance a student reads English if they already read Hindi, and the chance a student reads Hindi if they already read English.

🎯 Exam Tip: Conditional probability \( P(A|B) \) focuses on the reduced sample space of event B. Always ensure you correctly identify the intersection \( P(A \cap B) \) and the condition \( P(B) \).

Question 10. A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve the problem, selected at random from the book ?
Answer:Let A be the event that A solves the problem.
Let B be the event that B solves the problem.
Given:
\( P(A) = 90\% = \frac { 90 }{ 100 } = 0.9 \)
\( P(B) = 70\% = \frac { 70 }{ 100 } = 0.7 \)
We assume that A and B solving the problem are independent events.

We need to find the probability that at least one of them will solve the problem, which is \( P(A \cup B) \).
Using the formula for the union of two events:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Since A and B are independent, \( P(A \cap B) = P(A) \cdot P(B) \).
\( P(A \cap B) = 0.9 \cdot 0.7 = 0.63 \)

Now substitute this back into the union formula:
\( P(A \cup B) = 0.9 + 0.7 - 0.63 \)
\( P(A \cup B) = 1.6 - 0.63 \)
\( P(A \cup B) = 0.97 \)

Alternatively, we can use the complement rule:
\( P(\text{at least one solves}) = 1 - P(\text{neither solves}) \)
The event "neither solves" means both A does not solve AND B does not solve, which is \( P(\overline{A} \cap \overline{B}) \).
Since A and B are independent, \( \overline{A} \) and \( \overline{B} \) are also independent.
\( P(\overline{A}) = 1 - P(A) = 1 - 0.9 = 0.1 \)
\( P(\overline{B}) = 1 - P(B) = 1 - 0.7 = 0.3 \)
\( P(\overline{A} \cap \overline{B}) = P(\overline{A}) \cdot P(\overline{B}) \)
\( P(\overline{A} \cap \overline{B}) = 0.1 \cdot 0.3 = 0.03 \)
So, \( P(\text{at least one solves}) = 1 - 0.03 \)
\( P(\text{at least one solves}) = 0.97 \)
Both methods yield the same result, confirming the consistency of probability rules. The probability is quite high, which is good for problem-solving.
In simple words: A and B try to solve a math problem. We want the chance that at least one of them succeeds. It's like finding the chance that it's not the case that *both* of them fail.

🎯 Exam Tip: For "at least one" problems, using the complement method \( P(\text{at least one}) = 1 - P(\text{none}) \) is often simpler and less prone to error than directly calculating the union, especially with more than two events.

Question 11. A problem in mathematics is given to 3 students whose chances of solving it are \( \frac { 1 }{ 2 } \), \( \frac { 1 }{ 3 } \) and \( \frac { 1 }{ 4 } \). What is the probability that the problem is solved ?
Answer:Let A, B, and C be the events that the three students solve the problem, respectively.
Given probabilities:
\( P(A) = \frac { 1 }{ 2 } \)
\( P(B) = \frac { 1 }{ 3 } \)
\( P(C) = \frac { 1 }{ 4 } \)
We assume that the students solve the problem independently.

The problem is solved if at least one of the students solves it. This is \( P(A \cup B \cup C) \).
It is easier to calculate the probability that *none* of them solve the problem, and then subtract that from 1.
First, find the probabilities that each student *does not* solve the problem:
\( P(\overline{A}) = 1 - P(A) = 1 - \frac { 1 }{ 2 } = \frac { 1 }{ 2 } \)
\( P(\overline{B}) = 1 - P(B) = 1 - \frac { 1 }{ 3 } = \frac { 2 }{ 3 } \)
\( P(\overline{C}) = 1 - P(C) = 1 - \frac { 1 }{ 4 } = \frac { 3 }{ 4 } \)

Since the events are independent, the probability that none of them solve the problem is the product of their individual probabilities of not solving:
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = P(\overline{A}) \cdot P(\overline{B}) \cdot P(\overline{C}) \)
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = \frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } \cdot \frac { 3 }{ 4 } \)
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = \frac { 1 \cdot 2 \cdot 3 }{ 2 \cdot 3 \cdot 4 } \)
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = \frac { 6 }{ 24 } \)
\( P(\overline{A} \cap \overline{B} \cap \overline{C}) = \frac { 1 }{ 4 } \)

Finally, the probability that the problem is solved (at least one solves it) is:
\( P(\text{problem solved}) = 1 - P(\text{none solve it}) \)
\( P(\text{problem solved}) = 1 - \frac { 1 }{ 4 } \)
\( P(\text{problem solved}) = \frac { 3 }{ 4 } \)
It's highly likely that the problem will be solved when multiple individuals with different solving chances try independently.
In simple words: Three students try to solve a problem. We find the chance that none of them can solve it. Then, we subtract this from 1 to get the chance that at least one student solves it, which means the problem gets solved.

🎯 Exam Tip: For problems involving multiple independent events and "at least one" scenarios, calculating the probability of the complement (none of the events happening) is almost always the most efficient method.

Question 12. A bag contains 5 white and 3 black balls. Four balls are successively drawn out without replacement. What is the probability that they are alternately of different colours?
Answer:Given:
Number of white balls = 5
Number of black balls = 3
Total number of balls = \( 5 + 3 = 8 \)
Four balls are drawn successively without replacement.

We need to find the probability that the balls are alternately of different colours. There are two possible alternating sequences for 4 balls:
Case 1: White, Black, White, Black (WBWB)
Case 2: Black, White, Black, White (BWBW)

**Case 1: Probability of drawing White, Black, White, Black (WBWB)**
1. \( P(\text{1st White}) = \frac { 5 }{ 8 } \)
Remaining: 4 W, 3 B, Total 7
2. \( P(\text{2nd Black}) = \frac { 3 }{ 7 } \)
Remaining: 4 W, 2 B, Total 6
3. \( P(\text{3rd White}) = \frac { 4 }{ 6 } \)
Remaining: 3 W, 2 B, Total 5
4. \( P(\text{4th Black}) = \frac { 2 }{ 5 } \)
Remaining: 3 W, 1 B, Total 4
\( P(\text{WBWB}) = \frac { 5 }{ 8 } \cdot \frac { 3 }{ 7 } \cdot \frac { 4 }{ 6 } \cdot \frac { 2 }{ 5 } \)
\( P(\text{WBWB}) = \frac { 5 \cdot 3 \cdot 4 \cdot 2 }{ 8 \cdot 7 \cdot 6 \cdot 5 } \)
\( P(\text{WBWB}) = \frac { 120 }{ 1680 } = \frac { 1 }{ 14 } \)

**Case 2: Probability of drawing Black, White, Black, White (BWBW)**
1. \( P(\text{1st Black}) = \frac { 3 }{ 8 } \)
Remaining: 5 W, 2 B, Total 7
2. \( P(\text{2nd White}) = \frac { 5 }{ 7 } \)
Remaining: 4 W, 2 B, Total 6
3. \( P(\text{3rd Black}) = \frac { 2 }{ 6 } \)
Remaining: 4 W, 1 B, Total 5
4. \( P(\text{4th White}) = \frac { 4 }{ 5 } \)
Remaining: 3 W, 1 B, Total 4
\( P(\text{BWBW}) = \frac { 3 }{ 8 } \cdot \frac { 5 }{ 7 } \cdot \frac { 2 }{ 6 } \cdot \frac { 4 }{ 5 } \)
\( P(\text{BWBW}) = \frac { 3 \cdot 5 \cdot 2 \cdot 4 }{ 8 \cdot 7 \cdot 6 \cdot 5 } \)
\( P(\text{BWBW}) = \frac { 120 }{ 1680 } = \frac { 1 }{ 14 } \)

The events WBWB and BWBW are mutually exclusive (they cannot both happen at the same time), so we add their probabilities to find the total probability of getting alternating colours.
\( P(\text{alternating colours}) = P(\text{WBWB}) + P(\text{BWBW}) \)
\( P(\text{alternating colours}) = \frac { 1 }{ 14 } + \frac { 1 }{ 14 } \)
\( P(\text{alternating colours}) = \frac { 2 }{ 14 } = \frac { 1 }{ 7 } \)
This problem shows how to consider all possible sequences that fit the condition.
In simple words: We have a bag of white and black balls. We pick four balls one after another without putting them back. We want to find the chance that the colors switch with each pick, like white-black-white-black or black-white-black-white. We calculate the chance for each way and add them up.

🎯 Exam Tip: For "alternating" sequence problems, identify all possible valid sequences (e.g., WBWBWB... and BWBWBW...) and calculate the probability for each. If these sequences are mutually exclusive, sum their probabilities for the final answer. Remember to adjust counts for "without replacement" draws.

Question 13. Probabilities of solving a specific problem independently by A and B are \( \frac { 1 }{ 2 } \), and \( \frac { 1 }{ 3 } \) respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved,
(ii) exactly one of them solves the problem.
Answer:Let A be the event that A solves the problem.
Let B be the event that B solves the problem.
Given probabilities:
\( P(A) = \frac { 1 }{ 2 } \)
\( P(B) = \frac { 1 }{ 3 } \)
Since A and B work independently, their events are independent.

First, find the probabilities that each person *does not* solve the problem:
\( P(\overline{A}) = 1 - P(A) = 1 - \frac { 1 }{ 2 } = \frac { 1 }{ 2 } \)
\( P(\overline{B}) = 1 - P(B) = 1 - \frac { 1 }{ 3 } = \frac { 2 }{ 3 } \)

(i) **Probability that the problem is solved:**
The problem is solved if A solves it, or B solves it, or both solve it. This is \( P(A \cup B) \).
Using the complement rule, \( P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - P(\overline{A} \cap \overline{B}) \).
Since A and B are independent, \( \overline{A} \) and \( \overline{B} \) are also independent.
\( P(\overline{A} \cap \overline{B}) = P(\overline{A}) \cdot P(\overline{B}) \)
\( P(\overline{A} \cap \overline{B}) = \frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } \)
\( P(\overline{A} \cap \overline{B}) = \frac { 2 }{ 6 } = \frac { 1 }{ 3 } \)
Therefore, the probability that the problem is solved is:
\( P(\text{problem solved}) = 1 - \frac { 1 }{ 3 } = \frac { 2 }{ 3 } \)

(ii) **Probability that exactly one of them solves the problem:**
This means either A solves it and B does not, OR B solves it and A does not.
This can be written as \( P(A \cap \overline{B}) + P(\overline{A} \cap B) \).
Since A and B are independent, \( P(A \cap \overline{B}) = P(A) \cdot P(\overline{B}) \) and \( P(\overline{A} \cap B) = P(\overline{A}) \cdot P(B) \).
\( P(A \cap \overline{B}) = \frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } = \frac { 2 }{ 6 } \)
\( P(\overline{A} \cap B) = \frac { 1 }{ 2 } \cdot \frac { 1 }{ 3 } = \frac { 1 }{ 6 } \)
The probability that exactly one of them solves the problem is:
\( P(\text{exactly one solves}) = P(A \cap \overline{B}) + P(\overline{A} \cap B) \)
\( P(\text{exactly one solves}) = \frac { 2 }{ 6 } + \frac { 1 }{ 6 } \)
\( P(\text{exactly one solves}) = \frac { 3 }{ 6 } = \frac { 1 }{ 2 } \)
This breakdown shows how different combinations of events contribute to the overall probability.
In simple words: Two people try to solve a problem. First, we find the total chance the problem gets solved by at least one of them. Then, we find the chance that only one person solves it (either the first person solves and the second doesn't, or vice-versa).

🎯 Exam Tip: For "exactly one" problems, clearly define the two mutually exclusive scenarios that fit the condition (e.g., A solves and B doesn't, OR B solves and A doesn't). Then, calculate each probability and sum them up.