Samacheer Kalvi Class 11 Maths Solutions Chapter 12 Introduction to Probability Theory Exercise 12.2

Question 1. If A and B are mutually exclusive events P(A) = \( \frac{3}{8} \) and P(B) = \( \frac{1}{8} \), then find
(i) P(\( \overline{A} \))
(ii) P(A \( \cup \) B)
(iii) P(\( \overline{A} \) \( \cap \) B)
(iv) P(\( \overline{A} \) \( \cup \) \( \overline{B} \))
Answer:
Given that A and B are mutually exclusive events.
P(A) = \( \frac{3}{8} \)
P(B) = \( \frac{1}{8} \)

(i) To find P(\( \overline{A} \)):
We know that P(\( \overline{A} \)) = 1 - P(A)
P(\( \overline{A} \)) = \( 1 - \frac{3}{8} \)
\( \implies \) P(\( \overline{A} \)) = \( \frac{8-3}{8} \)
\( \implies \) P(\( \overline{A} \)) = \( \frac{5}{8} \)

(ii) To find P(A \( \cup \) B):
For mutually exclusive events, P(A \( \cup \) B) = P(A) + P(B)
P(A \( \cup \) B) = \( \frac{3}{8} + \frac{1}{8} \)
\( \implies \) P(A \( \cup \) B) = \( \frac{3+1}{8} \)
\( \implies \) P(A \( \cup \) B) = \( \frac{4}{8} \)
\( \implies \) P(A \( \cup \) B) = \( \frac{1}{2} \)

(iii) To find P(\( \overline{A} \) \( \cap \) B):
Since A and B are mutually exclusive, their intersection is an empty set (Φ), meaning P(A \( \cap \) B) = 0.
We know that P(\( \overline{A} \) \( \cap \) B) = P(B) - P(A \( \cap \) B)
\( \implies \) P(\( \overline{A} \) \( \cap \) B) = \( \frac{1}{8} - 0 \)
\( \implies \) P(\( \overline{A} \) \( \cap \) B) = \( \frac{1}{8} \)

(iv) To find P(\( \overline{A} \) \( \cup \) \( \overline{B} \)):
By De Morgan's Law, P(\( \overline{A} \) \( \cup \) \( \overline{B} \)) = P(\( \overline{A \cap B} \))
We know that P(\( \overline{A \cap B} \)) = 1 - P(A \( \cap \) B)
Since A and B are mutually exclusive, P(A \( \cap \) B) = 0.
So, P(\( \overline{A} \) \( \cup \) \( \overline{B} \)) = 1 - 0
\( \implies \) P(\( \overline{A} \) \( \cup \) \( \overline{B} \)) = 1
In simple words: When events cannot happen at the same time, we call them mutually exclusive. We use basic probability rules like finding the chance of an event not happening (complement) and the chance of either event happening (union) by adding their individual chances. De Morgan's Law helps simplify finding the probability of both events not happening.

🎯 Exam Tip: Always remember that for mutually exclusive events, P(A \( \cap \) B) is always 0. This is a key concept in probability problems.

Question 2. If A and B are two events associated with a random experiment for which P(A) = 0.35, P(A \( \cup \) B) = 0.85, and P(A \( \cap \) B) = 0.15, find
(i) P(only B)
(ii) P(B)
(iii) P(only A)
Answer:
Given:
P(A) = 0.35
P(A \( \cap \) B) = 0.15
P(A \( \cup \) B) = 0.85

(ii) To find P(B):
We use 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:
0.85 = 0.35 + P(B) - 0.15
Now, rearrange the equation to find P(B):
0.85 = 0.20 + P(B)
P(B) = 0.85 - 0.20
P(B) = 0.65

(i) To find P(only B):
P(only B) means the probability that event B occurs, but event A does not occur. This is written as P(\( \overline{A} \) \( \cap \) B).
We use the formula: P(\( \overline{A} \) \( \cap \) B) = P(B) - P(A \( \cap \) B)
We found P(B) = 0.65 and P(A \( \cap \) B) = 0.15
P(only B) = 0.65 - 0.15
P(only B) = 0.50

(iii) To find P(only A):
P(only A) means the probability that event A occurs, but event B does not occur. This is written as P(A \( \cap \) \( \overline{B} \)).
We use the formula: P(A \( \cap \) \( \overline{B} \)) = P(A) - P(A \( \cap \) B)
We are given P(A) = 0.35 and P(A \( \cap \) B) = 0.15
P(only A) = 0.35 - 0.15
P(only A) = 0.20
In simple words: We are given some probabilities about two events. We use a formula to find the probability of B, then use other formulas to find the chance of only A happening, or only B happening. It's like finding specific parts of how often things occur.

🎯 Exam Tip: Remember the basic probability formula P(A \( \cup \) B) = P(A) + P(B) - P(A \( \cap \) B) as it's fundamental. Also, P(only A) is P(A) - P(A \( \cap \) B).

Question 2. A die is thrown twice. Let A be the event, 'First die shows 5' and B be the event, 'second die shows 5'. Find P(A \( \cup \) B).
Answer:
When a die is thrown twice, the total number of possible outcomes (sample space S) is 6 \( \times \) 6 = 36.
The sample space S is:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
So, n(S) = 36.

Let A be the event that the first die shows 5.
A = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
The number of outcomes for event A is n(A) = 6.
The probability of event A is P(A) = \( \frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6} \).

Let B be the event that the second die shows 5.
B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
The number of outcomes for event B is n(B) = 6.
The probability of event B is P(B) = \( \frac{n(B)}{n(S)} = \frac{6}{36} = \frac{1}{6} \).

Now, let's find the intersection of A and B, which means both events happen (first die shows 5 AND second die shows 5).
A \( \cap \) B = {(5, 5)}
The number of outcomes for A \( \cap \) B is n(A \( \cap \) B) = 1.
The probability of A \( \cap \) B is P(A \( \cap \) B) = \( \frac{n(A \cap B)}{n(S)} = \frac{1}{36} \).

Finally, to find P(A \( \cup \) B), we use the formula:
P(A \( \cup \) B) = P(A) + P(B) - P(A \( \cap \) B)
P(A \( \cup \) B) = \( \frac{1}{6} + \frac{1}{6} - \frac{1}{36} \)
To add and subtract these fractions, find a common denominator, which is 36.
P(A \( \cup \) B) = \( \frac{6}{36} + \frac{6}{36} - \frac{1}{36} \)
\( \implies \) P(A \( \cup \) B) = \( \frac{6+6-1}{36} \)
\( \implies \) P(A \( \cup \) B) = \( \frac{11}{36} \)
In simple words: We list all possible results when rolling two dice. Then, we find all results where the first die is a 5 and all results where the second die is a 5. Since some results are in both lists (like 5,5), we count those only once when finding the chance of either event happening.

🎯 Exam Tip: When dealing with combined events like rolling two dice, always define the sample space carefully and clearly list the outcomes for each event and their intersection to avoid errors.

Question 4. The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of
(i) P(A \( \cup \) B)
(ii) P(\( \overline{A} \) \( \cap \) B)
(iii) P(A \( \cap \) \( \overline{B} \))
Answer:
Given:
P(A) = 0.5
P(B) = 0.3
Events A and B are mutually exclusive, which means they cannot happen at the same time. Therefore, P(A \( \cap \) B) = 0.

(i) To find P(A \( \cup \) B):
For mutually exclusive events, the probability of A or B occurring is the sum of their individual probabilities:
P(A \( \cup \) B) = P(A) + P(B)
P(A \( \cup \) B) = 0.5 + 0.3
P(A \( \cup \) B) = 0.8

(ii) To find P(\( \overline{A} \) \( \cap \) B):
This represents the probability that event B occurs, but event A does not occur. Since A and B are mutually exclusive, if B occurs, A cannot occur.
P(\( \overline{A} \) \( \cap \) B) = P(B) - P(A \( \cap \) B)
P(\( \overline{A} \) \( \cap \) B) = 0.3 - 0
P(\( \overline{A} \) \( \cap \) B) = 0.3

(iii) To find P(A \( \cap \) \( \overline{B} \)):
This represents the probability that event A occurs, but event B does not occur. Similar to the previous part, if A occurs, B cannot occur because they are mutually exclusive.
P(A \( \cap \) \( \overline{B} \)) = P(A) - P(A \( \cap \) B)
P(A \( \cap \) \( \overline{B} \)) = 0.5 - 0
P(A \( \cap \) \( \overline{B} \)) = 0.5
In simple words: If two events cannot happen together, we find the chance of either happening by just adding their individual chances. The chance of one happening but not the other is simply the chance of that one event happening, because the other one won't happen anyway.

🎯 Exam Tip: When events are mutually exclusive, their intersection (A \( \cap \) B) is always 0. This simplifies many calculations for union and complements.

Question 5. A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96.
(i) What is the probability that a fire engine is available when needed?
(ii) What is the probability that neither is available when needed?
Answer:
Let A be the event that the first fire engine is available.
Let B be the event that the second fire engine is available.
Given that the fire engines operate independently.
P(A) = 0.96 (Probability that the first engine is available)
P(B) = 0.96 (Probability that the second engine is available)

First, let's find the probability that an engine is *not* available.
P(\( \overline{A} \)) = 1 - P(A) = 1 - 0.96 = 0.04
P(\( \overline{B} \)) = 1 - P(B) = 1 - 0.96 = 0.04
Since the engines operate independently, the non-availability events are also independent.

(i) Probability that at least one engine is available:
This is found by 1 - (Probability that *no* engine is available).
P(at least one engine is available) = 1 - P(neither is available)
P(at least one engine is available) = 1 - P(\( \overline{A} \) \( \cap \) \( \overline{B} \))
Since \( \overline{A} \) and \( \overline{B} \) are independent, P(\( \overline{A} \) \( \cap \) \( \overline{B} \)) = P(\( \overline{A} \)) \( \times \) P(\( \overline{B} \))
P(\( \overline{A} \) \( \cap \) \( \overline{B} \)) = 0.04 \( \times \) 0.04 = 0.0016
P(at least one engine is available) = 1 - 0.0016 = 0.9984

(ii) Probability that neither is available when needed:
This is the probability that the first engine is not available AND the second engine is not available. This is P(\( \overline{A} \) \( \cap \) \( \overline{B} \)).
Since the events are independent, P(\( \overline{A} \) \( \cap \) \( \overline{B} \)) = P(\( \overline{A} \)) \( \times \) P(\( \overline{B} \))
P(\( \overline{A} \) \( \cap \) \( \overline{B} \)) = 0.04 \( \times \) 0.04
P(\( \overline{A} \) \( \cap \) \( \overline{B} \)) = 0.0016
In simple words: We are given the chance that a fire engine is ready to use. We first find the chance that an engine is *not* ready. Because the engines work on their own, we can multiply these chances to find the chance that *both* are not ready. Then, to find the chance that at least one is ready, we subtract the "neither is ready" chance from 1.

🎯 Exam Tip: For independent events, remember that the probability of both happening is the product of their individual probabilities: P(A \( \cap \) B) = P(A) \( \times \) P(B).

Question 6. The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that
(i) it will get at least one of the two awards
(ii) it will get only one of the awards.
Answer:
Let A be the event that the bridge gets an award for its design.
Let B be the event that the bridge gets an award for the efficient use of materials.
Given:
P(A) = 0.48
P(B) = 0.36
P(A \( \cap \) B) = 0.2 (Probability of getting both awards)

(i) Probability that it will get at least one of the two awards:
This is the probability of A or B occurring, 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)
P(A \( \cup \) B) = 0.48 + 0.36 - 0.2
P(A \( \cup \) B) = 0.84 - 0.2
P(A \( \cup \) B) = 0.64

(ii) Probability that it will get only one of the awards:
This means either A occurs but B does not (P(A \( \cap \) \( \overline{B} \))), OR B occurs but A does not (P(\( \overline{A} \) \( \cap \) B)). These two events are mutually exclusive.
So, P(only one award) = P(A \( \cap \) \( \overline{B} \)) + P(\( \overline{A} \) \( \cap \) B)
We know that: P(A \( \cap \) \( \overline{B} \)) = P(A) - P(A \( \cap \) B)
And: P(\( \overline{A} \) \( \cap \) B) = P(B) - P(A \( \cap \) B)
Substitute these into the "only one award" formula:
P(only one award) = (P(A) - P(A \( \cap \) B)) + (P(B) - P(A \( \cap \) B))
P(only one award) = (0.48 - 0.2) + (0.36 - 0.2)
P(only one award) = 0.28 + 0.16
P(only one award) = 0.44
In simple words: We know the chances of a bridge getting awards for design, materials, or both. To find the chance of at least one award, we add the individual chances and subtract the chance of getting both. To find the chance of getting only one award, we add the chances of getting design-only and materials-only awards.

🎯 Exam Tip: "At least one" often means using the union formula, while "only one" requires finding the probabilities of (A and not B) plus (B and not A).