Class 12 Mathematics Chapter 13 Probability

Key Concepts: Complex Chain, Multi-Step Problem

b) 0.069

[Solution Description] We need to find the probability that an item is defective or improperly labeled. This can be calculated using the formula:

$$P(\text{Defective or Improperly Labeled}) = P(D) + P(L|D) – P(D \cap L)$$

where:

\[
\begin{align*}
&P(D) = 0.05 \\
&P(\neg D) = 1 – P(D) = 0.95 \\
&P(L|D) = 0.10 \\
&P(L|\neg D) = 0.02
\end{align*}
\]

The probability of an item being improperly labeled regardless of defect status is calculated as:

$$P(L) = P(L|D)P(D) + P(L|\neg D)P(\neg D)$$

Calculating each term:

\[
\begin{align*}
P(L|D)P(D) &= 0.10 \times 0.05 = 0.005 \\
P(L|\neg D)P(\neg D) &= 0.02 \times 0.95 = 0.019
\end{align*}
\]

Therefore:

$$P(L) = 0.005 + 0.019 = 0.024$$

Now, using the addition rule for probabilities:

$$P(\text{Defective or Improperly Labeled}) = P(D) + P(L) – P(D \cap L)$$

Since $$P(D \cap L) = P(D) \cdot P(L|D)$$:

$$P(D \cap L) = 0.05 \times 0.10 = 0.005$$

Thus:

$$P(\text{Defective or Improperly Labeled}) = 0.05 + 0.024 – 0.005 = 0.069$$

Key Concept: Multiplication Rule

b) 0.15
[Solution Description] For independent events $$A$$ and $$B$$, the multiplication rule states that $$P(A \cap B) = P(A) \cdot P(B)$$. Here, $$P(A \cap B) = 0.3 \times 0.5 = 0.15$$.

Key Concepts: Complex Event Relationships, Conditional Probability with Total Probability

c) 0.036

[Solution Description]

To find $$P(A \cap B \cap C \cap D)$$, we need to multiply the probabilities together since these are conditional probabilities leading to the base event $$D$$.

We start by finding $$P(A \cap B \cap C \cap D)$$ using:

$$P(A \cap B \cap C \cap D) = P(A|B) \cdot P(B|C) \cdot P(C|D) \cdot P(D)$$

Substituting the given values:

$$P(A \cap B \cap C \cap D) = 0.3 \times 0.4 \times 0.5 \times 0.6$$

Calculating step-by-step:

$$0.3 \times 0.4 = 0.12$$

$$0.12 \times 0.5 = 0.06$$

$$0.06 \times 0.6 = 0.036$$

Thus, $$P(A \cap B \cap C \cap D) = 0.036$$.

Key Concept: Formula Application

b) 0.4
[Solution Description]
For independent events E and F, we find the probability of both events happening by multiplying their individual probabilities:

$$
P(E \cap F) = P(E) \cdot P(F) = 0.8 \times 0.5 = 0.4
$$

Therefore, the probability of both events occurring is 0.4.

Key Concepts: Conditional Probability

c) 0.275

[Solution Description] To find $$P(C)$$, use the law of total probability:

\[
\begin{aligned}
P(C) &= P(C|B)P(B) + P(C \cap B’) \\
&= 0.25 \times 0.7 + 0.1 \\
&= 0.175 + 0.1 \\
&= 0.275
\end{aligned}
\]

Therefore, $$P(C) = 0.275$$.

Key Concept: Real-World Scenarios

c) $$\frac{2}{9}$$
[Solution Description] Let $$R_1$$ be the event that the first ball is red, and $$R_2$$ be the event that the second ball is also red after drawing the first one. We need to find $$P(R_1|R_2)$$.

Since the problem involves conditional probability with dependent events, use:
$$
P(R_2|R_1) = \frac{P(R_1 \cap R_2)}{P(R_1)}
$$
$$
P(R_1 \cap R_2) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}
$$
Since $$P(R_1) = \frac{3}{10}$$,
$$
P(R_2|R_1) = \frac{\frac{1}{15}}{\frac{3}{10}}
$$
$$
= \frac{1}{15} \times \frac{10}{3} = \frac{2}{9}
$$
Thus, the probability is $$\frac{2}{9}$$.

Key Concept: Property Identification

b) $$P(S|F) = 1$$
[Solution Description] For any event F, we have $$P(F|F) = 1$$. This means the conditional probability of an event given itself is always 1.

Key Concept: Conditional Probability

b) 0.35
[Solution Description] To find $$P(E_1|A)$$, we use Bayes’ Theorem:

$$
P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}
$$

Substitute the given values:

$$
P(E_1|A) = \frac{0.3 \times 0.5}{0.3 \times 0.5 + 0.7 \times 0.4}
$$

Calculate the numerator and denominator:

$$
P(E_1|A) = \frac{0.15}{0.15 + 0.28} = \frac{0.15}{0.43} \approx 0.3488
$$

So, $$P(E_1|A) \approx 0.35$$.

Key Concept: Basic Application

a) Yes
[Solution Description] The outcome of rolling a die does not affect the outcome of flipping a coin. Hence, A and B are independent as the occurrence of one does not change the probability of the other.

Key Concepts: Real-world Scenario

b) 0.067

[Solution Description] Let $$D$$ be the event that the disease is present and $$T+$$ be the event that the test is positive.

Using the theorem of total probability:

$$P(T+) = P(D) \cdot P(T+|D) + P(D’) \cdot P(T+|D’)$$

We have $$P(D) = 0.02$$, $$P(T+|D) = 0.9$$, $$P(D’) = 0.98$$, and $$P(T+|D’) = 0.05$$.

Calculate:

$$P(T+) = 0.02 \times 0.9 + 0.98 \times 0.05 = 0.018 + 0.049 = 0.067$$

So, the probability that a randomly chosen person tests positive is 0.067.

Key Concept: Basic Real-World Application

b) 0.3
[Solution Description]
Let $$D$$, $$A$$, and $$M$$ represent membership in the Drama, Art, and Music clubs respectively. Let $$E$$ be the event that a member attended the event.

Find $$P(D|E)$$.

Given:
$$P(E|D) = \frac{30}{100} = 0.3$$, $$P(E|A) = \frac{50}{100} = 0.5$$, $$P(E|M) = \frac{20}{100} = 0.2$$

Assuming equal probability of being in any club, $$P(D) = P(A) = P(M) = \frac{1}{3}$$.

Calculate total probability of attending the event:

$$
P(E) = P(E|D)P(D) + P(E|A)P(A) + P(E|M)P(M)
$$

Substitute the values:

$$
P(E) = (0.3)(\frac{1}{3}) + (0.5)(\frac{1}{3}) + (0.2)(\frac{1}{3})
$$

$$
P(E) = \frac{0.3}{3} + \frac{0.5}{3} + \frac{0.2}{3} = \frac{1.0}{3} = \frac{1}{3}
$$

Now use Bayes’ theorem:

$$
P(D|E) = \frac{P(E|D)P(D)}{P(E)}
$$

$$
P(D|E) = \frac{(0.3)(\frac{1}{3})}{\frac{1}{3}} = 0.3
$$

Hence, the probability that a member belongs to the Drama club given they attended the event is 0.3.

Key Concepts: Real-World Application

b) 0.6667

[Solution Description]

Let $$E_1$$ and $$E_2$$ be the events of having the disease and not having the disease, respectively; and let $$A$$ be testing positive. We want $$P(E_1|A)$$:

$$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$$

Given: $$P(E_1)=0.02$$, $$P(E_2)=0.98$$, $$P(A|E_1)=0.98$$, $$P(A|E_2)=0.01$$.

$$P(E_1|A) = \frac{0.02 \times 0.98}{0.02 \times 0.98 + 0.98 \times 0.01} = \frac{0.0196}{0.0196 + 0.0098} = \frac{0.0196}{0.0294} \approx 0.6667$$

Therefore, the probability that the person actually has the disease is approximately 66.67%.

Key Concept: Independent Events

b) $$P(E \cap F) = P(E) \cdot P(F)$$
[Solution Description] For two events to be independent, their joint probability equals the product of their individual probabilities. Thus, the condition for independence is:
$$
P(E \cap F) = P(E) \cdot P(F)
$$

Key Concept: Basic Probability

b) $$\frac{1}{13}$$
[Solution Description]
There are 4 kings in a standard deck of 52 playing cards. The probability of drawing a king is calculated as the number of favorable outcomes divided by the total number of outcomes, which is $$\frac{4}{52}$$. Simplifying this fraction gives:

$$
\text{Probability} = \frac{4}{52} = \frac{1}{13}.
$$
So, the probability is $$\frac{1}{13}$$.

Key Concept: Conditional Probability

b) 0.161
[Solution Description]
Let $$A$$ be the event of having the disease and $$B$$ be the event of testing positive. We are given:

– The probability of testing positive given having the disease, $$P(B|A) = 0.95$$.
– Prevalence of the disease, $$P(A) = 0.01$$.
– The probability of not having the disease and testing positive, $$P(B|\bar{A}) = 0.05$$. Therefore, $$P(\bar{A}) = 0.99$$.

Using Bayes’ theorem:

$$
P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}
$$

where

$$
P(B) = P(B|A) \cdot P(A) + P(B|\bar{A}) \cdot P(\bar{A})
$$

Calculating $$P(B)$$:

$$
P(B) = 0.95 \times 0.01 + 0.05 \times 0.99 = 0.0095 + 0.0495 = 0.059
$$

Finally, calculating $$P(A|B)$$:

$$
P(A|B) = \frac{0.95 \times 0.01}{0.059} \approx 0.161
$$

Key Concept: Partition of Sample Space

b) $$\{E_1, E_2, E_3\}$$
[Solution Description] A set of events $$\{E_1, E_2, E_3\}$$ forms a partition of a sample space $$S$$ if every outcome in $$S$$ is in exactly one of these events and they are mutually exclusive. This means that their union covers all of $$S$$ and any intersection between two different events is empty. The condition $$E_1 \cup E_2 \cup E_3 = S$$ and $$E_i \cap E_j = \emptyset$$ for $$i \neq j$$ fulfills this definition.

Key Concepts: Addition Rule

b) 0.6

[Solution Description]

The addition rule for probability states:

$$P(C \cup D) = P(C) + P(D) – P(C \cap D)$$

Substituting the given probabilities:

$$P(C \cup D) = 0.3 + 0.4 – 0.1 = 0.6$$

Therefore, the probability $$P(C \cup D)$$ is 0.6.

Key Concept: Terminology Understanding

b) Posterior probability
[Solution Description]
In Bayes’ theorem, the “posterior probability” refers to the updated probability of an event occurring after taking into account new evidence or information. It’s the revised probability obtained by applying the theorem.