Class 12 Mathematics Chapter 13 Probability

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 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: 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: 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 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: 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: 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 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: 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 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 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 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 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 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: 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: 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 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.

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 $$P(A|B)$$:

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