Question

what is the probability that selected person is a female?

Answer 1

Concept:
• Probability of an event \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

Step 1: Identify the given data
Total number of smokers in the colony = 50
Number of male smokers = 30
Total number of females = Total smokers - Male smokers
\[ \text{Number of females} = 50 - 30 = 20 \]

Step 2: Calculate the probability
Let \( F \) be the event that the selected person is a female.
\[ P(F) = \frac{\text{Number of females}}{\text{Total smokers}} \]
\[ P(F) = \frac{20}{50} = \frac{2}{5} \]
In decimal form:
\[ P(F) = 0.4 \]

Answer 2

Concept:
• The probability of the complement of an event is \( P(E') = 1 - P(E) \).

Step 1: Find the probability of suffering from lung problems
Given that 170 in 1000 males develop complications.
Let \( C \) be the event of having lung complications given the person is male.
\[ P(C|M) = \frac{170}{1000} = 0.17 \]

Step 2: Calculate the probability of not suffering from lung problems
The probability of not having complications is the complement of having them:
\[ P(C'|M) = 1 - P(C|M) \]
\[ P(C'|M) = 1 - 0.17 \]
\[ P(C'|M) = 0.83 \]

Answer 3

Concept:
• Bayes' Theorem: \( P(A|B) = \frac{P(A) \cdot P(B|A)}{P(B)} \).

Step 1: Define the events and their probabilities
Let \( M \): Selected person is Male, \( F \): Selected person is Female.
Let \( C \): Selected person has lung complications.
\( P(M) = 30/50 = 0.6 \)
\( P(F) = 20/50 = 0.4 \)
Conditional probabilities given from the study:
\( P(C|M) = 170/1000 = 0.17 \)
\( P(C|F) = 120/1000 = 0.12 \)

Step 2: Calculate the total probability of complications \( P(C) \)
Using the theorem of total probability:
\[ P(C) = P(M)P(C|M) + P(F)P(C|F) \]
\[ P(C) = (0.6 \times 0.17) + (0.4 \times 0.12) \]
\[ P(C) = 0.102 + 0.048 = 0.150 \]

Step 3: Apply Bayes' Theorem to find \( P(F|C) \)
\[ P(F|C) = \frac{P(F) \cdot P(C|F)}{P(C)} \]
\[ P(F|C) = \frac{0.4 \times 0.12}{0.150} = \frac{0.048}{0.150} \]
\[ P(F|C) = \frac{48}{150} = \frac{8}{25} = 0.32 \]

Answer 4

Concept:
• Bayes' Theorem applied to the complement of the complications event.

Step 1: Define the events for 'no complications'
Let \( C' \) be the event that the person does not have lung complications.
\( P(C'|M) = 1 - 0.17 = 0.83 \)
\( P(C'|F) = 1 - 0.12 = 0.88 \)
Probabilities of selection:
\( P(M) = 0.6 \), \( P(F) = 0.4 \)

Step 2: Calculate total probability of no complications \( P(C') \)
\[ P(C') = P(M)P(C'|M) + P(F)P(C'|F) \]
\[ P(C') = (0.6 \times 0.83) + (0.4 \times 0.88) \]
\[ P(C') = 0.498 + 0.352 = 0.850 \]

Step 3: Apply Bayes' Theorem to find \( P(M|C') \)
\[ P(M|C') = \frac{P(M) \cdot P(C'|M)}{P(C')} \]
\[ P(M|C') = \frac{0.498}{0.850} = \frac{498}{850} = \frac{249}{425} \]