Let F be the event that a student is female, M be the event that a student is male, P be the event that a student passes the exam, and Fails be the event that a student fails the exam. Given probabili

Let's break down these probability and statistics problems.

Question 2.4

Let F be the event that a student is female, M be the event that a student is male, P be the event that a student passes the exam, and Fails be the event that a student fails the exam.

Given probabilities:

• $P(P|F) = 0.70$ (70% of female students pass)
• $P(\text{Fails}|F) = 1 - P(P|F) = 1 - 0.70 = 0.30$
• $P(P|M) = 0.50$ (50% of male students pass)
• $P(\text{Fails}|M) = 1 - P(P|M) = 1 - 0.50 = 0.50$
• $P(M) = 0.40$ (40% of students are male)
• $P(F) = 1 - P(M) = 1 - 0.40 = 0.60$ (60% of students are female)

2.4.1 Calculate the probability that a student selected at random fails the June BBS1115 examination.

Step 1: Use the law of total probability.

$$P(\text{Fails}) = P(\text{Fails}|F)P(F) + P(\text{Fails}|M)P(M)$$

Step 2: Substitute the given values.

$$P(\text{Fails}) = (0.30)(0.60) + (0.50)(0.40)$$ $$P(\text{Fails}) = 0.18 + 0.20$$ $$P(\text{Fails}) = 0.38$$

The probability that a student fails is $\boxed{0.38}$.

2.4.2 Calculate the probability that a student selected at random is a female student who fails the examination.

Step 1: Use the formula for the intersection of two events.

$$P(\text{Fails} \cap F) = P(\text{Fails}|F)P(F)$$

Step 2: Substitute the given values.

$$P(\text{Fails} \cap F) = (0.30)(0.60)$$ $$P(\text{Fails} \cap F) = 0.18$$

The probability that a student is female and fails is $\boxed{0.18}$.

2.4.3 If a student selected at random is found to have failed the examination, calculate the probability that the student is male.

Step 1: Use Bayes' theorem.

$$P(M|\text{Fails}) = \frac{P(\text{Fails}|M)P(M)}{P(\text{Fails})}$$

Step 2: Substitute the values from the given information and part 2.4.1.

$$P(M|\text{Fails}) = \frac{(0.50)(0.40)}{0.38}$$ $$P(M|\text{Fails}) = \frac{0.20}{0.38}$$ $$P(M|\text{Fails}) = \frac{20}{38} = \frac{10}{19}$$ $$P(M|\text{Fails}) \approx 0.5263$$

The probability that a student who failed is male is $\boxed{\frac{10}{19} \text{ or approximately } 0.5263}$.

Question 2.5

Given:

• $P(A) = 0.6$
• $P(B) = 0.5$
• $P(A \cup B) = 0.8$

Find $P(B|A)$.

Step 1: Find $P(A \cap B)$ using the formula for the union of two events.

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ $$0.8 = 0.6 + 0.5 - P(A \cap B)$$ $$0.8 = 1.1 - P(A \cap B)$$ $$P(A \cap B) = 1.1 - 0.8$$ $$P(A \cap B) = 0.3$$

Step 2: Find $P(B|A)$ using the conditional probability formula.

$$P(B|A) = \frac{P(A \cap B)}{P(A)}$$ $$P(B|A) = \frac{0.3}{0.6}$$ $$P(B|A) = 0.5$$

The probability $P(B|A)$ is $\boxed{0.5}$.

Question 3.1

Consider the probability distribution of the random variable X:

$x$	0	1	2	3	4
$p(x)$	0.3	0.01	0.4	0.2	$c$

3.1.1 Find the value of $c$ that makes $p(x)$ a probability distribution.

Step 1: The sum of all probabilities in a probability distribution must equal 1.

$$\sum p(x) = 1$$ $$0.3 + 0.01 + 0.4 + 0.2 + c = 1$$

Step 2: Sum the known probabilities.

$$0.91 + c = 1$$

Step 3: Solve for $c$.

$$c = 1 - 0.91$$ $$c = 0.09$$

The value of $c$ is $\boxed{0.09}$.

3.1.2 Find $E(X)$ and $Var(X)$.

Step 1: Calculate the expected value $E(X)$.

$$E(X) = \sum x \cdot p(x)$$ $$$$