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(Fails|F) = 1 - P(P|F) = 1 - 0.70 = 0.30 P(P|M) = 0.50 (50% of male students pass) P(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(Fails) = P(Fails|F)P(F) + P(Fails|M)P(M) Step 2: Substitute the given values. P(Fails) = (0.30)(0.60) + (0.50)(0.40) P(Fails) = 0.18 + 0.20 P(Fails) = 0.38 The probability that a student fails is 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(Fails F) = P(Fails|F)P(F) Step 2: Substitute the given values. P(Fails F) = (0.30)(0.60) P(Fails F) = 0.18 The probability that a student is female and fails is 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|Fails) = P(Fails|M)P(M)P(Fails) Step 2: Substitute the values from the given information and part 2.4.1. P(M|Fails) = ((0.50)(0.40))/(0.38) P(M|Fails) = (0.20)/(0.38) P(M|Fails) = (20)/(38) = (10)/(19) P(M|Fails) ≈ 0.5263 The probability that a student who failed is male is (10)/(19) or approximately 0.5263. Question 2.5 Given: P(A) = 0.6 P(B) = 0.5 P(A B) = 0.8 Find P(B|A). Step 1: Find P(A B) using the formula for the union of two events. P(A B) = P(A) + P(B) - P(A B) 0.8 = 0.6 + 0.5 - P(A B) 0.8 = 1.1 - P(A B) P(A B) = 1.1 - 0.8 P(A B) = 0.3 Step 2: Find P(B|A) using the conditional probability formula. P(B|A) = (P(A B))/(P(A)) P(B|A) = (0.3)/(0.6) P(B|A) = 0.5 The probability P(B|A) is 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. 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 0.09. 3.1.2 Find E(X) and Var(X). Step 1: Calculate the expected value E(X). E(X) = x · p(x) E(X) = (0 · 0.3) + (1 · 0.01) + (2 · 0