Probability #4 - Questions and Answers

Question 1 The probability of event Aᶜ is 0.4, the probability of (A∪B)ᶜ is 0.1, and the probability of the intersection of A and B is 0.2. What's the probability of event B? Answer Step 1 Given Data: The probability of event Aᶜ is: P(Aᶜ) = 0.4. Thus, P(A) = 1-P(Aᶜ) = 1-0.4 = 0.6 The probability of event (A∪B)ᶜ is: P((A∪B)ᶜ) = 0.1 The probability of event (A∩B) is: P(A∩B) = 0.2 The probability of event A is, P(A) = 1- P(Aᶜ) = 1 - 0.4 = 0.6 Step 2 The probability of event (A∪B) is, P(A∪B) = 1 - P((A∪B)ᶜ) = 1 - 0.1 = 0.9 The probability of event A union B is, P(A∪B) = P(A) + P(B) – P(A∩B) 0.9 = 0.6 + P(B) – 0.2 P(B) = 0.9 – 0.6 + 0.2 P(B) = 0.5 Thus, the probability of event B is 0.5. Question 2 Martha invited 4 friends to go with her to the movies. There are 120 different ways in which they can sit together in a row of 5 seats, one person per seat. In how many of those ways is Martha sitting in the middle seat? Answer There are totally 5 seats in which Martha sitting in the middle (third) seat then there is only one possible chance to fill the middle seat. Now, for first seat there are four possible chance, second seat has three possible chance, fourth seat has 2 possible chance and fifth seat has one possible chance. The number of ways Martha sitting in the middle seat is, Number of ways = 4 × 3 × 1 × 2 × 1 = 24 Thus, there are 24 possible ways Martha sitting in the middle seat. Question 3 Uniform distribution is defined over the interval from 2 to 5. Find the probability of a value more than 2.6. Answer Probability of a value more than 2.6 If the value is more than 2.6, it means that the value is between 2.6 and 5. So, the base is 5- 2.6 = 2.4 P(2.6 < value < 5) = height × base = (1 / (b - a)) * (5 - 2.6) = (1 / (5 - 2)) * 2.4 = (1/3) * 2.4 = 2.4 / 3 = 0.8 Question 4 Find the probability P(E or F) if E and F are not related, P(E) = 0.41, and P(F) = 0.47. The probability P(E or F) is nothing. Answer 1 Step 1: Given information As per the given information, we have Event E and F are not able to occur at the same time, P(E) = 0.41 P(F) = 0.47 Step 2: Answer is given by, The probability P(E or F) is given by, We know that, Event E and F are not able to occur at the same time P(E and F) = 0 and, P(E or F) = P(E) + P(F) - P(E and F) P(E or F) = 0.41 + 0.47 - 0 P(E or F) = 0.88 Therefore, the probability P(E or F) is 0.88. Answer 2 Result: 0.88 Solution: P(E or F) = P(E) + P(F) - P(E and F) Since E and F are not related, P(E and F) = 0. Thus, the formula simplifies to: P(E or F) = P(E) + P(F) - 0 Given that P(E) = 0.41 and P(F) = 0.47, we can substitute these values into the formula: P(E or F) = 0.41 + 0.47 - 0 Simplifying further, we get: P(E or F) = 0.41 + 0.47 Finally, evaluating the expression: P(E or F) = 0.88 Therefore, the probability P(E or F) when E and F are not related is 0.88. Answer 3 To solve the problem, we can use the formula for the probability of the union of two events: P(E∪F) = P(E) + P(F) - P(E∩F) Given that E and F are not related, P(E∩F) = 0. Substituting the given probabilities into the formula, we have: P(E∪ F) = 0.41 + 0.47 - 0 Simplifying further: P(E∪ F) = 0.88 Therefore, the probability P(E or F) is 0.88. Answer 4 In this case, since E and F are not related, the probability of both events occurring together is zero. Therefore, the probability P(E or F) can be calculated as follows: P(E or F) = P(E) + P(F) - P(E and F) Given that P(E) = 0.41 and P(F) = 0.47, and P(E and F) = 0 (since E and F are not related), we can substitute the values into the formula: P(E or F) = 0.41 + 0.47 - 0 = 0.88 Hence, the probability P(E or F) when E and F are not related is 0.88. Question 5 There are 3.5 commercial airliner aircraft crashes on average each month around the world. What is the likelihood that (a) there will be at least two of these accidents in the upcoming month; (b) there will only be one accident in the upcoming month? Describe your thinking! Answer 1 An average of 3.5 aircraft crashes occur per month. We must determine the likelihood that at least two accidents of this nature will occur in the upcoming month. Let n represent the number of monthly flights. Of course n is big. Also let Xᵢ be a Bernoulli random variable where P(Xᵢ = 1) = p represents a probability that a plane will crash. Obviously X₁ + X₂ + ... + Xₙ ~ B(n,p) The binomial distribution is approximated by a Poisson distribution since n is very high (the number of flights is supposed to be large) and we may assume that the likelihood of crash is very tiny (it is evident). Consequently, a random variable S = X₁ + X₂ + ... + Xₙ ~ P(λ) where λ = np = E[S]. We need to calculate P(S ≥ 2). Let's remember that P(X = k) = e⁻λ λᵏ / k! if X ~ P(λ) It follows: P(S ≥ 2) = 1 - P(S = 0) – P(S = 1) = 1 - e⁻³⋅⁵ - 3.5e⁻³⋅⁵ = 1 - 0.03 - 3.5 * 0.03 = 0.86 In the (b) part we are calculating the probability of at most 1 accident happening. Similar as before we have: P(S ≤ 1) = P(S = 0) + P(S = 1) = e⁻³⋅⁵ + 3.5e⁻³⋅⁵ = 0.136 We calculate the probability for the (a) part, which is 0.86 and for the (b) part, which is 0.136. Answer 2 Step-by-step explanation: We would apply the formula for poisson distribution which is expressed as P(x = r) = (e^(-μ) * μ^r) / r! Where μ represents the mean of the theoretical distribution. r represents the number of successes of the event. From the information given, μ = 3.5 a) For the probability that there will be at least 2 such accidents in the next month, it is expressed as P(x ≥ 2) = 1 – P(x < 2) P(x<2)= P(x = 0) + P(x = 1) P(x = 0) = (e^(-3.5) * 3.5⁰) / 0! P(x = 0) = 1 P(x=0) = 0.03 P(x = 1) = (e^(-3.5) * 3.5¹) / 1! P(x=1)= (e^(-3.5) * 3.5 ) / 1 P(x = 1) = 0.11 P(x < 2) = P(x = 0) + P(x = 1) = 0.03 + 0.11 = 0.14 P(x ≥ 2) = 1 – 0.14 = 0.86 b) For the probability that there will be at most 1 accident in the next month, it is expressed as P(x ≤ 1) = P(x = 0) + P(x = 1) P(x = 0) = 0.03 P(x=1) = 0.11 P(x ≤ 1) = 0.03 + 0.11 = 0.14