Probability #18 - Questions and Answers

Question 1 Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. What is the probability P(X>42)? Answer Step 1 We must standardise the Random Variable X with the Standardised Normal Distribution Z Variable using the relationship: Z = (X - μ) / σ And we will use Normal Distribution Tables of the function: Φ(z) = P(Z < z) And so we get: P(X>42) = P(Z > (42 - 50) / 7) = P(Z > -8/7) = P(Z > -1.1429) If we look at this graphically it is the shaded part of this Standardised Normal Distribution: (Graph) By symmetry of the Standardised Normal Distribution it is the same as this shaded part: (Graph) So: P(Z>42) = P(Z > -1.1429) = 1 − P(Z < 1.1429) = 1 – Φ(1.1429) = 1 -0.8729 (from tables) = 0.1271 Question 2 What is the mathematical definition of conditional probability? Answer A probability that can only be possible after another condition exists. See also "Bayesian Statistics". Explanation: An example would be that "B" has a probability of 40%, but ONLY if "A" occurs. If the probability of "A" occurring is 60% the conditional probability of B occurring is 0.60 * 0.40 = 0.24, or 24%. Question 3 What is the probability of drawing 2 cards in succession (without replacement) from a standard deck of playing cards and having them both be hearts? Answer Step 1 Because the results of the first draw have an impact on the second, the events are interconnected. Thus, we have: P(heart and heart) = P(heart) * P(heart then heart) The probability that a heart card is draw on the first draw is: P(heart) = 13/52 = 1/4 The probability that a heart card is draw on the second draw is: P(heart) = 12/51 = 4/17 Step 2 Therefore, the probability of drawing 2 hearts in succession without replacement is: P(heart and heart) = (1/4) * (12/51) = (1/4) * (4/17) = 1/17 P(heart and heart) = 1/17 Question 4 How likely is it that a seven-digit phone number will include the number 7? Answer We start by locating the complement of the number 7, which is not a phone number with the number 7, since we assume it must appear at least once. The probability of not having a 7 is 9/10 (there are 10 possible digits). Hence, P(no 7) = (9/10) * (9/10) * (9/10) * (9/10) * (9/10) * (9/10) * (9/10) = (9/10)^7 Hence, the probability that there is at least 1 number 7 is: P(at least one 7) = 1 - P(no 7) P(at least one 7) = 1 - (9/10)^7 P(at least one 7) ≈ 0.522 → 52.2% Question 5 A factory has three machines, A, B, and C, for producing items. Machine A produces 50%, B produces 30%, and C produces 20%. If a randomly selected item from the factory's output is found to be defective, what is the probability that B made it? Answer Step 1 Prior information Probability of an item produced by Machine A: P(A) = 50% = 0.5 Probability of an item produced by Machine B: P(B) = 30% = 0.3 Probability of an item produced by Machine C: P(C) = 20% = 0.2 Considering additional information The likelihood that the product made by A is flawed is: P(D|A) = 3% = 0.03 The likelihood that the product made by B is flawed is: P(D|B) = 2% = 0.02 The likelihood that the product made by C is flawed is: P(D|C) = 1% = 0.01 Given the item is defective, probability of it being produced by B P(B|D) = (P(B) * P(D|B)) / ((P(B) * P(D|B)) + (P(A) * P(D|A)) + (P(C) * P(D|C))) = (0.3 * 0.02) / ((0.3 * 0.02) + (0.5 * 0.03) + (0.2 * 0.01)) = 0.006 / (0.006 + 0.015 + 0.002) = 0.006 / 0.023 = 0.26 Given the item is defective, probability of it being produced by B = 0.26 Question 6 What is the sample space of rolling a 6-sided die? Answer Step 1 Its {1,2,3,4,5,6} which is actually a set of all the possible outcomes as the definition of sample space specifies. The number of dots on the top face of a six-sided die is the result. Now, the outcome of rolling a die is one of 1, 2, 3, 4, or 6 dots appearing on the topmost face. So experiment here is "Rolling a 6 faced dice" and list of possible outcomes is {1,2,3,4,5,6} Sample space by its definition is list of all the possible outcomes of an experiment. So answer to your question is S = {1, 2, 3, 4, 5, 6}