Probability #16 - Questions and Answers

Question 1 If you roll 5 standard six-sided dice, what's the probability that you get at least two 4s? Answer Let X denote number of die that show face '4' in n = independent rolls of die, where each has a probability p = 0.166667. We denote this by: X~ Binomial(n = 5, p = 0.166667) Probability of x number of die that show face '4' is: P(X = x) = (n choose x) * p^x * (1 - p)^(n-x) where (n choose x) is combinatorial expression: (n choose x) = n! / (x! * (n - x)!) P(X ≥ 2), n = 5, p = 0.166667 P(X ≥ 2) = 1 – P(X < 2) We first calculate: P(X < 2) = P(0 ≤ X ≤ 1) = P(X = 0, 1) = P(X = 0) + P(X = 1) P(X = 0) = (5 choose 0) * (0.166667)⁰ * (1 – 0.166667)⁵ = 1 * (0.166667)⁰ * (0.833333)⁵ ≈ 0.401878 P(X = 1) = (5 choose 1) * (0.166667)¹ * (1 – 0.166667)⁴ = 5 * (0.166667)¹ * (0.833333)⁴ ≈ 0.401878 P(X = 0) + P(X = 1) = 0.401878 + 0.401878 = 0.803756 ≈ 0.804 P(X < 2) = 0.804 P(X ≥ 2) = 1 - P(X < 2) = 1 - 0.804 = 0.196 P(X ≥ 2) = 0.196 Question 2 Spinning a wheel, the probability of landing in area A is 0.99 probability of landing in area B is 0.01. In 60 spins what is the probability of landing in area B? Answer 1 Step 1 There are only two possible sets of outcomes: all 60 spins land on A at least one spin lands on B Therefore, we can say that: P(all 60 A) + P(at least 1 B) = 1 So in order to find the chance of at least 1 spin landing on B, lets Answer 2 Step 1 First we convert the probability of Area B from dec To find the probability of landing in Area B, we multiply the probability of area B by the number of spins, so it looks like P = 0.01 * 60 where P is the probability of landing in Area B. Do the math, and the answer becomes P = 0.6 This doesn't look like the answer until you realize that it means P = 0.6 / 1 Knowing that 6/10 = 0.6, 6/10 converted to percentage = 6%, and that 6 / 100 = 6% is 6% divided by 10, one can see how 6 / 10 = 6% which makes 0.6 / 100 = 0.6% Question 3 How do you find probabilities of compound events? Answer Depends on the relation between the events. Explanation: Examplel: probability of rolling a 3 or a 5 with one throw. P(3) = 1/6 P(5) = 1/6 Since this is an exclusive OR situation we ADD the probabilities: P(3 or 5) = P(3) + P(5) = 1/6 + 1/6 = 2/6 = 1/3 Example2: probability of first rolling a 3, and then a 5 with one die. 1st roll: P(3) = 1/6 2nd roll: P(5) = 1/6 Since this is an AND situation, we MULTIPLY: P(3,5) = 1/6 * 1/6 = 1/36 Note: if the question were about a Question 4 How do you compute the variance of the probability distribution in the table provided? Outcome | Probability 50 | 0.5 51 | 0.2 52 | 0.1 53 | 0.2 Answer Var (X) = 1.4 Explanation: Let X be the random variable that represents a potential result: x | 50 | 51 | 52 | 53 P(X = x) | 0.5 | 0.2 | 0.1 | 0.2 First we quickly check that ∑P(x) = 1 which is indeed the case. The, we calculate xP(x), and x²P(x): x | 50 | 51 | 52 | 53 P(X = x) | 0.5 | 0.2 | 0.1 | 0.2 xP(X = x) | 25 | 10.2 | 5.2 | 10.6 x² | 2500 | 2601 | 2704 | 2809 x²P(X = x) | 1250 | 520.2 | 270.4 | 561.8 So then the Expectation is calculated using: E(X) = Σ xP(x) = 25 + 10.2 + 5.2 + 10.6 = 51 Next prior to calculating the Variance we calculate E(x²): E(x²) = Σ x²P(x) = 1250 + 520.2 + 270.4 + 561.8 = 2602.4 After that, we can determine the variance: Var (X) = E(X²) – E²(X) = 2602.4 - (51)² = 2602.4 - 2601 = 1.4 We can also calculate the Standard Deviation (if required); as σ² = Var (X) ⇒ σ = 1.18 (3sf)