Questions and Answers #5 Probability

Questions and Answers Sheet 5 Probability Question #1 Suppose A and B are independent events. If P(A)=0.4 are P(B)=0.1, what is 𝑃(𝐴 ∩ 𝐵)? Answer: If events A and B' are independent then: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵) Thus 𝑃(𝐴 ∩ 𝐵) = (0.4)(0.1) = 0.04 Question #2 What does it mean about events A and B if P(A|B)=P(B|A)? Answer: Note that P(A|B)=P(B|A) ⇒ 𝑃(𝐴∩𝐵) 𝑃(𝐵) = 𝑃(𝐴∩𝐵) 𝑃(𝐴) ⇒ 𝑃(𝐴) = 𝑃(𝐵) So we can conclude probabilities of event A and B are same. Question #3 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: 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 On the basis of additional information Given the item produced by A, probability of it being defective 𝐷 𝑃 ( ) = 3% = 0.03 𝐴 Given the item produced by B, probability of it being defective 𝐷 𝑃 ( ) = 2% = 0.02 𝐵 Given the item produced by C, probability of it being defective 𝐷 𝑃 ( ) = 1% = 0.01 𝐶 Given the item is defective, probability of it being produced by B 𝐵 𝑃( ) = 𝐷 𝐷 𝑃(𝐵)×𝑃(𝐵) 𝐷 𝐷 𝐷 [𝑃(𝐵)×𝑃( )]+[𝑃(𝐴)×𝑃( )]+[𝑃(𝐶)×𝑃( )] 𝐵 𝐴 𝐶 0.3×0.02 = [0.3×0.02]+[0.5\𝑡𝑖𝑚𝑒0.03]+[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 #4 I have two servers at a grocery store. P(server 1) is 0.4. P(server 2) is 0.6. P(time is less than 5 min | 4 server 1) is 1. P(time is less than 5 min | server 2) is . What is P(server 1 | time is less than 5 min)? 9 Answer: Suppose that, the events Server 1=A, Serve 2=B, &, Time less than 5 minutes =C We are given that. 4 𝑃(𝐴) = 0.4, 𝑃(𝐵) = 0.6, 𝑃(𝐶|𝐴) = 1, 𝑃(𝐶|𝐵) = 9 Then, 𝑃(𝐶) = 𝑃(𝐴)𝑃(𝐶|𝐴) + 𝑃(𝐵)𝑃(𝐶|𝐵) 4 6 4 = ×1+ × 10 ⇒ 𝑃(𝐶) = 10 2 9 3 Now, The Reqd. Prod. = 𝑃(𝐴|𝐶) = = = 𝑃(𝐴∩𝐶) 𝑃(𝐶) ⋯ [𝐷𝑒𝑓𝑛. ] 𝑃(𝐶 |𝐴)×𝑃(𝐴) 𝑃(𝐶) (1)(0.4) 2 3 =0.6 Question #5 What is the value of 𝑛(𝐴 ∩ 𝐵) if 𝑛(𝐴) = 7, 𝑛(𝐵) = 9, and 𝑛(𝐴 ∪ 𝐵) = 13? Answer: We use the following 𝑛(𝐴 ∪ 𝐵) = 𝑛(𝐴) + 𝑛(𝐵) − 𝑛(𝐴 ∩ 𝐵) Therefore, 13 = 7 + 9 − 𝑛(𝐴 ∩ 𝐵) 𝑛(𝐴 ∩ 𝐵) = 7 + 9 − 13 = 3 Question #6 Your statistics teacher tells you that there is a 50% chance that a coin will land heads up. How would you state this chance in terms of a probability? Answer: 0.5 or 1 2 IF we have a fair coin there are two possibilities: heads or tails Both have an equal chance. So you divide the favorable chances ("success") S by the total number of chances T: 𝑆 1 = = 0.5 = 50% 𝑇 2 Another example: What is the chance of rolling less than three with a normal die? S (success)=(1 or 2)=2 possibilities T (total)=6 possibilities, all equally probable 𝑆 2 1 Chance = = 𝑇 6 3 Question #7 What is the mean, median, mode, and range of the following numbers?: {1,2,3,4,5,6,7,8,9} Answer: Mean =5 and Median =5 There is no mode Range =9-1=8 Solution: We have the distribution: 1, 2, 3, 4, 5, 6, 7, 8, 9 The Range is the difference between the highest and the lowest - an indication of how far the numbers are spread out. range = 9 − 1 = 8 Mode, median and mean are all regarded as different types of average, they give different information. The Mode is the value with the highest frequency - the one that occurs the most often. There is no mode in this case because each number occurs exactly once. The median is the number exactly in the middle of a set of numbers arranged in order. There are 9 numbers, the middle one is 5 There are 4 numbers on each side. 1, 2, 3, 4, 5, 6, 7, 8, 9 The mean is what is generally understood as 'average' Mean = Mean = = 45 9 sum of the values the number of values 1+2+3+4+5+6+7+8+9 9 =5 Question #8 If the probability of events A & B occurring are 𝑃(𝐴) = 0.3 and 𝑃(𝐵) = 0.8, what is the probability that both A and B occur? Answer: The explanation is given below. Explanation: For any two events, A and B we have the product rule or the multiplication theorem of probability as follows. 𝑃𝑟{𝐴 ∩ 𝐵] = 𝑃(𝐴). 𝑃(𝐵) if A and B are independent. 𝑃𝑟[𝐴 ∩ 𝐵] = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∪ 𝐵) otherwise. Hence if A and B are independent, then we have the answer as 0.3 × 0.8 = 0.24 Otherwise, we cannot obtain 𝑃[𝐴 ∩ 𝐵] as the information is incomplete. Question #9 For random variables X and Y. What is Var(X+Y) in terms of 𝑉𝑎𝑟(𝑋) and 𝑉𝑎𝑟(𝑌) ? Answer: Var(X+Y)=Var(X)+Var(Y)+2Cov(X, Y) Explanation: Var(X+Y)=Var(X)+Var(Y)+2Cov(X, Y) You can write Var(X+Y)=Var(X)+Var(Y) Only if Cov(X, Y)=0 Question #10 Three people go to the market in the morning with the following probabilities: likelihood that two go to the market on a given morning? 1 2 3 2 3 4 , , and . What is the Answer: We'll have to work out the probabilities for all three combinations of two, where one stays home. Let's call the people A, B, C Explanation: We'll define the probabilities by the letter P (the sign means "not", ∧ means "and") P( A) = 1 2 1 2 P( B) = 2 3 1 3 → P( B) = P(C ) = 3 4 1 → P( C ) = 4 → P( A) = Then we get (remember "and" means multiply) 1 2 1 2 P( A ? B C ) =   = 2 3 4 24 1 1 3 3 P( A ? B ? C ) =   = 2 3 4 24 1 2 3 6 P( A ? B ? C ) =   = 2 3 4 24 Since these combinations are mutually exclusive we add: 2 3 6 11 𝑃(𝑡𝑜𝑡𝑎𝑙) = + + = 24 24 24 24 Note: If the question had been "at least two" we would have to add: 1 2 3 6 17 𝑃(𝐴 ∧ 𝐵 ∧ 𝐶) = ⋅ ⋅ = for a total of 2 3 4 24 24 If you had just added the probabilities of P(A ∧ B)+P(A ∧ C)+P(B ∧ C) without the NOT's, then P(A ∧ B ∧ C) would have been included three times. In fact P(total) would then be greater than 1 (a capital sin).