m358k-hw-four-solns

0 10 20 30 40 100 150 200 Time plot of daily sales Day indices Sales Problem 3. (2 points) What is the maximum number of wraps sold on a Tuesday? Use R commands, please, do not just look it up in the spreadsheet - that defies the purpose! Be careful about the type of data your number of wraps is recorded as. Hint: Typing ?numeric in the RStudio console might help. tue.wraps<- as.numeric (cafe.data $ Wraps.Sold[cafe.data $ Day.of.Week == "Tue" ]) max (tue.wraps) ## [1] 25 Problem 4. (3 points) Create a stacked dot plot of the number of sandwiches sold daily. Be careful about the type of data your number of sandwiches is recorded as. Hint: Typing ?numeric in the RStudio console might help. sands<- as.numeric (cafe.data $ Bread.Sand.Sold) ## Warning: NAs introduced by coercion stripchart (sands, at= 0 , pch= 20 , method= "stack" , col= "steelblue" , main= "Stacked dotplot of the daily sandwich sales" ) 3

Then, F c = H 0 ∪ H 1 , and H 0 and H 1 are mutually exclusive. Also, with G 0 = { there are exactly 12 heads in last 15 tosses } , G 1 = { there are exactly 11 heads in last 15 tosses } , we have, due to independence of trials, P [ E ∩ H 0 ] = P [ H 0 ∩ G 0 ] = P [ H 0 ] P [ G 0 ] , P [ E ∩ H 1 ] = P [ H 1 ∩ G 1 ] = P [ H 1 ] P [ G 1 ] . Moreover, the number of heads in the first five trials is Binomial ( n = 5 , p = 1 / 2) , and the number of heads in the remaining 15 trials is Binomial ( n = 15 , p = 1 / 2) . So, P [ E ∩ H 0 ] = 5 0 15 12 · 1 2 20 , P [ E ∩ H 1 ] = 5 1 15 11 · 1 2 20 . and P [ F | E ] = 1 - ( 15 12 ) ( 20 12 ) - 5 · ( 15 11 ) ( 20 12 ) . Note that the fact that the coin is (by default) fair is completely irrelevant in the above calculation. Problem 7. (5 points) A pair of dice is thrown. Find the probability that the sum of the outcomes is 10 or greater if a 5 appears on the first die. Solution: Let A i denote the event that i was the outcome on the first die for i = 1 , 2 , . . . 6 . Let E denote the event that the sum of the outcomes on both of the dies was greater than or equal to 10. Formally, E = { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j ≥ 10 } = { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j = 10 } ∪ { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j = 11 } ∪ { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j = 12 } . (1) We want to find the probability P [ E | A 5 ] . Directly from the definition of conditional probability, we get P [ E | A 5 ] = P [ E ∩ A 5 ] P [ A 5 ] . From the representation in (1), we get that P [ E ∩ A 5 ] = P [ { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j = 10 } ∩ { ( i, j ) : i = 5 and 1 ≤ j ≤ 6 } ] + P [ { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j = 11 } ∩ { ( i, j ) : i = 5 and 1 ≤ j ≤ 6 } ] + P [ { ( i, j ) : 1 ≤ i, j ≤ 6 and i + j = 12 } ∩ { ( i, j ) : i = 5 and 1 ≤ j ≤ 6 } ] = P [ { ( i, j ) : i = 5 and i + j = 10 } ] + P [ { ( i, j ) : i = 5 and i + j = 11 } ] + P [ { ( i, j ) : i = 5 and i + j = 12 } ] = P [ { (5 , 5) } ] + P [ { (5 , 6) } ] + P [ ∅ ] = 1 36 + 1 36 + 0 = 1 18 . On the other hand, P [ A 5 ] = 1 6 , and so P [ E | A 5 ] = 1 3 . Note: The above solution is deliberately ridiculously formal. You could have solved this problem correctly by straightforward counting of “good” outcomes quite fast and in many ways. 6