Section 4.2 Solutions

Section 4.2 Solutions: 1) In her wallet, Anne Kelly has 14 bills. Seven are $1 bills, two are $5 bills, four are $10 bills and one is a $20 bill. She passes a volunteer seeking donations for the Salvation Army and decides to select one bill at random from her wallet and give it to the Salvation Army. Determine: a) The probability she selects a $5 bill The numerator will be the number of 5 dollar bills, (2). The denominator will be the 14 bills in the wallet. Answer: 2/14 = 1/7 b) The probability she does not select a $5 bill The numerator will be the number of bills that aren’t fives (12) The denominator will be the 14 bills in the wallet. Answer: 12/14 = 6/7 c) The odds in favor of her selecting a $5 bill The long way to do this is: 𝑃(𝑜𝑓 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑎 5) 1⁄ = 6⁄7 = 𝑃(𝑛𝑜𝑡 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑎 5) 7 1 7 6 1 7 1 ÷7=7∗6=6 The easier way: Number of 5’s : Number of bills that are not 5’s 2 : 12 Answer: 2:14 which is better written 1:6 d) The odds against her selecting a $5 bill Odds against and odds in favor of the same event are opposite. I will just switch my answer to part c. Answer: 6:1 #3 – 14: A card is picked from a deck of cards. Find the odds against and odds in favor of selecting: 3) a heart Odds in favor: # hearts : # non hearts 13:39 (reduce by 3) Odds against, just flip the odds in favor Odds in favor: 1:3 Odds against: 3:1 5) a seven Odds in favor: # of sevens: # non sevens 4:48 (reduce by 4) Odds against, just flip the odds in favor Odds in favor 1:12 Odds against 12:1 7) a seven or a queen Odds in favor: number of cards that are 7’s or queens: number of cards that are not 7’s or queens 8:44 Reduce to 2:11 Odds against, just flip odds in favor Odds in favor 2:11 Odds against 11:2 9) a seven and a heart Odds in favor: number of cards that are 7’s and a heart at the same time: number of cards that are not 1:51 (this only counts the 7 of hearts) Odds in favor 1:51 Odds against 51:1 11) the three of spades Odds in favor: Number of cards that are the 3 of spades: Number of cards that are not the 3 of spades 1:51 Flip for odds against. Odds in favor 1:51 Odds against 51:1 13) a red king Odds in favor Number of cards that are red kings (2): Number of cards that are not red kings (50) 2: 50 Reduce to 1:25 Flip for odds against. Odds in favor: 1:25 Odds against: 25:1 15) A pair of dice is rolled and the sum of the dice is recorded. Here is the sample space. DICE 1 → 1 2 3 4 5 6 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) Sum 2 Sum 3 Sum 4 Sum 5 Sum 6 Sum 7 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) Sum 3 Sum 4 Sum 5 Sum 6 Sum 7 Sum 8 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) Sum 4 Sum 5 Sum 6 Sum 7 Sum 8 Sum 9 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) Sum 5 Sum 6 Sum 7 Sum 8 Sum 9 Sum 10 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) Sum 6 Sum 7 Sum 8 Sum 9 Sum 10 Sum 11 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) Sum 7 Sum 8 Sum 9 Sum 10 Sum 11 Sum 12 Dice 2 ↓ 1 2 3 4 5 6 a) Find the probability of rolling a sum of 7. 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑟𝑜𝑙𝑙 𝑎 7 6 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑟𝑜𝑙𝑙𝑖𝑛𝑔 𝑎 7 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 = 36 = 1/6 Answer: 1/6 b) Find the probability of not rolling a sum of 7 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑟𝑜𝑙𝑙 𝑎 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑡ℎ𝑒𝑟 𝑡ℎ𝑎𝑛 7 30 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑛𝑜𝑡 𝑟𝑜𝑙𝑙𝑖𝑛𝑔 𝑎 7 = = 36 = 5/6 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 Answer: 5/6 c) Find the odds in favor of the sum being 7. The long way to do this is: 𝑃(𝑠𝑢𝑚 𝑏𝑒𝑖𝑛𝑔 7) 𝑃(𝑛𝑜𝑡 𝑏𝑒𝑖𝑛𝑔 7) 1⁄ = 5⁄6 = 6 1 6 5 1 6 1 ÷6= 6∗5= 5 Easier way to do this: # of outcomes that are 7: # of outcomes that are not 7 6:30 reduce to 1:5 Answer: 1/5 or 1 to 5 or 1:5 d) Find the odds against the first dice showing a 5. I will do this the easier way. Number of elements in the sample space where the first dice not 5: everything else in the sample space. The number to the right of the colon will be 6 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) The number to the left of the colon will be 30. There are 36 items in the sample space and 6 of them have a 5 first, the other 30 don’t have a 5 first. Answer: 30:6 reduce to 5:1 e) Find the probability the sum is less than 7. The numerator will be the 15 elements in the sample space that have sums between 2 and 6 inclusive. The denominator will be the 36 elements in the sample space. Answer: 15/36 = 5/12 f) Find the odds against of the sum being less than 7. I will write the number of elements whose sum is greater than or equal to 7 first. This will be the 21 elements in the sample space that are 7 or larger. I will write the number of elements whose sum is less than 7 second. This will be the 15 I counted in part “e”. Answer: 21:15 = 7:5 g) Find the probability of rolling a double (both dice have the same number). The numerator will be 6. (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) The denominator will be the 36 elements in the sample space. Answer 6/36 = 1/6 h) Find the odds against rolling a double. The number to the left of the colon will be the 30 elements in the sample space that are not doubles. The number to the right of the colon will be the 6 doubles in the sample space. Answer: 30:6 = 5:1 17) One person is selected at random from a class of 16 men and 14 women. Find the odds against selecting: a) A woman The number of men will go to the left of the colon as this is an odds against. The number of women will go to the right of the colon. Answer: 16:14 = 8:7 b) A man The number of women will go to the left of the colon as this is an odds against. The number of men will go to the right of the colon. Answer: 14:16 = 7:8 18 – 21: A dart is thrown at this target and the color it lands on is noted. Find the requested odds. 19) Odds in favor of landing in the blue region. I have to break this up into 4 regions. There will be 2 purple, 1 red and 1 blue region. My answer will be of the form: number of blue regions: number of regions that are not blue Answer: 1:3 21) Odds against landing in the purple region. My answer will be in the form: number of regions that are not purple: number of purple regions. Answer 2:2 or 1:1