3. The probability of Jim hitting the bull's-eye on a dart board is 0.04 . What are the odds in favour of Jim not hitting the bull's-eye? 4. What are the odds in favour of a total greater than 9 in a given roll of two standard dice? 5. Suppose you simultaneously roll a standard die and spin a spinner that is divided into 10 equal sectors, numbered 1 to 10. What is the probability of getting a 4 on both the die and the spinner? 6. Carrie is a goalie on her soccer team. She estimates that her chances of stopping a penalty kick during a game are \( 80 \% \) when this happens in the first half, but only \( 65 \% \) in the second half. If the other team has a \( 58 \% \) probability of getting a penalty kick awarded in the first half, what is the probability of Carrie stopping a penalty kick in a match this afternoon?

Below are the solutions to each question: 1. (Question 3)  • Jim’s probability of hitting the bull’s-eye is 0.04. Therefore, his probability of not hitting it is 1 – 0.04 = 0.96.  • “Odds in favour” are given by the ratio of the probability of the desired outcome to the probability of its complement. In this case, the odds in favour of Jim not hitting the bull’s-eye are 0.96 to 0.04.  • Dividing both by 0.04 gives 24:1.  • Answer: The odds in favour of Jim not hitting the bull's-eye are 24:1. 2. (Question 4)  • When rolling two dice, you want the total to be greater than 9 (i.e. sums of 10, 11, or 12).   – For a sum of 10, the favorable pairs are: (4,6), (5,5), (6,4) → 3 outcomes.   – For a sum of 11, the favorable pairs are: (5,6), (6,5) → 2 outcomes.   – For a sum of 12, the only pair is: (6,6) → 1 outcome.  • Total favorable outcomes = 3 + 2 + 1 = 6.  • There are 36 possible outcomes when rolling two dice.  • Thus, the probability is 6/36 = 1/6.  • To find the odds in favour, compare the probability of the desired event (1/6) with the probability of its complement (1 – 1/6 = 5/6). This gives the ratio 1/6 : 5/6, which simplifies to 1:5.  • Answer: The odds in favour of a total greater than 9 are 1:5. 3. (Question 5)  • When rolling a die and spinning a 10-sector spinner:   – The probability of getting a 4 on the die is 1/6.   – The probability of getting a 4 on the spinner is 1/10.  • Because the two events are independent, multiply the probabilities: (1/6) × (1/10) = 1/60.  • Answer: The probability of getting a 4 on both the die and the spinner is 1/60. 4. (Question 6)  • Carrie’s chance of stopping a penalty kick depends on which half of the game it is awarded:   – First half: her stopping probability is 80%.   – Second half: her stopping probability is 65%.  • The problem states that there is a 58% probability that the other team is awarded a penalty in the first half. It is natural to assume that if a penalty isn’t awarded in the first half, then it comes in the second half (with a probability of 1 – 0.58 = 0.42).  • We use the law of total probability to combine the two cases:   – Overall probability of stopping a penalty = (Probability of first half penalty × 0.80) + (Probability of second half penalty × 0.65)   – = (0.58 × 0.80) + (0.42 × 0.65)   – = 0.464 + 0.273 = 0.737.  • Answer: The probability of Carrie stopping a penalty kick in the match is approximately 73.7%. Summary of Answers: 3. Odds in favour of Jim not hitting the bull’s-eye: 24:1. 4. Odds in favour of a total greater than 9: 1:5. 5. Probability of a 4 on both the die and the spinner: 1/60. 6. Probability of Carrie stopping a penalty kick: approximately 73.7%.