Question 6 (Essay Worth 10 points) \( (05.02 \mathrm{HC} \) ) A football quarterback has 2 more chances to throw a touchdown before his team is forced to punt the ball. He misses the receiver on the first throw \( 30 \% \) of the time. When his first throw is incomplete, he misses the receiver on the second throw \( 10 \% \) of the time. Part A: What is the probability of not throwing the ball to a receiver on either throw? (5 points) Part B: What is the probability of making at least 1 successful throw? ( 5 points) Ber

**Step 1. Define the events** Let: - \( M_1 \) be the event that the quarterback misses on the first throw. - \( M_2 \) be the event that the quarterback misses on the second throw (given that the first throw was a miss). We are given: - \( P(M_1) = 0.30 \) - \( P(M_2 \mid M_1) = 0.10 \) **Step 2. Part A: Probability of Missing on Both Throws** The probability of missing on both throws is: \[ P(M_1 \cap M_2) = P(M_1) \times P(M_2 \mid M_1) \] Substitute the given values: \[ P(M_1 \cap M_2) = 0.30 \times 0.10 = 0.03 \] Thus, the probability of not throwing the ball to a receiver on either throw is \(0.03\) (or \(3\%\)). **Step 3. Part B: Probability of At Least One Successful Throw** A successful throw means throwing the ball to a receiver. The quarterback can succeed on the first throw or, if he misses the first throw, succeed on the second throw. The probability of a successful first throw is: \[ P(\text{Success on 1st}) = 1 - P(M_1) = 1 - 0.30 = 0.70. \] If the first throw is a miss (which happens with probability \(0.30\)), the probability of a successful second throw is: \[ P(\text{Success on 2nd} \mid M_1) = 1 - P(M_2 \mid M_1) = 1 - 0.10 = 0.90. \] Thus, the probability of succeeding on the second throw (given a miss on the first) is: \[ P(\text{Miss first and Success on 2nd}) = P(M_1) \times P(\text{Success on 2nd} \mid M_1) = 0.30 \times 0.90 = 0.27. \] Now, the overall probability of at least one successful throw is: \[ P(\text{At least one success}) = P(\text{Success on 1st}) + P(\text{Miss first and Success on 2nd}) = 0.70 + 0.27 = 0.97. \] Thus, the probability of making at least one successful throw is \(0.97\) (or \(97\%\)).