3. On his high school’s football team, Jeffrey makes the extra point made 60% of the time. In a particular game versus the team’s biggest rivals, he had 6 extra point attempts. Assume independence between kicks.
a. Based on his average, what was the probability that all 6 attempts are successful?
b. What is the probability that at least 1 was successful?
c. What is the probability that at least 1 is not successful?
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a. probability that all six attempts are successful: (0.6)^6, because he is throwing six times and the probability of each success is 0.6
b. to find the probability that at least one was successful, consider the opposite scenario. the probability that all failed is (0.4)^6, because the probability of failure is 0.4 and he is throwing six times.
the probability that at least one was successful = 1 - (0.4)^6
c. using the same reasoning, we conclude that the probability that at least 1 is not successful = 1 - (0.6)^6
tag me if you are still confused
@NandagopalPV also, I see you have posted this question 3 times, would you mind closing the other two?