Answer:
Explanation:
This is a binomial problem where the basketball player has a probability of success of 0.725 (making a free throw) and is attempting 60 shots. The number of successful shots is a binomial random variable X ~ Bin(60, 0.725). We need to find the probability that she makes at least 15 of her next 60 shots, which is P(X ≥ 15). This can be computed using the binomial cumulative distribution function (CDF) as follows:
P(X ≥ 15) = 1 - P(X < 15) = 1 - binom.cdf(14, 60, 0.725) ≈ 0.998
Therefore, the probability that she makes at least 15 of her next 60 shots is approximately 0.998.
This is also a binomial problem where the probability of success is 0.14 (a smartphone requiring servicing) and the number of trials is 100 (smartphones in a shipment). We need to find the probability that no more than 10 smartphones in the shipment will require servicing, which is P(X ≤ 10). This can be computed using the binomial CDF as follows:
P(X ≤ 10) = binom.cdf(10, 100, 0.14) ≈ 0.289
Therefore, the probability that no more than 10 smartphones in the shipment will require servicing is approximately 0.289.
This is also a binomial problem where the probability of success is 0.06 (a scratch-off lottery ticket having a prize of a free ticket) and the number of trials is 200 (tickets at the local convenience store). We need to find the probability that this prize is on 15 or more of the tickets, which is P(X ≥ 15). This can be computed using the binomial CDF as follows:
P(X ≥ 15) = 1 - P(X < 15) = 1 - binom.cdf(14, 200, 0.06) ≈ 0.021
Therefore, the probability that the prize is on 15 or more of the 200 tickets at the local convenience store is approximately 0.021.