Carolyn usually only misses about 10% of her free throws. If she misses fewer than 2 of her next 4 free throws, then she will hold the school record for free throws. To estimate the probability of this happening, Carolyn uses a computer to randomly select 4 numbers from 0 to 9. She lets 0 represent a missed free throw and 1 through 9 represent a made free throw. She repeats this for a total of 18 trials and records the results in the table below.8100 1347 0693 9869 3399 1376 1273 7905 34717987 7604 4587 2968 5064 2697 4248 4545 3308What is the best estimate, based on this simulation, of the probability that she will miss fewer than 2 of her next 4 free throws?6%33%67%94%

The correct option will be :  94%Explanation8100  1347   0693  9869   3399   1376   1273   7905    3471    7987  7604   4587   2968  5064   2697   4248   4545   3308Here, '0' represent a missed free throw and '1 through 9' represent a made free throw. So, the number of results in which there are fewer than two '0' is  17. That means she will miss fewer than 2 throws in 17 trials and there are total 18 trials. Thus, the probability that she will miss fewer than 2 throws [tex]= \frac{17}{18}=0.9444.....[/tex]For finding the probability in percentage, we need to multiply by 100% . So, (0.9444..... × 100%) = 94.44.... % = 94% (Approximately)