If she is a 41% shooter she will probably score a goal 41 times out of a hundred, but she will probably miss the other 59 times out of a hundred. So the probability that she will miss is 59 out of 100 = 59% = 0.59
You want the probability of miss and miss and miss which is .67 * .67 * .67 = 0.301.
We'll if they are 89% it's pretty high
There is a 10 percent chance that you are gonna throw a 1.
The probability of four out of six shots is the probability of four successes, times the probability of two failures, times the number of permutations of successes and failures: P(four out of six free throws) = 15 * (2/5)4 * (3/5)2 = (15 * 16 * 9) / (625 * 25) = 432 / 3125
should be 1/6 chance for one. So 1/192 .5 percent
You want the probability of miss and miss and miss which is .67 * .67 * .67 = 0.301.
We'll if they are 89% it's pretty high
The probability the shooter makes both shots is .7 * .7 = .49, and the probability of making neither is .3 * .3 = .09. So the probability of making exactly 1 out of 2 is 1 - .49 - .9 = .42, or 42 percent.
There is a 58% chance they are going to miss.
In Division 1, Markus Howard from Marquette and Fletcher Magee from Wofford both have 100% free throw shooting
With a single throw of a normal die, the probability is 0.With a single throw of a normal die, the probability is 0.With a single throw of a normal die, the probability is 0.With a single throw of a normal die, the probability is 0.
There is a 10 percent chance that you are gonna throw a 1.
Ben Wallace
yes
The probability of four out of six shots is the probability of four successes, times the probability of two failures, times the number of permutations of successes and failures: P(four out of six free throws) = 15 * (2/5)4 * (3/5)2 = (15 * 16 * 9) / (625 * 25) = 432 / 3125
No, not that I ever heard of. It maybe if you are talking to the free throw shooter.
should be 1/6 chance for one. So 1/192 .5 percent