We can brute-force this problem. If each man -- okay, or woman, so just pipe down, Annie Oakley -- takes one shot at the target, he'll either hit it or miss it, so there are 23 = 8 states that the target can be in when they're done shooting. Let's take a look at them. Let a one represent a hit and a zero a miss. A B C 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Only three of those states meet the criteria stated in the question, the 4th state, the 6th one, and the 7th one. The 8th state does not count because the question said "exactly two," not "at least two." If each of the eight states had an equal probability of occurrence, we could say that the probability of any of the three two-hit states occurring would be three in eight, or 0.375. But we cannot say that because each state does not have an equal likelihood. The question says that person A will hit the target six out of seven times, so his hit probability is 6/7, and his miss probability is 1/7. B's hit probability is 4/5, and his miss probability is 1/5. C's hit probability is 3/4, and his miss probability is 1/4. So, what is the probability that the target will be in the fourth state (0 1 1) after the shooting's over? It's (1/7) * (4/5) * (3/4) = 12/140. And how about the sixth state (1 0 1)? It's (6/7) * (1/5) * (3/4) = 18/140. And the seventh (1 1 0)? It's (6/7) * (4/5) * (1/4) = 24/140. The sum of the three probabilities is 54/140 = 0.3857. So, there is about a 38.6 percent chance that the target will be struck exactly twice if each of the three men takes one shot at the target. Note that the probability that all three men will hit the target is 6*4*3/140 = 72/140 = 0.5143, so it's more likely that there will be THREE bullet holes in the target than just two. Can you figure out the probability of there being FEWER than two holes in the target?
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.
The answer depends on the probability that Alex makes a single shot.
Probabilities are calculated by the binomial distribution. Ans: 0.1536 and 0.9728 Discussion: Prob of exactly two times = 4!/(2! x 2! ) x (0.8)^2 x (0.2)^2 = 0.1536 Prob of at least twice = hits target 2 times + hits target 3 times + hits target 4 times Prob of at least twice = 1 - Prob of no hits - Prob of one hit Prob of one hit = 4!/(3! x 1!) x (0.8)^1 x (0.8)^1 x (0.2)^3 = 0.0256 Prob of no hits= 4!/(0! x 4!) x (0.8)^0 x (0.2)^4 = (0.2)^4 = 0.0016 Prob of 2 or more hits = 1-0.0256-0.0016 = 0.9728 Note: You can calculate these values using Excel, where Prob of 2 hits = binom(2,4,0.8,false) and Prob of 2 or more hits = 1 - binom(1,4,0.8,false) - binom(0,4,0.8,false) or: Prob of 2 > hits = 1- binom(1,4,0.8,true) , as false is requesting the PMF value and true is requesting CDF value. See help in Excel for further explanation on this function. Also, there might be an issue of independent events in this problem, in that the probability is given as a constant. If a rifleman missed three times, do you think he would learn and might do better than his average on the fourth shot?
You would make 25 * 0.68 shots; or 17 shots made.
In basketball, the free throw is an important shot; it can highly increase points for a team if the player should make the shot. In 2010, Stephen Curry was the NBA player who has the highest percentage of free throw shots.
It is 0.3857
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.
No, penalties do not count as shots on target in soccer. Shots on target refer to shots that are heading towards the goal and would go in if not saved by the goalkeeper. Penalties are a separate category in soccer statistics.
The answer depends on the probability that Alex makes a single shot.
by using shot feeling from target
Jay would make 4 out of 5 shots.
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
Seems like the probability for a single shot is 0.6 So the probability of success with shot 1 is 0.6 and the probability of failing with the first and winning with the second is 0.4x0.6=0.24, same for the other way around, and the probability if winning both times is 0.6x0.6=0.36, and the probability of failing both times is 0.4x0.4=0.16, which is the same as (1-2x.24-.36) Multiply the probability by 100 to get percent.
Are you Aiming carefully
a bartender
I'm going to have to make a few assumptions in order to answer your question. 1. you are shooting pellets not BB's. (USE Match Pellets) 2. you are asking about open sights not a scope 3. you have a fairly decent rifle with an adjustable rear sight. Set up a Target 33 feet from where you plan to stand. (it about 10 meters) Bench rest the rifle and ALWAYS hold it exactly in the same place for every shot. aim and fire 3 shots at the target. If the strikes are to the left of the target then move your rear sight to the right a few clicks. Move them left if you are striking the target to the right. Strikes to high, then move the sight down. or visa versa. Do it again with 3 more shots and keep it up until you are consistently on target. remember their will always be a few wild shots.
Nope! And drinking 40 shots quickly could be fatal.