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?
The probability is 0.0035
i think since they are mutually exclusive events the probability would by 9/10*7/10 = 63/100
the question asks probability of at least one gun hitting the target. = 1 - no gun hitting the target = 1 - (1/10) x (3/10) = 97/100 or 97 %
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?
P(at least two hits out of 6)=1-P(1 or fewer hits out of 6)=1-binomcdf(6, .3, 1)=1-.420175=0.579825
The probability is 0.0035
The probability is 71/489 = 0.145, approx.
i think since they are mutually exclusive events the probability would by 9/10*7/10 = 63/100
the question asks probability of at least one gun hitting the target. = 1 - no gun hitting the target = 1 - (1/10) x (3/10) = 97/100 or 97 %
Theres a probability that they do?
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?
You would multiply the probabilities. The probability the first marksmen hits the target x the probability the second marksmen hits the target x the probability the third marksmen hits the target x the probability the fourth marksmen hits the target. So you take .80 x .80 x .80 x .80 = .4096 or about 41%. So if they all fire at the target with each having an 80% probability of hitting, there will be about a 41% chance they will all hit. If you actually think about this question, though, you would be wise to hesitate about the answer. What causes marksmen to miss? In general it would be many things, often acting in combination: trembling of the hands, distractions, shifts of the wind, variations in the ammunition being fired, and so on. If all four marksmen are shooting simultaneously, or nearly so, then some of these causes will be acting in the same way on all four. These would include the wind and the environmental distractions, for instance. It's therefore conceivable that when one marksman misses, so will all the others, and (usually) when one marksman hits the target, the others will be able to as well. In this situation the probability that four marksmen will all hit the target will be close to 80%, not 41%. The question itself couches some ambiguities. For instance, at a tournament of 100 marksmen, the probability that some four will all hit their target is likely close to 100%. (Problems like this in interpreting the intended meaning of probability questions go all the way back to the very first book on probability by Christian Huygens in 1657. There was argument among very good mathematicians for a long time about the answer to one of his problems because it had three distinct interpretations.)
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P(at least two hits out of 6)=1-P(1 or fewer hits out of 6)=1-binomcdf(6, .3, 1)=1-.420175=0.579825
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