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10 Points!! Can You Please Help?

10 POINTS / CAN SOMEBODY PLEASE HELP ME WITH THIS EASY MATHTASK ?

  • So a guy shoots arrows in a circle. from the 10 times he gets 6 hits on average. a. so the guy does 8 shots. whats the chance he has 5 hits. I came out with a chance of 16.66% I just did 6/10 x 6/9 x 6/8 etc etc etc 6/1 and came out with 16.66 % Somehow it just doesnt seem right though. a chance of 6/1 :s So I did it just 6/10 x 6/10 x 6/10 etc etc etc x 6/10 and that 10 times and came out wiiith 0.006046 but that seems so damn low. Maybe I did something wrong typed it in wrong. Can someone give some feedback on this. What do I need to do ? Im very bad at math and english is not my first language, so explain to me like im a 10 year old..

  • Answer:

    ok so the probability of hitting is p = 0.6 (6 out of ten) and the probability of missing is 0.4 (4 out of ten) so to find the probability of hitting 5 out of 8 n = 8 (total number of shots) x = 5 (number of successful shots) the formula for the probability is p(x) = n! / [x!(n-x)!] * p^x * (1-p)^(n-x) plug in your numbers p(5) = 8! / (5! * 3!) * 0.6^5 * 0.4^3 p(5) = 0.2787 = 27.87%

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ok so the probability of hitting is p = 0.6 (6 out of ten) and the probability of missing is 0.4 (4 out of ten) so to find the probability of hitting 5 out of 8 n = 8 (total number of shots) x = 5 (number of successful shots) the formula for the probability is p(x) = n! / [x!(n-x)!] * p^x * (1-p)^(n-x) plug in your numbers p(5) = 8! / (5! * 3!) * 0.6^5 * 0.4^3 p(5) = 0.2787 = 27.87%

Big Orange

Two concepts are involved here: 1. Probability 2. Combinations Solution: Probability of success P(s) = 0.6 Probability of failure P(f) = 0.4 Desired pattern is 8 shots with 5 success and 3 failures So each such desired pattern has probability of 0.6^5 * 0.4^3 = 0.004977 (Probability of one desired pattern) Now let us decide how many patters out of all possible patters when 8 shots are taken would conform to the above requirement of (Success = 5 & Failure = 3) Possible number of patters matching with the desired patters= 8! / (5! - 3!) = 56 Therefore, final answer is Probability of one desired pattern * possible number of desired patterns 0.004977 * 56 27.9% Hope this explains the logic behind the formula, however, it is better to use the formula already cited in previous answers.

Ketul

herrro im yerrrro

derekshiisgayy

Answer: 27.9% This is a binomial probability distribution hence you can calculate with the following formula (see Note 2): P(X) = nCx . p^x . (1-p)^(n-x) where C denotes a Combination p = probability of successful hits = 0.60 (see Note 1) n = number of shots = 8 x = number of shots hit = 5 So input these values into the formula using a calculator and you get.. P(5 successful hits out of 8) = (8C5) (0.60)^5 (1-0.60)^(8-5) = 0.2787 (to 4 decimal places) = 27.9% Note 1: You have the probability (p) given by 6 out of 10 times he hits.. p = 6/10 = 0.60 Note 2: A binomial probability distribution occurs - when the outcome of each shot is classified into one or two mutually exclusive categories - either you hit or you don't. - the number of successful hits can be counted in a fixed number of shots - the probability of successful hits stays the same for each attempt (60% from the average of 6 out of 10) - each shot is independent, meaning one shot does not affect the result of the next.

Aundrea

Answer: 27.9% This is a binomial probability distribution hence you can calculate with the following formula (see Note 2): P(X) = nCx . p^x . (1-p)^(n-x) where C denotes a Combination p = probability of successful hits = 0.60 (see Note 1) n = number of shots = 8 x = number of shots hit = 5 So input these values into the formula using a calculator and you get.. P(5 successful hits out of 8) = (8C5) (0.60)^5 (1-0.60)^(8-5) = 0.2787 (to 4 decimal places) = 27.9% Note 1: You have the probability (p) given by 6 out of 10 times he hits.. p = 6/10 = 0.60 Note 2: A binomial probability distribution occurs - when the outcome of each shot is classified into one or two mutually exclusive categories - either you hit or you don't. - the number of successful hits can be counted in a fixed number of shots - the probability of successful hits stays the same for each attempt (60% from the average of 6 out of 10) - each shot is independent, meaning one shot does not affect the result of the next.

Aundrea

Two concepts are involved here: 1. Probability 2. Combinations Solution: Probability of success P(s) = 0.6 Probability of failure P(f) = 0.4 Desired pattern is 8 shots with 5 success and 3 failures So each such desired pattern has probability of 0.6^5 * 0.4^3 = 0.004977 (Probability of one desired pattern) Now let us decide how many patters out of all possible patters when 8 shots are taken would conform to the above requirement of (Success = 5 & Failure = 3) Possible number of patters matching with the desired patters= 8! / (5! - 3!) = 56 Therefore, final answer is Probability of one desired pattern * possible number of desired patterns 0.004977 * 56 27.9% Hope this explains the logic behind the formula, however, it is better to use the formula already cited in previous answers.

Ketul

herrro im yerrrro

derekshiisgayy

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