这是伯努利试验。设 p 为射击的可能性。因此，4 中 2 的概率为 C(4, 2)p^2(1-p)^2 + ... + C(4, 4)p^4，其中 C(n, k) = n！ /(k!*(nk)!)。 6 中的 3 也是如此。您将得到两个必须进行比较的多项式。进一步简单的数学计算 - 哪个 p 使它正确或错误。您将得到一个六次方程，因此最好绘制图形并找到必要的 p 值。
It is Bernoulli trials. Let p be the possibility of a shot. So the probability of 2 out of 4 is C(4, 2)p^2(1-p)^2 + ... + C(4, 4)p^4, where C(n, k) = n!/(k!*(n-k)!). The same with 3 out of 6. You will get two polynomials which you have to compare. Simple math further - which p makes it true or wrong. You will have an equation of six degree, so better plot the graphic and find necessary values of p.
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这是伯努利试验。设 p 为射击的可能性。因此，4 中 2 的概率为 C(4, 2)p^2(1-p)^2 + ... + C(4, 4)p^4，其中 C(n, k) = n！ /(k!*(nk)!)。 6 中的 3 也是如此。您将得到两个必须进行比较的多项式。进一步简单的数学计算 - 哪个 p 使它正确或错误。您将得到一个六次方程，因此最好绘制图形并找到必要的 p 值。
It is Bernoulli trials. Let p be the possibility of a shot. So the probability of 2 out of 4 is C(4, 2)p^2(1-p)^2 + ... + C(4, 4)p^4, where C(n, k) = n!/(k!*(n-k)!). The same with 3 out of 6. You will get two polynomials which you have to compare. Simple math further - which p makes it true or wrong. You will have an equation of six degree, so better plot the graphic and find necessary values of p.
编辑
选项 1
连续投掷是独立的 - 就像 Deshene 所说：它是 伯努利试验。
你的论点是错误的，因为失去 x 投掷的概率与失去 y 投掷的概率不同。
选项 2
连续抛出是相关的 - 我们需要构建一个更深刻的模型。
尽管如此，失去 x 次投掷的概率与失去 y 次投掷的概率并不相同。
我相信在现实生活中，连续投掷是相关的（这只是我的信念 - 它没有得到经验证明或反驳）。
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Option 1
Consecutive throws are independent - Like Deshene said: it is Bernoulli trials.
Your argument is wrong since the probability of losing x throw is not the same as the probability of losing y throws.
Option 2
Consecutive throws are dependent - We'll need to build a more profound model.
Still, the probability of losing x throw is not the same as the probability of losing y throws.
I believe that in real-life, consecutive throws are dependent (this is only my belief - it was not empirically proved nor disproved).