英文:
What is the probability?
问题
The translation of the provided content is as follows:
如果我们有一间有多人的房间，并且我们知道一个人有手机的概率是30%（P(B) = 30%），一个人没有手机的概率是70%（P(不是B) = 70%），我们选择3个人，事件D的概率是多少：“至少有1个人中的3个人有手机”我想计算不考虑D的逆事件和考虑D的逆事件。
我尝试了 p(D) = 0.3 × 0.7 × 0.7 + 0.3 × 0.3 × 0.7 + 0.3 × 0.3 × 0.3
英文:
If we have a room with a number of people and we know that the probability that a person has a phone is 30% p(B) = 30% and the probability that a person does not have a phone is 70% P(not B) = 70%, we choose 3 people, What is the probability of event D: “that at least 1 person out of 3 has a phone” I want to calculate without the inverse event of D, and with this inverse event.
I tried p(D) = 0.3×0.7×0.7 + 0.3×0.3×0.7 + 0.3×0.3×0.3
答案1
得分: 0
计算三人都没有手机的概率...
P_0 = (0.7)^3
然后至少有一人有手机的概率是...
1 - P_0
或以百分比表示，100*(1 - (0.7)^3) = 65.7
你说你也想计算概率，但不使用'逆事件' P_0。将概率 P_1 + P_2 + P_3 相加只是对 P_1 或 P_2 或 P_3 的实际概率的一种一阶近似，这只有在所有 P_is 都非常小的情况下才能以相当高的准确度工作。我不知道如何计算不使用你所说的'逆事件'的结果，甚至不知道是否可能完全做到。
英文:
Calculate the probability that none of the three have a phone...
P_0 = (0.7)^3
then the probability that at least one person has a phone must be...
1 - P_0
or as a percentage, 100*(1 - (0.7)^3) = 65.7
You say you want to also calculate the probability without using the 'inverse event' P_0. Adding probabilities P_1 + P_2 + P_3 is only a first order approximation to the actual probability of P_1 or P_2 or P_3 which only really works with decent accuracy if all the P_is are very small. I don't know how to calculate the result without using what you call the 'inverse event' or even if it's possible to do exactly.
答案2
得分: 0
Simon的答案是执行此操作的规范方式，但如果您必须在不计算逆的情况下解决它，请计算分配k台手机的k台手机的概率。
对于n个人，k台手机，p = 拥有手机的概率，然后至少1人拥有手机的概率为：
选择(n, 1) * p^1 * (1-p)^(n-1) +
选择(n, 2) * p^2 * (1-p)^(n-2) +
... +
选择(n, n) * p^n * (1-p)^(n-n)
在您的示例中，n=3，p=0.3，因此我们有：
选择(3,1) * (0.3)^1 * (0.7)^2 +
选择(3,2) * (0.3)^2 * (0.7)^1 +
选择(3,3) * (0.3)^3 * (0.7)^0
3 * 0.3 * 0.49 +
3 * 0.09 * 0.7 +
1 * 0.027 * 1
= 0.657
= 65.7%
因此，通过更多的工作，我们得到与Simon相同的答案。
英文:
Simon's answer is the canonical way to do this, but if you must solve it without calculating the inverse, calculate the probability of k phones times the ways of allocating k phones for k >= 1.
For n people, k phones, p = probability of having a phone, then the probability that at least 1 person has a phone is:
choose(n, 1) * p^1 * (1-p)^(n-1) +
choose(n, 2) * p^2 * (1-p)^(n-2) +
... +
choose(n, n) * p^n * (1-p)^(n-n)
In your example, n=3, p=0.3, so we have:
choose(3,1) * (0.3)^1 * (0.7)^2 +
choose(3,2) * (0.3)^2 * (0.7)^1 +
choose(3,3) * (0.3)^3 * (0.7)^0
=
3 * 0.3 * 0.49 +
3 * 0.09 * 0.7 +
1 * 0.027 * 1
= 0.657
= 65.7%
So with a lot more work, we get the same answer as Simon.
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