Magoosh hard 求解答

sunshinehanUID：3762313

A weighted coin has a probability p of showing heads. If successive
flips are independent, and the probability of getting at least one head
in two flips is greater than 0.5, then what could p be?
Indicate all possible values.
0.1
0.2
0.3
0.4
0.6
0.7

答案CDEF
https://magoosh.com/gre/2011/gre ... ity-of-a-coin-toss/
这里有个解答，没看懂

Probability of Heads = p
Probability of Tails = (1-p)

Therefore the four possible outcomes have the following probabilities

Head, Head = p.p
Head, Tail = p(1-p)
Tail, Head = (1-p) 这里是不是写错了，应该是 (1-p)p ?
Tail, Tail = (1-p)^2

If the probability of getting atleast one head is greater than 0.5, then the probability of getting all tails is less than 0.5

Therefore we can come up with the following inequality.

(1-p)^2 < 0.5

We know that 0.5 is a little bigger than 0.7^2 which is 0.49, therefore

(1-p) 0.29 这里是不是写错了，应该是 (1-p) < 0.29 ?

Therefore p can be 0.3, 0.4, 0.6, 07
c,d,e,f

如果 (1-p) < 0.29 则 p > 0.71 ，CDEF都不符合？

sunshinehanUID：3762313

或者，我有另一种解法：

假设：
head head = 1/4
head tail= 1/4
tail head= 1/4
tail tail= 1/4

所以包含head的就是3/4，两次都是head只有1/4
1/4 <= x <= 1/2+1/4
0.25 <= x <= 0.75

yeyeshaneUID：3153575

今天正好做到了這題，也糾結了一會兒。看了Magoosh的官方視頻看懂了。

遇到這種“at least”的題，主要找出現零次HEAD的概率。用1減去出現零次H的概率也就是出現至少一次的概率了。

題中P為出現H的概率，所以1-P為出現T的概率（拋一次），拋兩次則為（1-P）的平方，即出現零次Head的概率。

題目問的是出現至少一次H的概率，所以應該就是1-(1-p)^2

一一帶進去，P=0.1時，數值為0.19<0.5不符合；
P=0.2時，數值為0.36<0.5不符合；
P=0.3時，數值為0.51>0.5符合；
而且，我們知道被減數越小，值就越大，所以後面的數也都符合，所以就是CDEF。