Answer:
The right response is "0.0085". A further explanation is given below.
Step-by-step explanation:
Let,
P (across each pair, this same null hypothesis would be wrongly rejected) be "a".
Therefore,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
To solve the question we use the following analogy:
Given that the odds of winning is a/b, then the probability is b/(a+b)
hence to solve the question we proceed as follows:
a]
P(E)=11/36
odds against E=(36-11)/11
simplifying this gives us:
25/11
b] P(E)=31/36
odds against E=(36-31)/31
simplifying this we get:
5/31
hence odds in favor of E=31/5
c]P(E)=32/36
odds in favor of E will be:
36/(36-32)
simplifying the above we get:
36/4
=9/1
d]
P(E)=30/36
odds against E will be:
(36-30)/36
simplifying this we obtain:
6/36
=1/5
e] P(E)=13/36
odds against E will be:
(36-13)/13
simplifying this we obtain:
23/13
![\begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \underset{\stackrel{\uparrow }{\textit{let's use this one}}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7BLogarithm%20of%20rationals%7D%20%5C%5C%5C%5C%20%5Clog_a%5Cleft%28%20%5Cfrac%7Bx%7D%7By%7D%5Cright%29%5Cimplies%20%5Clog_a%28x%29-%5Clog_a%28y%29%20%5Cend%7Barray%7D~%5Chfill%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%20%5C%5C%5C%5C%20log_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cunderset%7B%5Cstackrel%7B%5Cuparrow%20%7D%7B%5Ctextit%7Blet%27s%20use%20this%20one%7D%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

notice, -2 is a valid value for the quadratic, however, the argument value for a logarithm can never 0 or less, it has to be always greater than 0, so for the logarithmic expression with (x-2), using x = -2 will give us a negative value, so -2 is no dice.