Find the probability of the outcome which has highest chance of occurring. A biased die has numbers $1, 2, 3, 4, 5, 6$. The probability of obtaining one of the numbers is greater
than ${1\over 6}$, whereas the probability of obtaining a number opposite to it is less than ${1\over 6}$ . The remaining four numbers each have a probability of ${1\over 6}$ of being obtained. Given that any two opposite faces add up to $7$. When two such dice are rolled, probability of obtaining a sum $7$ is ${47\over 288}$. If the number which has the highest probability of being obtained has probability ${p\over q}$, for ($p, q$) = $1$, find $p + q$.
Well I know that $P(E)$ = ${No.\;of\;Favourable\;Outcomes\over Total\;Outcomes}$ but I am confused as to how to find the Probability in this case. Because the formula which I wrote, I guess, is applicable only when all the possible outcomes have equal chances of occurring which is not happening in this case.
 A: You may obtain the sum $7$ with $1+6,2+5,3+4,4+3,5+2,6+1$. Among these sums, two are biased and the other $4$ have probability $\frac{1}{6}\times\frac{1}{6}$. The other two pairs have faces with probabilities $\frac16+x$ and $\frac16-x$ for some positive $x$.
Hence, the probability of obtaining the sum $7$ is
$$\frac{4}{36}+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}$$
$$\frac{1}{36}-x^2=\frac12(\frac{47}{288}-\frac{4}{36})=\frac{5}{192}$$
$$x^2=\frac{1}{36}-\frac{5}{192}=\frac{1}{576}=(\frac{1}{24})^2$$
Hence, the highest probability is $\frac16+\frac1{24}=\frac{5}{24}$ and $p+q=5+24=29$.
A: Call a number "good" if the probability that it comes up is $\frac14$, and "bad" otherwise.  Let $P$ be the bad number whose probability is $p$ and let $Q$ be the other bad number.
The first die lands on a good number with probability $\frac16$ and then the number we need on the other die is also good, so the probability of rolling $7$ with two good numbers is $\frac46\cdot\frac16=\frac19.$
If the first die lands on $P$ (probability $p$) then we need $Q$ on the other die (probability $q$) so the probability of this event is $pq$.  Similarly, if the first dies shows $Q$, we need $P$ on the second die, and the probability is again $pq$.
This accounts for all the probabilities, so the probability of rolling $7$ is $$\frac19+2pq= {47\over 288}$$
Take it from here.
A: Let $P(A)$ be the event of getting the number 'A' on the dice
$2P(1)P(6) + 2P(2)P(5) + 2P(3)P(4) = \dfrac{47}{288}$
Let $P$(biased number) $= x$
Therefore,
$2x(\frac{1}{3} - x) + \dfrac{1}{18} + \dfrac{1}{18} = \dfrac{47}{288}$
