What's bad in S

Let $S'$ be the event of non-occurrence of S, $T$ & $H$ be the events of getting a tail & head respectively. Then using traditional Bayes' theorem, we have:

$P(T/S') = \displaystyle \frac{P(S'/T) \times P(T)}{P(S'/T) \times P(T) + P(S'/H) \times P(H)} = \frac{6}{11}$ after substituting values.

Let $T$ be the event that the coin is tails and $H$ be the event that it is heads.

We know that $P(T) = \dfrac{2}{3}$ and $P(H) = \dfrac{1}{3}$

We have to find

$P(T | \text{Letters had no S})$ .

Now, $P(T | \text{Letters had no S}) = \dfrac{ P(\text{Letters had no S} | T) \times P(T)}{P(\text{Letters had no S})}$

$P(\text{Letters had no S} | T) = \dfrac{\dbinom{7}{5}}{\dbinom{9}{5}} = \dfrac{21}{126}$

$P(\text{Letters had no S} | H ) = \dfrac{\dbinom{7}{4}}{\dbinom{9}{4}} = \dfrac{35}{126}$

And $P(\text{Letters had no S}) = P(\text{Letters had no S} | T) \times P(T) + P(\text{Letters had no S} | H) \times P(H)$

Thus finally we get

$P(\text{Letters had no S} | T) = \dfrac{\dfrac{21}{126} \times \dfrac{2}{3}}{\dfrac{21}{126} \times \dfrac{2}{3} + \dfrac{35}{126} \times \dfrac{1}{3}} = \dfrac{6}{11}$ which gives us $6 + 11 = \boxed{17}$

Let $S$ be the event that the chosen letters contain no S, $h$ the event that head appeared on the coin, and $t$ the event that tail appeared on the coin. By Bayes' formula, the probability of $t$ given $S$ is $\frac{P(t\cap S)}{P(S)}$ . We calculate the numerator and denominator separately.

First we calculate $P(t\cap S)$ . The probability of getting tails is $\frac{2}{3}$ . Once we have flipped tails, we must only get the $7$ letters that are not S. We can choose these letters in $\binom{7}{5}=21$ ways. Thus we have $P(t\cap S)=\frac{2}{3}\cdot21=14$ .

Now we calculate $P(S)$ . Note that $h$ and $t$ are complementary events, so that $P(S)=P(h\cap S)+P(t\cap S)$ . We have already calculated $P(t\cap S)$ to be $14$ . We can count $P(h\cap S)$ in a similar way. There is a $\frac{1}{3}$ chance of getting heads. Once we have flipped heads, we must choose only the $7$ non-S letters in our $4$ pickings; this can be done in $\binom{7}{4}=35$ ways. Thus $P(h \cap S)=\frac{1}{3}\cdot35$ .

Now we have that $P(t|S)=\frac{14}{14+\frac{35}{3}}=\frac{\frac{42}{3}}{\frac{42}{3}+\frac{35}{3}}=\frac{42}{77}=\frac{6}{11}$ . Thus $a+b=6+11=\boxed{17}$ .

If there are no $S$ 's then the number of ways of choosing $4$ from $7$ is $7\choose4$ = 35 and the number of ways of choosing $5$ letters from $7$ is $7\choose5$ = $21$

So if we get tails we have $21$ ways of getting a combination with no $S$

And if we get heads we have $35$ ways of getting a combination with no $S$

We expect the ratio to be $21*2:35$ however because a tails is twice as likely.

i.e. a ratio of $6: 5$ so the probability is $\frac{6}{11}$ so $a+b=6+11=\boxed{17}$