not from a c section what is the probability that her baby survives?
$P(E) =$ percent that survives minus the percent that survive a c section (should give the percentage of survival that are not c section)
$P(E) = .98-.96.15$
$P(E) = 97.856$
However the book says its .9835. THIS MAKES NO SENSE how could the probability that a baby is not a c section and survive be higher than the overall survival rate of all births even including those that result from a c section?
Of course it makes sense; in fact, it must be greater than the overall proportion.
To understand why, suppose I said that half of all flowers are white. Of those flowers with five petals, only a fourth of these are white. Then of those flowers that do not have five petals, you must have more than half of them being white, otherwise there is no way to have half of the overall population of flowers be white. The extent to which non-five-petaled flowers are white depends on the relative incidence of five-petaled flowers versus non-five-petaled flowers.
Things look better if we assign actual events and probabilities. Let $C$ represent the event that a birth is a c-section. Let $S$ represent the event that a baby survives. Denote by $\bar C$ (the birth is not c-section) and $\bar S$ (the baby does not survive) the complementary events. Now we have $$\Pr[S] = 0.98, \quad \Pr[C] = 0.15, \quad \Pr[S \mid C] = 0.96,$$ as given by the question. Your calculation was: $$\Pr[S \mid \bar C] = \Pr[S] - \Pr[S \mid C]\Pr[C].$$ This is not correct. The RHS can be written $$\Pr[S] - \Pr[S \cap C] = \Pr[S \cap \bar C] \ne \Pr[S \mid \bar C].$$ You can now see where you went wrong: a joint probability is not the same as a conditional probability. To get the right answer, then, you must observe that $$\Pr[S \mid \bar C] = \Pr[S \mid \bar C] \Pr[\bar C],$$ so that you need to divide your answer by the probability of a non c-section birth; i.e., $\Pr[\bar C] = 1 - \Pr[C] = 1 - 0.15 = 0.85$. Therefore, the correct answer is $$\Pr[S \mid \bar C] = \frac{\Pr[S] - \Pr[S \mid C]\Pr[C]}{\Pr[\bar C]} = \frac{0.98 - (0.15)(0.96)}{0.85} = 0.983529,$$ as claimed.
Another way to reason is to use a frequentist model. Suppose we have a cohort of $8500$ births. Then the number of survivors overall is $(0.98)(8500) = 8330$, the number of non-survivors is $170$; the number of c-sections is $(0.15)(8500) = 1275$, the number of non c-sections is $7225$; and the number of survivors among those with c-sections is $(0.96)(1275) = 1224$. We then create the following table: $$\begin{array}{c|c|c|c} & C & \bar C & \\ \hline S & 1224 & ? & 8330 \\ \hline \bar S & ? & ? & 170 \\ \hline & 1275 & 7225 & 8500 \end{array}$$ It is now simple to fill in the blanks, since the columns and rows must add up to their respective totals: $$\begin{array}{c|c|c|c} & C & \bar C & \\ \hline S & 1224 & 7106 & 8330 \\ \hline \bar S & 51 & 119 & 170 \\ \hline & 1275 & 7225 & 8500 \end{array}$$ Now what is the desired probability? Count the proportion of survivors among non c-section births: this is simply $7106/7225 = 418/425 \approx 0.983529$.
Let $S$ be the event of surviving a birth, and $C$ the event that the birth is a C-section.
We are given $\mathsf P(S)=0.98, \mathsf P(C)= 0.15, \mathsf P(S\mid C)=0.96$
The Law of Total Probability states, $\mathsf P(S) = \mathsf P(S\mid C)\mathsf P(C)+\mathsf P(S\mid C^\complement)\mathsf P(C^\complement)$
So what we seek is: $\mathsf P(S\mid C^\complement) = \dfrac{P(S)-P(S\mid C)P(C)}{1-P(C)} = \dfrac{0.98-0.96\cdot0.15}{1-0.15} \approx 0.9835$
THIS MAKES NO SENSE how could the probability that a baby is not a c section and survive be higher than the overall survival rate of all births even including those that result from a c section?
Because the probability we seek is for the baby to survive when given that it is not a C-section.
The probability that a baby survives and the birth is not a C-section is $\mathsf P(S\cap C^\complement) = 0.98-0.96\cdot0.15 = 0.836$. Not the same thing at all.