Assume I'm thinking of investing in a small biotech company. In a phase 2 study with 120 patients sp

It seems reasonable to assume that the tests involved in both Phase 2 and Phase 3 are tests that compare 'survival' fractions $p_1=X_1/n_1$ (new drug) with $p_2=X_2/n_2$ (old drug) to see if the former significantly exceeds the latter.
The test statistic $z=(p_1-<!--\; -\; -->p_2)/d$, where$d^2=p_1(1-<!--\; -\; -->p_1)/n_1+p_2(1-<!--\; -\; -->p_2)/n_2.$. We would say that the new drug is significantly better than the old at if z>z∗, where the critical value is z∗=2.3263 for a test at the 1% level and z∗=1.6449 at the 5% level.
For the Phase 2 test, we had $n_1=80$ and $n_2=40.$. The surviving numbers $X_1$ and $X_2$ with the new and old drugs, respectively, are unknown. To start, we assume that $X_2=20$ so that $p_2=0.5$. Then we ask what X1 must have been in order to achieve significance at the 1% level. Algebra would be a little messy, but a straightforward grid search for the smallest X1 that would give the desired result is $X_1=58$ so that $p_1=58/80=0.725,$,$z=2.4064>\mathrm{2.3263.}$
Then for a the Phase 3 test (assuming the same protocol for choosing patients, drug dosages, etc.) we suppose we would be choosing $X_2\sim <!--\; \sim \; -->Binom(100,0.5).$ Finding the corresponding $p_1,p_2,$ and z, we wonder what proportion of the z's would exceed 1.6449 in order to have significance at the 5% level, and (we hope) approval by the FDA. A simple simulation shows that proportion to be about 98.4%.
Fortunately, the outlook is about as promising, if we assume $p_2=0.3$ in the the Phase 2 trial; and even more promising (proportion about 99.5%), if we assume the the old drug has p2=0.8.
By making additional, possibly realistic, assumptions, this problem might be solved satisfactorily without using a computer. If this is a homework problem, I will leave that to you.
In case you are interested in my computation and simulation in R, the code is shown below. To avoid confusion in programming, I used X's and p's in Phase 2 and Y's and q's in Phase 3. (I do not claim my code is optimal or elegant.)
# Phase 2: Grid search for required p1
n1 = 80; n2 = 40; x2 = 20; z = numeric(n1)
for(x1 in 1:80) {
p1 = x1/n1; p2 = .5
den = sqrt( p1*(1-p1)/n1 + p2*(1-p2)/n2 )
z[x1] = (p1 - p2)/den }
crit = qnorm(.99)
i = 1:n1
pp1 = min(i[z > crit])/n1 ; pp1
## 0.725
Phase 3: Proportion of expts signif at 5%
m = 10^5; n1 = 200; n2 = 100
y1 = rbinom(m, n1, pp1); y2 = rbinom(m, n2, p2)
q1 = y1/n1; q2 = y2/n2
den = sqrt( q1*(1-q1)/n1 + q2*(1-q2)/n2 )
z = (q1 - q2)/den
mean(z > qnorm(.95))
## 0.98436