There are 50 people in a room.Each person flips a unbiased coin only two times.If someone flips two heads in a row goes in the next room otherwise stays in the current room.What is the probability that at least 25 people will pass in the next room?
I am thinking that the solution is $P(X \geq 25)$ where $X$ is binomial distribution with parameters $X \sim B(50,1/4)$?
Am I correct ?
You are practically done. Depending on the problem context, you can do several things
Compute numerically the binomial sum.
Use the fact that a Binomial with large $N$ approximates a normal distribution of same mean and variance, and compute the (approximate) probability as a gaussian integral (tail).
Bound it using for example https://en.wikipedia.org/wiki/Chebyshev%27s_inequality
For the later: the Binomial has $\mu= N p = 12.5$ and $\sigma^2=Np(1-p)=9.375$, hence $\sigma=3.06$. Now $(25-\mu)/\sigma \approx 4$. Hence, you can readily assert that the probability will be very small. In particular, the (not tight) Chebyshev bound says that $p<1/16$