The probability of recovering from a disease

Question

It is known that the probability of recovering from a disease is $0.23$. In a community of 6 people with this disease, what is the probability that at least $3$ people will survive?

Hello guys, today our teacher had asked this question and I found and answer. I used binomial distribution to solve this problem and my answer is $0.1391$, in the system the answer $0,89$.

How can this would be possible? I am thinking that answer of teacher is wrong and mine is correct.

Answer 1

This seems to be resolved, but following the site recommendations to put answers in Answer format instead of Comments, I offer the following:

Using R to compute $P(X \ge 3) = \sum_{i=3}^6 {6\choose i}(.23)^i(.77)^{6-i}= 1 - P(X\le 2)$ for $X \sim \mathsf{Binom}(n=6, p=.23).$ In this computation, we assume that 'recover from the disease' and 'survive' mean the same thing. [In R, pbinom and dbinom are binomial CDF and PDF, respectively.]

1 - pbinom(2,6,.23)
[1] 0.139102
sum(dbinom(3:6, 6, .23))
[1] 0.139102