Statistics probability question a treatment for a specific disease is composed of two steps. If a patient gets a successful treatment for the first step, then he can receive the second step treatment. A patient who gets a successful treatment for both steps can be determined to complete recovery. 1st and 2nd step are statistically independent and their successful probability is 0.5 and 2/3 respectively. Find the probability that only 2 patients are cured completely after new treatment when there are 4 patietns. $$2 \cdot \frac 2 3 \cdot (0.5\cdot2+0.5\cdot2+0.5\cdot2)$$ Is this right?

The probability a specific patient goes successfully through both treatment phases is $(0.5)(2/3)$, which is $1/3$.

It follows that the number $X$ of patients who are cured has _binomial distribution_ , with the number of trials equal to $4$ and the probability of success on any trial equal to $1/3$.

The probability that $X=2$ is therefore $\binom{4}{2}(1/3)^2(2/3)^2$. Note that $\binom{4}{2}=6$.

**Remarks:** $1.$ If the binomial distribution is not yet familiar to you, let the patients be A, B, C, D. The probability A and B are cured but C and D are not is $(1/3)(1/3)(2/3)(2/3)$, which is $(1/3)^2(2/3)^2$. The probability A is cured, B is not, C is cured, D is not, is $(1/3)(2/3)(1/3)(2/3)$, which again is $(1/3)^2(2/3)^2$. In total there are $6$ ways to choose the two people who will be cured, and each way has probability $(1/3)^2(2/3)^2$, so our required probability is $6(1/3)^2(2/3)^2$.

$2.$ The expression you wrote down turns out to be equal to $1$. This cannot be right.