Question

The probability that a patient recovers from a rare disease is 0.6. Calculate the
probability that out of 5 patients suffering from this disease, at least two would
recover.

Answer 1

It is Binomial distribution with $p=0.6, n=5.$

$$P(X\geq2)=1-(P(X=0)+P(X=1))=$$

$$=1-\begin{pmatrix} 5 \\ 0 \end{pmatrix}0.6^0(1-0.6)^{5-0}-\begin{pmatrix} 5 \\ 1 \end{pmatrix}0.6^1(1-0.6)^{5-1}=$$

$$=1-0.01024-0.0768=0.91296$$

$P(X\geq2)=0.91296$