Answer to Question #231652 in Statistics and Probability for Jamilah

Question

Hospital records show that of patients suffering from a certain disease, 25% die
of it.
(i) What is the probability that of 6 randomly selected patients, 4 will recover?
[3 marks]
(ii) What is the most probable number of recoveries out of 6 randomly selected
patients?
[

Answer 1

Let $X=$ the number of recovered patients: $X\sim Bin(n, p).$

Given $n=6, q=0.25, p=1-q=1-0.25=0.75.$

(i)

$$P(X=4)=\dbinom{6}{4}(0.75)^4(0.25)^{6-4}$$

$$=0.296630859375$$

(ii) The mode is an integer $M$  that satisfies

$$(n+1)p-1\leq M<(n+1)p$$

$$(6+1)\cdot0.75-1\leq M<(6+1)\cdot0.75$$

$$4.25\leq M<5.25$$

$$M=5$$

$$P(X=5)=\dbinom{6}{5}(0.75)^5(0.25)^{6-5}$$

$$=0.35595703125$$

The most probable number of recoveries out of 6 randomly selected patients is $5$ patients.