Question

1. Hospital records show that of patients suffering

from a chronic disease 75% die of it.

What is the probability that of 6 randomly

selected patient 4 will recover.

2. A die is tossed 3 times. What is the probability of:

a. One five turning up

b. 3 fives turning up

c. No five turning up.

Answer 1

1. P(die) = 75 % = 0.75

P(recover) = 1 - 0.75 = 0.25

$P(4 \;will \;recover) = C^n_xp^xq^{n-x} \\ = C^6_4(0.25)^4(0.75)^2 \\ = 0.0329$

2. a. x =1

$P(X=1) = C^n_xp^xq^{n-x} \\ =C^3_0(\frac{1}{6})^1(\frac{5}{6})^2 \\ = \frac{75}{216} \\ = 0.34722$

b. x=3

$P(X=3) = C^n_xp^xq^{n-x} \\ =C^3_3(\frac{1}{6})^3(\frac{5}{6})^0 \\ = \frac{1}{216} \\ = 4.6296 \times 10^{-3}$

c. x= 0

$P(X=0) = C^n_xp^xq^{n-x} \\ = C^3_0(\frac{1}{6})^0(\frac{5}{6})^3 \\ = \frac{125}{216} \\ = 0.5787$