#### Question

The probability that a person who undergoes kidney operation will recover is 0.5. Find the probability that of the six patients who undergo similar operations,

(a) None will recover

(b) Half of them will recover.

#### Solution 1

Probability of recovery=P(R)= 0.5

Probability of non-recovery = `P(barR)=1-0.5=0.5`

(a) If there are six patients, the probability that none recovers

`=^6C_0xx[P(R)]^0xx[P(barR)]^6=(0.5)^6=1/64`

(b) Of the six patients, the probability that half will recover

`=^6C_3xx[P(R)]^3xx[P(barR)]^3=(6!)/(3!3!)xx0.5^3xx0.5^3=20xx1/64=5/16`

#### Solution 2

Let X be the number of patients who recovered out of 6.

P(patient recovers) = p = 0.5

∴ q = 1 − p = 1 − 0.5 = 0.5

Given, n = 6

∴ X ~ B(6, 0.5)

The p.m.f. of X is given by

P(X = x) = p(x) = `""^6C_x`(0.5)^{x}(0.5)^{6−x}, x = 0, 1, 2, ...., 6

a)P(none will recover) = P (X = 0)

= `""^6C_0`(0.5)^{0}(0.5)^{6}

= (1) (1) (0.5)^{6}

= 0.015625

(b) P(half of the patients will recover) = P (X = 3)

= `""^6C_3`(0.5)^{3}(0.5)^{3}

= 20 (0.5)6

= 20 × 0.015625

= 0.3125