F.5 probability(15marks)

14(a) P(Disease) = 0.05, P(No disease) = 0.95

P(Test positive |Disease) = 0.9 P(Test positive |No Disease) = 0.1P(positive) = 0.05*0.9 + 0.95*0.1 = 0.14(b) 0.095/0.14 = 0.6786(c) P(negative result) = 1 - 0.14 = 0.86P(Disease | negative result)P(negative result) = P(negative result | Disease)P(Disease)P(Disease | negative result)*0.86 = 0.1*0.05P(Disease | negative result) = 0.00581420(a) (6/15)*0.75 + (9/15)0.9 = 0.84(b) (6/15)*0.75/0.84 = 0.3571(c) Since the case is replacement (strange !), the answer is just square the value of (b)0.3571 = 0.1276(d) P(one trainee, one experienced typist | two error free)= 2(6/15)(9/15)(0.75)(0.9)/0.84^2= 0.4592

14a. Required probability
= 5% x 90% + 95% x 10%
= 0.14
b. Required probability
= (95% x 10%)/0.14
= 0.6786
c. Required probability
= (5% x 10%)/(5% x 10% + 95% x 90%)
= 0.005814

20a. Required probability
= (6/15)(0.75) + (9/15)(0.9)
= 0.84
b. Required probability
= (6/15)(0.75)/0.84
= 0.3571
c. Required probability
= [(6/15)(0.75)]^2/0.84^2
= 0.1276
d. Required probability
= 2 x (6/15)(0.75) x (9/15)(0.9) / 0.84^2
= 0.4592