Answer:
Step-by-step explanation:
The persons who have diseases = 19%
The persons who do not have diseases = 81%
Probability for the test showing positive = Prob of actual disease and shows positive + Prob of no disease but test shows wrong
= [tex]0.7*0.19+0.04*0.81\\=0.133+0.0324\\=0.1654[/tex]
Prob that the person actually has the disease given that the test administered to an individual is positive= [tex]\frac{0.133}{0.1654} \\=0.0804[/tex]
30) A B C total
2x 7x x 10x
P(A) = 0.2, P(B) = 0.7, P(C) = 0.1
Let D be the event that the product is defective
P(D/A) = 0.032, P(D/B) = 0.034 and P(D/C) = 0.056
Required probability =P(C/D)
= [tex]\frac{P(D/C) P(C)}{P(D/A) P(a)+P(D/B) P(B)+P(D/C) P(C)}[/tex]
by using Bayes theorem for conditional probability
= [tex]\frac{0.1*0.056}{0.2*0.032+0.7*0.034+0.1*0.056} \\=\frac{0.0056}{0.0064+0.0238+0.0056} \\\\=0.156425[/tex]
32) 7 digit phone numbers with repitition including 0 will be
= [tex]10^7[/tex]
Numbers which have atleast one 7 = total number - numbers which do not have a single 7
= [tex]10^7 -9^7[/tex]
the probability that a 7-digit phone number contains at least one 7
=[tex]\frac{10^7-9^7}{10^7}[/tex]