Answer to Question #193815 in Statistics and Probability for Jessica

Question

A pharmacy buys one box of 50 thermal scanners each from four

different companies. There are 8 defective scanners in the box from the first

company, 4 defective scanners from the second company, 3 defective

scanners from the third company, and 5 defective scanners from the fourth

company. If one scanner is taken from each box, what is the probability

that the scanners from the first and second companies are defective and

the scanners from the third and fourth companies are not?​

Answer 1

Total scanner = 50

Probability of defective scanner from A i.e. $p(A)=\dfrac{8}{50}=0.16$

Probability of defective scanner from B i.e. $p(B)=\dfrac{4}{50}=0.08$

Probability of defective scanner from C i.e. $p(C)=\dfrac{3}{50}=0.06$

Probability of defective scanner from D i.e. $p(D)=\dfrac{5}{50}=0.1$

Let X be the event of drawing scanner from box.

Probability of drawing defective scanner from A

$P(X/A)=\dfrac{P(A)}{P(A)+P(B)+P(C)+P(D)}\\[9pt]=\dfrac{0.16}{0.16+0.08+0.06+0.1}\\[9pt]=\dfrac{0.16}{0.4}=0.4$

Probability of drawing defective scanner from B

$P(X/B)=\dfrac{P(B)}{P(A)+P(B)+P(C)+P(D)}\\[9pt]=\dfrac{0.08}{0.16+0.08+0.06+0.1}\\[9pt]=\dfrac{0.08}{0.4}=0.2$

Probability of drawing defective scanner from C

$P(X/C)=\dfrac{P(C)}{P(A)+P(B)+P(C)+P(D)}\\[9pt]=\dfrac{0.06}{0.16+0.08+0.06+0.1}\\[9pt]=\dfrac{0.06}{0.4}=0.15$

Probability of drawing defective scanner from D

$P(X/D)=\dfrac{P(D)}{P(A)+P(B)+P(C)+P(D)}\\[9pt]=\dfrac{0.1}{0.16+0.08+0.06+0.1}\\[9pt]=\dfrac{0.1}{0.4}=0.25$

Required Probability $= P(X/A)*P(X/B)*P(X/C)*'P(X/D)'$

$=0.4*0.2*(1-0.15)*(1-0.25)\\=0.08*0.85*075\\=0.051$