Using the table attached answer part A and B. Let us use the following notation for events: A = aggressive approach, Pa = passive approach, S = sale, N = no sale. So, P(A) is the probability that an aggressive approach was used, and so on. A) P(N | A) = B) P(A or S) =

Using the table attached answer part A and B. Let us use the following notation for events: A = aggressive approach, Pa = passive approach, S = sale, N = no sale. So, P(A) is the probability that an aggressive approach was used, and so on. A) P(N | A) = B) P(A or S) =

Mathematics For Machine Technology

8th Edition

ISBN:9781337798310

Author:Peterson, John.

Publisher:Peterson, John.

Chapter84: Binary Numeration System

Section: Chapter Questions

Problem 10A: Analyze the following numbers. 0.802

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Question

A) P(N | A) =

B) P(A or S) =

Sale
No Sale
Row Total
Aggressive
Passive
270
310
580
490
90
580
Column Total
760
400
1160

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Expert Solution

Step 1

A).

$P (N) =$ Probability that there is n " no sale" of automobile

$= \frac{N u m b e r o f " n o s a l e s "}{T o t a l n u m b e r o f c u s t o m e r s p a r t i c i p a t e d i n t h e s e m i n a r}$

$= \frac{400}{1160}$

$= 0.3448$

$P (N | A) =$ $\frac{N u m b e r o f a g r e s s i v e a p p r o a c h e s u s e d i n " n o s a l e " o f a u t o m o b i l e}{T o t a l n u m b e r o f a g r e s s i v e a p p r o a c h u s e d}$

$= \frac{310}{580}$

$= 0.5344$

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