Question

A salesman normally makes a sale​ (closes) on ​% of his presentations. Assuming the presentations are​ independent, find the probability of the following. ​a) He fails to close for the first time on his attempt. ​b) He closes his first presentation on his attempt. ​c) The first presentation he closes will be on his second attempt. ​d) The first presentation he closes will be on one of his first three attempts.

Answer 1

Answer:

$(a)\ 0.0864$

$(b)\ 0.096$

$(c)\ 0.24$

$(d)\ 0.936$

Step-by-step explanation:

Given

$p \to$ close

$q \to$ fail to close

$p = 60\%$

$p = 0.60$

First, calculate the value of q

Using complement rule

$q = 1 - p$

$q = 1 - 0.60$

$q = 0.40$

So, we have:

$p = 0.60$ and $q = 0.40$

Solving (a): Fails to close on the 4th attempt

This means that he closes the first three attempts. The event is represented as: p p p q

So, we have:

$Pr = p*p*p*q$

$Pr = p^3*q$

$Pr = 0.60^3*0.40$

$Pr = 0.0864$

Solving (b): He closes for the first time on the 3rd attempt

This means that he fails to close the first two attempts. The event is represented as: q q p

So, we have:

$Pr = q * q * p$

$Pr = q^2 * p$

$Pr = 0.40^2 * 0.60$

$Pr = 0.096$

Solving (c): First he closes is his 2nd attempt

This means that he fails to close the first. The event is represented as: q p

So, we have:

$Pr = q * p$

$Pr = 0.40 * 0.60$

$Pr = 0.24$

Solving (d): The first he close is one of his 3 attempts

To do this, we make use of complement rule

The event that he does not close any of his first three attempts is: q q q

The probability is:

$Pr = q*q*q$

$Pr = q^3$

The opposite is that the first he closes is one of the first three

So, we have:

$Pr' = 1- Pr$ --- complement rule

$Pr' = 1- q^3$

$Pr' = 1- 0.40^3$

$Pr' = 1- 0.064$

$Pr' = 0.936$