Answer:
[tex](a)\ 0.0864[/tex]
[tex](b)\ 0.096[/tex]
[tex](c)\ 0.24[/tex]
[tex](d)\ 0.936[/tex]
Step-by-step explanation:
Given
[tex]p \to[/tex] close
[tex]q \to[/tex] fail to close
[tex]p = 60\%[/tex]
[tex]p = 0.60[/tex]
First, calculate the value of q
Using complement rule
[tex]q = 1 - p[/tex]
[tex]q = 1 - 0.60[/tex]
[tex]q = 0.40[/tex]
So, we have:
[tex]p = 0.60[/tex] and [tex]q = 0.40[/tex]
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:
[tex]Pr = p*p*p*q[/tex]
[tex]Pr = p^3*q[/tex]
[tex]Pr = 0.60^3*0.40[/tex]
[tex]Pr = 0.0864[/tex]
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:
[tex]Pr = q * q * p[/tex]
[tex]Pr = q^2 * p[/tex]
[tex]Pr = 0.40^2 * 0.60[/tex]
[tex]Pr = 0.096[/tex]
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:
[tex]Pr = q * p[/tex]
[tex]Pr = 0.40 * 0.60[/tex]
[tex]Pr = 0.24[/tex]
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:
[tex]Pr = q*q*q[/tex]
[tex]Pr = q^3[/tex]
The opposite is that the first he closes is one of the first three
So, we have:
[tex]Pr' = 1- Pr[/tex] --- complement rule
[tex]Pr' = 1- q^3[/tex]
[tex]Pr' = 1- 0.40^3[/tex]
[tex]Pr' = 1- 0.064[/tex]
[tex]Pr' = 0.936[/tex]
