Answer:
The probability of Thomas inviting Madeline to the party over the phone is 0.143.
Step-by-step explanation:
Consider the tree diagram below.
The events are denoted as follows:
A = Thomas bumps into Madeline at school
B = Thomas call Madeline on the phone
X = Thomas asks Madeline to the party
The information provided is:
P (A) = 0.80
P (B) = 1 - P (A) = 1 - 0.80 = 0.20
P (X|A) = 0.90
⇒ P (X'|A) = 1 - P (X|A) = 1 - 0.90 = 0.10
P (X|B) = 0.60
⇒ P (X'|B) = 1 - P (X|B) = 1 - 0.60 = 0.40
The conditional probability of event U given that another events V has already occurred is:
[tex]P(U|V)=\frac{P(V|U)P(U)}{P(V)}[/tex]
The law of total probability states that:
[tex]P(V)=P(V|U)P(U)+P(V|U')P(U')[/tex]
In this case we need to determine the probability that Thomas invites Madeline to the party over the phone, i.e. P (B|X).
Use the law of total probability to determine the value of P (X) as follows:
[tex]P(X) = P(X|A)P(A)+P(X|B)P(B)[/tex]
[tex]=(0.90\times 0.80)+(0.60\times 0.20)\\=0.72+0.12\\=0.84[/tex]
Compute the value of P (B|X) as follows:
[tex]P(B|X)=\frac{P(X|B)P(B)}{P(X)}[/tex]
[tex]=\farc{0.60\times 0.20}{0.84}\\\\=0.14286\\\\\approx 0.143[/tex]
Thus, the probability of Thomas inviting Madeline to the party over the phone is 0.143.
