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:
<em>A</em> = Thomas bumps into Madeline at school
<em>B</em> = Thomas call Madeline on the phone
<em>X</em> = 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 <em>U</em> given that another events <em>V</em> has already occurred is:
![P(U|V)=\frac{P(V|U)P(U)}{P(V)}](https://tex.z-dn.net/?f=P%28U%7CV%29%3D%5Cfrac%7BP%28V%7CU%29P%28U%29%7D%7BP%28V%29%7D)
The law of total probability states that:
![P(V)=P(V|U)P(U)+P(V|U')P(U')](https://tex.z-dn.net/?f=P%28V%29%3DP%28V%7CU%29P%28U%29%2BP%28V%7CU%27%29P%28U%27%29)
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:
![P(X) = P(X|A)P(A)+P(X|B)P(B)](https://tex.z-dn.net/?f=P%28X%29%20%3D%20P%28X%7CA%29P%28A%29%2BP%28X%7CB%29P%28B%29)
![=(0.90\times 0.80)+(0.60\times 0.20)\\=0.72+0.12\\=0.84](https://tex.z-dn.net/?f=%3D%280.90%5Ctimes%200.80%29%2B%280.60%5Ctimes%200.20%29%5C%5C%3D0.72%2B0.12%5C%5C%3D0.84)
Compute the value of P (B|X) as follows:
![P(B|X)=\frac{P(X|B)P(B)}{P(X)}](https://tex.z-dn.net/?f=P%28B%7CX%29%3D%5Cfrac%7BP%28X%7CB%29P%28B%29%7D%7BP%28X%29%7D)
![=\farc{0.60\times 0.20}{0.84}\\\\=0.14286\\\\\approx 0.143](https://tex.z-dn.net/?f=%3D%5Cfarc%7B0.60%5Ctimes%200.20%7D%7B0.84%7D%5C%5C%5C%5C%3D0.14286%5C%5C%5C%5C%5Capprox%200.143)
Thus, the probability of Thomas inviting Madeline to the party over the phone is 0.143.