Question

Thomas wants to invite Madeline to a party. He has an 80% chance of bumping into her at school. Otherwise, he'll call her on the phone. If he talks to her at school, he's 90% likely to ask hem
to a party. However, he's only 60% likely to ask her over the phone.
This is the tree diagram that represents the probability of Thomas inviting Madeline to a party.

What is the probability of Thomas inviting Madeline to the party over the phone?

Answer 1

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:

$P(U|V)=\frac{P(V|U)P(U)}{P(V)}$

The law of total probability states that:

$P(V)=P(V|U)P(U)+P(V|U')P(U')$

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:

$P(X) = P(X|A)P(A)+P(X|B)P(B)$

         $=(0.90\times 0.80)+(0.60\times 0.20)\\=0.72+0.12\\=0.84$

Compute the value of P (B|X) as follows:

$P(B|X)=\frac{P(X|B)P(B)}{P(X)}$

             $=\farc{0.60\times 0.20}{0.84}\\\\=0.14286\\\\\approx 0.143$

Thus, the probability of Thomas inviting Madeline to the party over the phone is 0.143.