Question

1. thomas wants to invite madeline to a party. he has an 80% chance of bumping into her at school. otherwise, hell call her on the phone. if he talks to her at school, hes 90% likely to ask her to a party. however, hes 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. tree diagram image what is the value of x? 0.2 0.6 0.8 0.4

Explanation:

Step1: Recall tree diagram rules

In a tree diagram, the sum of probabilities from a single node should be 1. The node with 0.2 and X's branch: the other branch from that node (after 0.2) should satisfy \( 0.2 + \text{other probability} = 1 \)? Wait, no, wait. Wait, the first split: 0.8 (bumping at school) and 0.2 (not bumping, so calling). Then, for the "not bumping" (prob 0.2) branch, the two sub - branches should sum to 1. Wait, the two sub - branches from the node with 0.2 are X and 0.6? Wait, no, wait. Wait, the problem: when he doesn't bump into her (prob 0.2), he calls. Then, when he calls, the probability of asking is 0.6? Wait, no, the tree: the first level is bump (0.8) or not bump (0.2). Then, for bump (0.8), the next level is ask (0.9) or not ask (0.1). For not bump (0.2), the next level is ask (let's say X) or not ask? Wait, no, the problem says "if he talks to her at school, 90% likely to ask; over phone, 60% likely to ask". Wait, maybe the node with 0.2: the two branches are ask (X) and not ask? Wait, no, the sum of probabilities from a node in a tree diagram must be 1. So for the node where the probability of not bumping is 0.2, the two sub - probabilities (ask and not ask) should sum to 1? Wait, no, wait. Wait, the "otherwise, he'll call her on the phone. If he talks to her at school, 90% likely to ask; over the phone, 60% likely to ask". Wait, maybe the node with 0.2: the two branches are "ask over phone" (0.6) and "not ask over phone" (X)? Wait, no, that can't be. Wait, no, the sum of probabilities from a single node (the second - level node for not bumping) should be 1. So if one branch is 0.6 (ask over phone), the other branch (not ask over phone) should be \( 1 - 0.6=0.4 \)? Wait, no, wait the options are 0.2, 0.6, 0.8, 0.4. Wait, maybe I got the tree wrong. Wait, the first split: bump (0.8) or not bump (0.2). Then, for bump (0.8), the two branches are ask (0.9) and not ask (0.1) (since 0.9 + 0.1 = 1). For not bump (0.2), the two branches are ask (let's say X) and not ask (Y), and X+Y = 1. But the problem says "he's only 60% likely to ask her over the phone", so X = 0.6? Wait, no, the options don't have 0.6 as the answer? Wait, no, the options are 0.2, 0.6, 0.8, 0.4. Wait, maybe the node with 0.2: the two branches are "ask" (0.6) and "not ask" (X), so X=1 - 0.6 = 0.4? Wait, but 0.4 is an option. Wait, but let's re - examine. Wait, the tree: the first level is bump (0.8) or not bump (0.2). Then, for bump (0.8), the next level is ask (0.9) and not ask (0.1). For not bump (0.2), the next level is ask (0.6) and not ask (X). Then, since the sum of probabilities from the not - bump node must be 1, X = 1 - 0.6 = 0.4? But 0.4 is an option. Wait, but the question is "What is the value of X?". Wait, maybe the node with 0.2: the two branches are "ask over phone" (0.6) and "not ask over phone" (X), so X = 1 - 0.6 = 0.4. But let's check the options. The options are 0.2, 0.6, 0.8, 0.4. So 0.4 is an option. Wait, but maybe I made a mistake. Wait, no, the sum of probabilities from a node in a tree diagram must be 1. So for the node where the probability of not bumping is 0.2, the two sub - probabilities (ask and not ask) must sum to 1. If the probability of asking over phone is 0.6, then the probability of not asking over phone is \( 1 - 0.6 = 0.4 \). So X is 0.4? Wait, but let's check the options. The options include 0.4. So:

Step1: Identify the node's probability sum rule

In a tree diagram, the sum of probabilities from a single node is 1. For the node corresponding to the "not bumping into her" (probability 0.2) branch, the two sub - probabilities (asking over phone and not asking over phone) must sum to 1.

Step2: Calculate X

We know one of the sub - probabilities (asking over phone) is 0.6. So the other sub - probability (not asking over phone, which is X) is \( 1 - 0.6=0.4 \).

Answer:

0.4