For each of the last 100 days at a certain factory, Light A was either on or off for that...
GMAT Two Part Analysis : (TPA) Questions
For each of the last 100 days at a certain factory, Light A was either on or off for that day, and likewise for Light B. Each day that Light A was off, Light B was also off. Each day that Light B was on, Light A was also on. In addition, for exactly 40 of the last 100 days, at least one of the lights was on, and for exactly 20 of the last 100 days, both of the lights were on.
Assume one of the last 100 days is chosen at random. Select for \(\mathrm{P(A)}\) the probability that on the chosen day Light A was on, and select for \(\mathrm{P(B)}\) the probability that on the chosen day Light B was on. Make only two selections, one in each column.
Phase 1: Owning the Dataset
We have 100 days with two lights (A and B) that can each be on or off. Let's create a table to organize the possible states:
| Light State | Number of Days |
| Both off | ? |
| Only A on | ? |
| Only B on | ? |
| Both on | ? |
| Total | 100 |
Now let's apply our constraints:
- Constraint 1: If A is off → B is off
- Constraint 2: If B is on → A is on
These constraints tell us that B can ONLY be on when A is also on. This means "Only B on" is impossible!
Phase 2: Understanding the Question
Given information:
- For exactly 40 of 100 days: at least one light was on
- For exactly 20 of 100 days: both lights were on
Let's update our table with what we can deduce:
| Light State | Number of Days | Reasoning |
| Both off | 60 | \(100 - 40 = 60\) |
| Only A on | 20 | See below |
| Only B on | 0 | Impossible! |
| Both on | 20 | Given |
| Total | 100 | ✓ |
Key insight: Since B can only be on when A is on:
- Days with "at least one light on" = Days with "A on" = 40
- These 40 days split into: 20 days (only A) + 20 days (both A and B)
Phase 3: Finding the Answer
Now we can calculate the probabilities:
For \(\mathrm{P(A)}\): Light A was on for:
- 20 days (only A on) + 20 days (both on) = 40 days
- \(\mathrm{P(A)} = \frac{40}{100} = 0.40\)
For \(\mathrm{P(B)}\): Light B was on for:
- 0 days (only B on) + 20 days (both on) = 20 days
- \(\mathrm{P(B)} = \frac{20}{100} = 0.20\)
Phase 4: Solution
Final Answer:
- \(\mathrm{P(A)} = 0.40\)
- \(\mathrm{P(B)} = 0.20\)
These values satisfy all our constraints and match the given information perfectly.