Question

Room\Reading <= 49 50 51 52 53 54<=
Bedroom 0.33 0.41 0.13 0.08 0.05 0
Kitchen 0.07 0.13 0.53 0.14 0.08 0.05
Living Room 0.06 0.09 0.13 0.28 0.33 0.11
a. What are the probabilities (that we are inside) for each room given that we obtain a reading of 53 (and assuming that we spend about equal amounts of time in each room – i.e., that the sample comes equally likely from any of the three rooms)?


b. For each reading we have our best guess (highest probability) determining which room we are in. What is our chance for being wrong when the reading is 53?


c. How would the result in B change if we knew that we are spending twice as much time in the bedroom than in the kitchen or living room (but the same in these two)?

Answer 1

$a.\text{ Events:}\\ D\text{ --- "we obtain a reading of 53"}\\ A\text{ --- "we are in a bedroom"}\\ B\text{ --- "in a kitchen"}\\ C\text{ --- "in a living-room"}\\ \text{We should find } P(A|D),\ P(B|D),\ P(C|D)\\ \text{We have:}\\ P(D|A)=0.05\\ P(D|B)=0.08\\ P(D|C)=0.33\\ P(A)=P(B)=P(C)=\frac{1}{3}\\ P(D)=0.05+0.08+0.33=0.46\\ P(A\cdot D)=P(D|A)\cdot P(A)=0.05\cdot \frac{1}{3}\\ P(A|D)=\frac{P(A\cdot D)}{P(D)}=\frac{0.05\cdot \frac{1}{3}}{0.46}\approx 0.036\\ \text{Similiarly:}\\ P(B\cdot D)=P(D|B)\cdot P(B)=0.08\cdot \frac{1}{3}\\ P(B|D)=\frac{P(B\cdot D)}{P(D)}=\frac{0.08\cdot \frac{1}{3}}{0.46}\approx 0.058\\ P(C\cdot D)=P(D|C)\cdot P(C)=0.33\cdot \frac{1}{3}\\ P(C|D)=\frac{P(C\cdot D)}{P(D)}=\frac{0.33\cdot \frac{1}{3}}{0.46}\approx 0.24\\ b.\ P=P(A|D)+P(B|D)=0.05+0.08=0.13\\ c.\text{ Let } P(B|D)=P(C|D)=x.\text{ Then } P(A|D)=2x\\ x+x+2x=0.46\\ 4x=0.46\\ x=0.115\\ 2x=0.23\\ \text{Then our } P=0.23$