Math  /  Data & Statistics

Question4.13. According to Nielsen Media Research, approximately 86%86 \% of all U.S. households have high-definition television (HDTV). In addition, 49\% of all U.S. households own digital video recorders (DVR). Suppose 40\% of all U.S. households have HDTV and have DVR. A U.S. household is randomly selected. 1) What is the probability that the household has HDTV or has DVR? 2) What is the probability that the household does not have HDTV or does have DVR? 3) What is the probability that the household does have HDTV or does not have DVR? 4) What is the probability that the household does not have HDTV or does not have DVR?

Studdy Solution

STEP 1

What is this asking? What are the chances of a US household having different combinations of HDTV and DVR, given the percentages of households that own each? Watch out! Don't mix up "or" and "and", and remember that "not having something" is different from "having something"!
Also, be careful with how you visualize the overlapping percentages.

STEP 2

1. Define events and probabilities
2. Calculate the probability of HDTV or DVR
3. Calculate the probability of not HDTV or DVR
4. Calculate the probability of HDTV or not DVR
5. Calculate the probability of not HDTV or not DVR

STEP 3

Let's **define** some events! HH represents the event that a household has HDTV, and DD represents the event that a household has a DVR.
We're given some **key probabilities**: P(H)=0.86P(H) = 0.86 (**probability of having HDTV**) P(D)=0.49P(D) = 0.49 (**probability of having DVR**) P(HD)=0.40P(H \cap D) = 0.40 (**probability of having both HDTV and DVR**)

STEP 4

We want to find P(HD)P(H \cup D), the probability that a household has HDTV *or* a DVR.
Remember the **inclusion-exclusion principle**: P(HD)=P(H)+P(D)P(HD)P(H \cup D) = P(H) + P(D) - P(H \cap D) This prevents us from **double-counting** the households that have both!

STEP 5

Let's **plug in** our values: P(HD)=0.86+0.490.40P(H \cup D) = \textbf{0.86} + \textbf{0.49} - \textbf{0.40} P(HD)=1.350.40P(H \cup D) = 1.35 - 0.40P(HD)=0.95P(H \cup D) = \textbf{0.95}

STEP 6

Now we want P(HcD)P(H^c \cup D), the probability of *not* having HDTV *or* having a DVR.
We can use the **inclusion-exclusion principle** again: P(HcD)=P(Hc)+P(D)P(HcD)P(H^c \cup D) = P(H^c) + P(D) - P(H^c \cap D)

STEP 7

First, P(Hc)=1P(H)=10.86=0.14P(H^c) = 1 - P(H) = 1 - \textbf{0.86} = \textbf{0.14}.
This is the probability of *not* having HDTV.

STEP 8

Next, P(HcD)P(H^c \cap D) is the probability of *not* having HDTV *and* having a DVR.
Think of it as the part of DD that *doesn't* overlap with HH.
So, P(HcD)=P(D)P(HD)=0.490.40=0.09P(H^c \cap D) = P(D) - P(H \cap D) = \textbf{0.49} - \textbf{0.40} = \textbf{0.09}.

STEP 9

Putting it all together: P(HcD)=0.14+0.490.09P(H^c \cup D) = \textbf{0.14} + \textbf{0.49} - \textbf{0.09} P(HcD)=0.630.09P(H^c \cup D) = 0.63 - 0.09P(HcD)=0.54P(H^c \cup D) = \textbf{0.54}

STEP 10

We're looking for P(HDc)P(H \cup D^c).
Similar to the previous step, P(Dc)=1P(D)=10.49=0.51P(D^c) = 1 - P(D) = 1 - \textbf{0.49} = \textbf{0.51}, and P(HDc)=P(H)P(HD)=0.860.40=0.46P(H \cap D^c) = P(H) - P(H \cap D) = \textbf{0.86} - \textbf{0.40} = \textbf{0.46}.

STEP 11

Using the **inclusion-exclusion principle**: P(HDc)=P(H)+P(Dc)P(HDc)P(H \cup D^c) = P(H) + P(D^c) - P(H \cap D^c) P(HDc)=0.86+0.510.46P(H \cup D^c) = \textbf{0.86} + \textbf{0.51} - \textbf{0.46}P(HDc)=1.370.46P(H \cup D^c) = 1.37 - 0.46P(HDc)=0.91P(H \cup D^c) = \textbf{0.91}

STEP 12

Finally, we want P(HcDc)P(H^c \cup D^c).
Using the inclusion-exclusion principle: P(HcDc)=P(Hc)+P(Dc)P(HcDc)P(H^c \cup D^c) = P(H^c) + P(D^c) - P(H^c \cap D^c) We already know P(Hc)=0.14P(H^c) = \textbf{0.14} and P(Dc)=0.51P(D^c) = \textbf{0.51}.

STEP 13

P(HcDc)P(H^c \cap D^c) is the probability of *neither* having HDTV *nor* DVR.
This is the complement of having at least one, which we calculated in the first step as P(HD)=0.95P(H \cup D) = \textbf{0.95}.
So, P(HcDc)=1P(HD)=10.95=0.05P(H^c \cap D^c) = 1 - P(H \cup D) = 1 - \textbf{0.95} = \textbf{0.05}.

STEP 14

Therefore: P(HcDc)=0.14+0.510.05P(H^c \cup D^c) = \textbf{0.14} + \textbf{0.51} - \textbf{0.05} P(HcDc)=0.650.05P(H^c \cup D^c) = 0.65 - 0.05P(HcDc)=0.60P(H^c \cup D^c) = \textbf{0.60}

STEP 15

1) 0.95 2) 0.54 3) 0.91 4) 0.60

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