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Answer: Â Probability that either had Internet access or had cable television is 94%.
Step-by-step explanation:
Since we have given that
Probability of households had internet access P(I) = 85% = 0.85
Probability of households had cable television P(C) = 81% = 0.81
Probability of households had both Internet and cable television P(I ∩ C) = 72% = 0.72
We need to find the probability that either had Internet access or had cable television.
As we know the formula:
P(I ∪ C)=P(I) + P(C) - P(I ∩ C)
[tex]P(I\cup C)=0.85+0.81-0.72\\\\P(I\cup C)=0.94=94\%[/tex]
Hence, our required probability is 94%.
Probability of some event indicates its change of occurrence as an outcome. The probability of selecting a household that had either internet access or cable television is 0.94 = 94%
For such cases when two events and their probabilities (of occurring together or occurrence of either of them) are discussed, we can use the addition rule of two events.
What is the addition rule of probability for two events?
For two events A and B, we have:
Probability that event A or B occurs = Probability that event A occurs + Probability that event B occurs - Probability that both the event A and B occur simultaneously.
This can be written symbolically as:
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
How to convert percent to probability?
Percent counts the number compared to 100 whereas probability counts it compare to 1.
So, if we have a%, that means for each 100, there are 'a' parts. If we divide each of them with 100, we get:
For each 1, there are a/100 parts.
Thus, 50% = 50/100 = 0.50 (in probability)
Let we take
A = A household having internet access
B = A household having cable television
Then, we have:
[tex]P(A) = 85\% = 0.85\\P(B) = 81\% = 0.81\\P(A \cap B) = P(\text{A household having both internet access and cable television})\\P(A \cap B) = 72\% = 0.72[/tex]
Using the addition rule of probability, we get:
[tex]P(\text{A household having either internet access or cable television}) = P(A \cup B)\\\\P(A \cup B)= P(A) + P(B) + P(A \cap B)\\P(A \cup B) = 0.85 + 0.81 - 0.72 = 0.94[/tex]
Thus,
The probability of selecting a household that had either internet access or cable television is 0.94
Learn more about addition rule of probability here:
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