fionanamweyi

2022-07-21

83 households have desktop computers 47 have both desktop and laptop 3 percent have neither .what is the probability that a random choice made a laptop household will be choosen

Don Sumner

To solve the problem, we can use the concept of probability.
We are given the following information:
- There are 83 households that have desktop computers.
- 47 households have both desktop and laptop computers.
- 3 percent of households have neither desktop nor laptop computers.
Let's denote the event choosing a laptop household as A. We want to find the probability of event A occurring, denoted as $P\left(A\right)$.
To calculate $P\left(A\right)$, we can use the principle of inclusion-exclusion. The principle states that for two events, A and B:
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
In our case, we want to find $P\left(A\right)$, which is the probability of choosing a laptop household. We know that 47 households have both desktop and laptop computers, and 3 percent have neither.
Using the principle of inclusion-exclusion, we can write:

Let's calculate each probability separately:
1. Probability of a laptop household with a desktop:
We are given that 47 households have both desktop and laptop computers. Out of the 83 households with desktop computers, the probability of choosing a household with a laptop is:

2. Probability of a laptop household without a desktop:
We know that 3 percent of households have neither desktop nor laptop computers. The probability of choosing a laptop household without a desktop is:

Now, let's substitute the values into our formula for $P\left(A\right)$:

Calculating the numerical value, we find:
$P\left(A\right)\approx \overline{)0.5422}$
Therefore, the probability of randomly choosing a laptop household is approximately 0.5422, or 54.22%.

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