Question

according to the national telecom unction and information administration, 56.5% of us households owned a computer in 2001. what is the porbabilty that of three randomly selected us households at least one pwned a computer in 2001

Answer 1

The probability that at least one of three randomly selected US households owned a computer in 2001 is approximately 0.918, or 91.8%. ​​

How to find probability?

Probability of a household not owning a computer

`\[ P(\text{No Computer}) = 1 - 0.565 = 0.435 \]`

Probability that all three households did not own a computer:

`\[ P(\text{All three No Computer}) = (0.435)^3 = 0.082313 \]`

This represents the likelihood that each of the three households, independently, did not own a computer.

Probability that at least one household owned a computer:

`\[ P(\text{At least one Computer}) = 1 - P(\text{All three No Computer}) \\= 1 - 0.082313 \\= 0.917687 \]`

Therefore, the probability that at least one of three randomly selected US households owned a computer in 2001 is approximately 0.918, or 91.8%.