Given:
Percentage of numbers unlisted = 40% = 0.40
Sample size, n = 5
Let's find the probability that all 5 of the selected houses have unlisted numbers.
To find the probability, we have:
p(numbers unlisted) = 0.40
Hence, we have:
q = 1 - p = 1 - 0.40 = 0.60
For the probability, apply the binomial probability:
0 .
[tex]P(x=5)=(^5C_5)(0.40)^5(0.60)^{5-5}[/tex]Solving further:
[tex]\begin{gathered} P(x=5)=(\frac{5!}{5!(5-5)!})(0.01024)(0.60)^0 \\ \\ P(x=5)=(\frac{5!}{5!(0)!})(0.01024)(1) \\ \\ p(x=5)=1*0.01024*1 \\ \\ p(x=5)=0.01024 \end{gathered}[/tex]Therefore, the probability that all 5 numbers have unlisted numbers is 0.01024
• ANSWER:
0.01024
