In a particular suburb 30% of housholds have installed electronic security systems. If 2 households are chosen at random from this area, what is the probability that neither has installed a security system?
Using the binomial probability function for this problem:
P(x) = (nCx)(p^x)[q^(n-x)]
p = .3
q = 1 - p = .7
x = 0
n = 2
P(0) = (2C0)(.3^0)(.7^2)
Finish the calculation for your probability.
To calculate the probability that neither household has installed a security system, we need to substitute the given values into the binomial probability formula:
P(x) = (nCx) * (p^x) * (q^(n-x))
Here:
p = 0.3 (probability of a household having installed a security system)
q = 1 - p = 0.7 (probability of a household not having installed a security system)
x = 0 (number of households without a security system)
n = 2 (total number of households being chosen)
Now we can substitute these values into the formula and solve for P(0):
P(0) = (2C0) * (0.3^0) * (0.7^(2-0))
= (2C0) * 1 * 0.7^2
= 1 * 1 * 0.7^2
= 0.7^2
= 0.49
Therefore, the probability that neither household has installed a security system is 0.49 or 49%.