Answer: 0.0123
Step-by-step explanation:
Given : Customers leaving a subway station can exit through any one of three gates.
The probability of selecting any particular gate = [tex]\dfrac{1}{3}[/tex]
Sample size : n= 4
Using binomial distribution,
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where P(x) is the probability of getting success inn x trials , n is the number of trails and p is the probability of getting success in each trial.
Then , the probability that among a sample of 4 customers, they all exit through the same gate.
[tex]P(4)=^4C_4(\dfrac{1}{3})^4(\dfrac{2}{3})^{0}=(1)(\dfrac{1}{3})^4=0.0123456790123\approx0.0123[/tex]
Hence, the probability that among a sample of 4 customers, they all exit through the same gate = 0.0123