An experiment was conducted where rats could choose to enter one of two corridors, A or B. A random sample of three rats is selected. Let x = number of rats that select corridor B. Assuming the rats select their favorite corridor independently of one another and that the two corridors are equally likely to be selected, find the probability distribution of x.
d: P(3) =
To find the probability distribution of x, we need to calculate the probability of each possible outcome (values of x) for the given experiment.
In this case, x represents the number of rats that select corridor B. The possible values for x are 0, 1, 2, and 3.
To calculate the probability of each outcome, we can use the binomial probability formula:
P(x) = (nCx) * (p^x) * (1-p)^(n-x)
Where:
- n is the total number of trials (number of rats selected in this case)
- x is the number of successful outcomes (number of rats that select corridor B)
- p is the probability of success (probability of a rat selecting corridor B)
- nCx is the binomial coefficient, which represents the number of ways to choose x items from a set of n items
Given the information, we know that the probability of success, p, is 0.5 (since both corridors are equally likely to be selected) and the total number of rats, n, is 3.
Now we can calculate the probability distribution for x:
P(0) = (3C0) * (0.5^0) * (1-0.5)^(3-0)
= (1) * (1) * (0.5^3)
= 0.125
P(1) = (3C1) * (0.5^1) * (1-0.5)^(3-1)
= (3) * (0.5) * (0.5^2)
= 0.375
P(2) = (3C2) * (0.5^2) * (1-0.5)^(3-2)
= (3) * (0.5^2) * (0.5)
= 0.375
P(3) = (3C3) * (0.5^3) * (1-0.5)^(3-3)
= (1) * (0.5^3) * (1)
= 0.125
So, the probability distribution for x is:
P(0) = 0.125
P(1) = 0.375
P(2) = 0.375
P(3) = 0.125
Therefore, P(3) = 0.125