Answer:
A) P(Betty is first in line and mary is last) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))
B) The method used is Relative frequency approach.
Step-by-step explanation:
From the question, we are told a sample of n kids line up for recess.
Now, the order in which they line up is random with each ordering being equally likely. Thus, this means that the probability of each kid to take a position is n(total of kids/positions).
Since we are being asked about 3 kids from the class, let's assign a letter to each kid:
J: John
B: Betty
M: Mary
A) Now, we want to find the probability that Betty is first in line or Mary is last in line.
In this case, the events are not mutually exclusive, since it's possible that "Betty is first but Mary is not last" or "Mary is last but Betty is not first" or "Betty is the first in line and Mary is last". Thus, there is an intersection between them and the probability is symbolized as;
P(B₁ ∪ Mₙ) = P(B₁) + P (Mₙ) - P(B₁ ∩ Mₙ) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))
Where;
The suffix 1 refers to the first position while the suffix n refers to the last position.
Also, P(B₁ ∩ Mₙ) = P(B₁) × P(Mₙ/B₁)
This is because the events "Betty" and "Mary" are not independent since every time a kid takes his place the probability of the next one is affected.
B) The method used is Relative frequency approach.
In this method, the probabilities are usually assigned on the basis of experimentation or historical data.
For example, If A is an event we are considering, and we assume that we have performed the same experiment n times so that n is the number of times A could have occurred.
Also, let n_A be the number of times that A did occur.
Now, the relative frequency would be written as (n_A)/n.
Thus, in this method, we will define P(A) as:
P(A) = lim:n→∞[(n_A)/n]
