Given a Koi population that is 55 male and that 1 out of eve

Given a Koi population that is 55% male and that 1 out of every 3 males is yellow, while only 1 out of every 5 females is yellow, find the following probabilities.

M = A Male Koi is Drawn.

F= A Female Koi is Drawn.

Y = A Yellow Koi is Drawn.

What is the probability that a randomly drawn Koi will be yellow? Select the probability equation from the table below that will best answer this question.

a.) P(Y) = P(M) + P(F) - P(M Y)

b.) P(Y) = P(M) + P(F) - P(M U Y)

c.) P(Y|F) = P(Y) + P(F) - P(Y U F)

d.) P(Y|F) = P(Y) + P(F) - P(Y F)

e.) P(F|Y) = P(Y) + P(F) - P(Y U F)

f.) P(F Y) = P(F) + P(Y)

g.) P(F U Y) = P(F) + P(Y)

h.) P(Y|F) = P(F|Y)·P(Y) + P(M|Y)·P(Y)

i.) P(F|Y) = P(F|Y)·P(Y) + P(M|Y)·P(Y)

j.) P(Y|M) = P(F|Y)·P(Y) + P(M|Y)·P(Y)

k.) P(M|Y) = P(F|Y)·P(Y) + P(M|Y)·P(Y)

l.) P(M|Y) = P(F|Y)·P(F) + P(M|Y)·P(M)

m.) P(F|Y) = P(F|Y)·P(F) + P(M|Y)·P(M)

n.) P(F) = P(F|M)·P(M) + P(F|M)·P(M)

o.) P(F) = P(F|Y)·P(F) + P(F|Y)·P(M)

p.) P(Y) = P(F|Y)·P(Y) + P(M|Y)·P(Y)

q.) P(Y) = P(M|Y)·P(M) + P(F|Y)·P(F)

r.) P(Y) = P(Y|M)·P(M) + P(Y|F)·P(F)

Solution

Probability that a Koi is yellow can be found out using the theorem of total probability:

P(Y) = P(Male) * P(Yellow, given that person is male) + P(Female) * P(Yellow, given that person is female)

or, P(Y) = P(M) * P(Y | M) + P(F) * P(Y | F)

So equation in Option (r) is correct.