on a farm, there are a variety of animals.\n• 30% of the animals on the farm are chickens.\n• 24% of the…

on a farm, there are a variety of animals.\n• 30% of the animals on the farm are chickens.\n• 24% of the animals on the farm are chickens that lay eggs.\ngiven that a randomly selected animal from the farm is a chicken, what is the probability that it lays eggs?\nenter your answer in the space provided. enter only your answer. click save after entering your answer.\nclick here to enter your answer.

on a farm, there are a variety of animals.\n• 30% of the animals on the farm are chickens.\n• 24% of the animals on the farm are chickens that lay eggs.\ngiven that a randomly selected animal from the farm is a chicken, what is the probability that it lays eggs?\nenter your answer in the space provided. enter only your answer. click save after entering your answer.\nclick here to enter your answer.

Answer

Answer:

0.8

Explanation:

Step1: Define events

Let $A$ be the event of being a chicken, $B$ be the event of being an egg - laying chicken. We know $P(A)=0.3$ and $P(B) = 0.24$.

Step2: Use conditional - probability formula

The formula for conditional probability is $P(B|A)=\frac{P(A\cap B)}{P(A)}$. Since $B\subseteq A$, $P(A\cap B)=P(B)$.

Step3: Calculate the probability

Substitute $P(A) = 0.3$ and $P(B)=0.24$ into the formula: $P(B|A)=\frac{0.24}{0.3}=0.8$.