The probability that a randomly selected 2 year old feral cat will live to be 3 years old is 0.73877.
A) what is the probability that 2 randomly selected 2 year old cats will live to be 3?
B) what is the probability that 7 will live to be 3?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
A) .73877^2 = ?
.5457811129
To calculate the probability of events, we can use the concept of independent events. If two events are independent, the probability of both events occurring is the product of their individual probabilities.
A) To find the probability that two randomly selected 2-year-old cats will live to be 3 years old, we can multiply the individual probabilities. Since both events are assumed to be independent, we multiply the probability of one cat living to be 3 by the probability of the other cat living to be 3:
Probability of both cats living to be 3 = Probability of the first cat living to be 3 * Probability of the second cat living to be 3
Let's assume the probability of a randomly selected 2-year-old cat living to be 3 years old is 0.73877, as stated in the question. Therefore, the probability is:
Probability of both cats living to be 3 = 0.73877 * 0.73877 = 0.5451015129 (rounded to 9 decimal places)
So, the probability that two randomly selected 2-year-old cats will live to be 3 is approximately 0.5451.
B) If you are asking for the probability of exactly 7 out of a group of 2-year-old cats living to be 3 years old, we need to use the concept of binomial probability. Binomial probability calculates the probability of obtaining a specific number of successes in a fixed number of independent Bernoulli trials.
For this case, we have a probability of success (living to be 3) for each trial (each cat) as 0.73877.
To calculate the probability of exactly 7 cats living to be 3 out of a group of randomly selected 2-year-old cats, we can use the binomial probability formula:
Probability = (Number of ways to choose 7 cats out of the total cats) * (Probability of success)^7 * (Probability of failure)^(total cats - 7)
Let's assume you have a total of 10 cats.
Number of ways to choose 7 cats out of 10 = 10C7 = (10!)/(7! * (10-7)!) = 120
Probability of exactly 7 cats living to be 3 = 120 * (0.73877)^7 * (1 - 0.73877)^(10 - 7)
Calculating this gives us:
Probability of exactly 7 cats living to be 3 = 0.3104451981 (rounded to 10 decimal places)
So, the probability that exactly 7 out of 10 randomly selected 2-year-old cats will live to be 3 is approximately 0.3104.