Question

the probability that a randomly-selected 2 year old male salamander will live to be three years old is 0.9 567 what is the possibility probability that two randomly selected two year old male salamanders with live to be three years old? what is the probability that nine randomly selected 2-year old male salamanders will live to be 3-years old? what is the probability that at least one of the nine randomly selected 2-year old male salamanders will not live to be 3 years old? would it be unusual if at least one of nine randomly selected 2 year old male salamanders did not live to be 3 years old?
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Answer 1

Answer:

a. $Probability = 0.9153$

b. $Probability = 0.6714$

c. $P(atleast\ one) = 0.99999999999$

d. It is not unusual

Step-by-step explanation:

Represent the given probability with P(T)

$P(T) = 0.9567$

Solving (a): Probability two selected will live to be three years old

This is calculated as:

$Probability = P(T) * P(T)$

$Probability = 0.9567 * 0.9567$

$Probability = 0.91527489$

$Probability = 0.9153$ --- Approximated

Solving (b): The probability that nine selected will live to be three years old

This is calculated as:

$Probability = (P(T))^9$

$Probability = 0.9567^9$

$Probability = 0.67140097202$

$Probability = 0.6714$   --- Approximated

Solving (c): Probability that at least one of the nine selected will not live to be three years old

To solve this, we need to determine the probability that none of the 9 selected will live to be 3 years old.

But first, we need to solve the probability that a selected male will not live to be 3 years old.

This is represented as: P(T') and is calculated as thus:

$P(T) + P(T') = 1$

$P(T') = 1 - P(T)$

$P(T') = 1 - 0.9567$

$P(T') = 0.0433$

The probability that none of the nine will live to be 3 years old is:

$P(none) = (P(T'))^9$

$P(none) = 0.0433^9$

So, the required probability is calculated as thus;

$P(atleast\ one) = 1 - P(none)$

$P(atleast\ one) = 1 - 0.0433^9$

$P(atleast\ one) = 0.99999999999$

Solving (d): Would (c) be unusual

This probability is calculated in (c) above.

$P(atleast\ one) = 0.99999999999$

Because it approximates to 1, we can conclude that it is not unusual