Answer:
a. [tex]Probability = 0.97735[/tex]
b. [tex]Probability = 0.92294[/tex]
c. [tex]P(At\ Least\ One) = 1[/tex]
No, it is not unusual if at least 1 lives up to 3.
Step-by-step explanation:
Given
Represent the probability that a 2 year old snake will live to 3 with P(Live);
[tex]P(Live) = 0.98861[/tex]
Solving (a): Probability that two selected will live to 3 years.
Both snakes have a chance of 0.98861 to live up to 3 years.
So, the required probability is:
[tex]Probability = P(Live)\ and\ P(Live)[/tex]
[tex]Probability = 0.98861 * 0.98861[/tex]
[tex]Probability = 0.9773497321[/tex]
[tex]Probability = 0.97735[/tex] --- Approximated
Solving (b): Probability that seven selected will live to 3 years.
All 7 snakes have a chance of 0.98861 to live up to 3 years.
So, the required probability is:
[tex]Probability = P(Live)^n[/tex]
Where [tex]n = 7[/tex]
[tex]Probability = 0.98861^7[/tex]
[tex]Probability = 0.92294324145[/tex]
[tex]Probability = 0.92294[/tex] --- Approximated
Solving (c): Probability that at least one of seven selected will not live to 3 years.
In probabilities, the following relationship exist:
[tex]P(At\ Least\ One) = 1 - P(None).[/tex]
So, first we need to calculate the probability that none of the 7 lived up to 3.
If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861
This gives:
[tex]P(Not\ Live) = 0.01139[/tex]
The probability that none of the 7 lives up to 3 is:
[tex]P(None) = P(Not\ Live)^7[/tex]
[tex]P(None) = 0.01139^7[/tex]
Substitute this value for P(None) in
[tex]P(At\ Least\ One) = 1 - P(None).[/tex]
[tex]P(At\ Least\ One) = 1 - 0.01139^7[/tex]
[tex]P(At\ Least\ One) = 0.99999999999997513055642436060443621[/tex]
[tex]P(At\ Least\ One) = 1[/tex] ---- Approximated
No, it is not unusual if at least 1 lives up to 3.
This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years
