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Sagot :
Answer:
0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they pass, or they do not. The probability of an student passing is independent of other students, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Probability that all students pass in a class:
Class of 8 students, which means that [tex]n = 8[/tex]
Each student has a 95% chance of passing their class independent of the other students, which means that [tex]p = 0.95[/tex]
This probability is P(X = 8). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 8) = C_{8,8}.(0.95)^{8}.(0.05)^{0} = 0.6634[/tex]
Find the probability that, in exactly one of the two classes, all 8 students pass.
Two classes means that [tex]n = 2[/tex]
0.6634 probability all students pass in a class, which means that [tex]p = 0.6634[/tex].
This probability is P(X = 1). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,1}.(0.6634)^{1}.(0.3366)^{1} = 0.4466[/tex]
0.4466 = 44.66% probability that, in exactly one of the two classes, all 8 students pass.
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