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Sagot :
Answer:
a) 0.0287 = 2.87% probability that three non-defective components in a batch of seven components.
b) 0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.
Step-by-step explanation:
A sequence of Bernoulli trials composes the binomial distribution.
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.
The probability that a selected component is non-defective is 0.8
This means that [tex]p = 0.8[/tex]
a) Three non-defective components in a batch of seven components.
This is P(X = 3) when n = 7. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{7,3}.(0.8)^{3}.(0.2)^{4} = 0.0287[/tex]
0.0287 = 2.87% probability that three non-defective components in a batch of seven components.
b) 8 non-defective components are drawn before the first defective component is chosen.
Now the order is important, so the we just multiply the probabilities.
8 non-defective, each with probability 0.8, and then a defective, with probability 0.2. So
[tex]p = (0.8)^8*0.2 = 0.0336[/tex]
0.0336 = 3.36% probability that 8 non-defective components are drawn before the first defective component is chosen.
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