Respuesta :
Answer:
0.784 = 78.4% probability that there will be at least one failed graft in the next three done
Step-by-step explanation:
To solve this question, we need to understand the binomial probability distribution and conditional probability.
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.
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: No failed grafts in the first seven
Event B: At least one fail in the next three.
Intersection of events A and B:
Since the probability of a graft failling is independent of other grafts, we have that:
[tex]P(A \cap B) = P(A)*P(B)[/tex]
So
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(A)*P(B)}{P(A)} = P(B)[/tex]
So we just have to find the probability of one fail in three trials.
Three trials means that [tex]n = 3[/tex].
The probability is
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{3,0}.(0.4)^{0}.(0.6)^{3} = 0.216[/tex]
Then
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.216 = 0.784[/tex]
0.784 = 78.4% probability that there will be at least one failed graft in the next three done
