Respuesta :
Answer:
46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups
Step-by-step explanation:
We use two binomial trials 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 of at least nine participants finishing the study in a group.
0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that [tex]p = 0.8[/tex].
10 students, so [tex]n = 10[/tex]
We have to find:
[tex]P(X \geq 9) = P(X = 9) + P(X = 10)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684[/tex]
[tex]P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074[/tex]
[tex]P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758[/tex]
0.3758 probability that at least nine participants complete the study in a group.
Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?
0.3758 probability that at least nine participants complete the study in a group. This means that [tex]p = 0.3758[/tex]
Two groups, so [tex]n = 2[/tex]
We have to find P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691[/tex]
46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups
Answer:
Step-by-step explanation:
Assume that X and Y are the number of people who the survey in first and second groups respectively.
The probability of at least 9 participants complete the survey in one of the two groups , but not in both is
[X ≥ 9, Y ≤ 8 or X ≤ 8, Y ≥ 9] = P ( X ≥ 9) P(Y ≤ 8) + P ( X ≤ 8) P(Y ≥ 9)
Since the distribution in the two groups are same, so
P(X ≤ 8) = P(Y ≤ 8) and P(X ≥ 9) = P(Y ≥ 9)
Thus,
P(X ≥ 9) P(Y ≤ 8) + P(X ≤ 8) P(Y ≥ 9) = P(X ≥ 9)P(Y ≤ 8) + P( X ≤ 8)P(X ≥9)
= 2 P(X ≥ 9)P(X ≤ 8)
P(X ≥ 9)P(Y ≤ 8) + P(X ≤ 8)P(Y ≥ 9) = 2 P(X ≥ 9)(1 - P(X ≥ 9)) -----(1)
Since,
P(X ≤ 8) = 1 - P(X ≥ 9)
Here,n = 10, P = 0.8 and X = 9 and 10
P(x ≥ 9) = P(a) + P(10)
[tex]= \left(\begin{array}{ccc}10\\9\end{array}\right)(0.8)^9(0.2)^1+\left(\begin{array}{ccc}10\\10\end{array}\right)(0.8)^1^0(0.2)^0\\\\=0.3758[/tex]
Substitute P(X ≥ 9) = 0.3758 in eqn(1)
P( X ≥ 9 )P(Y ≤ 8) + P(X ≤ 8)P(Y ≥ 9) = 2 * 0.3758(1 - 0.3758)
= 0.7516(0.6242)
= 0.469
Therefore, the probability of at least 9 participants complete the study in one of the two groups but not in both the group is
(0.3758)(0.6242) + (0.6242)(0.3758)
= 0.469g