Respuesta :
Answer:
The answer is "0.338"
Step-by-step explanation:
We have 133-57+1 = 77 balls to pocket, essentially. See the number of words around 57 and 133 and separable by 3 (132):
[tex]a_o = 57\\d = 3\\A_n = 132\\\to 132 = 57 + (n-1) \times 3\\\\\to 132= 57+3n-3\\\\\to 135-57 =3n\\\\\to 3n=78\\\\\to n=\frac{78}{3}\\\\\to n= 26[/tex]
Therefore, the number for a number around 57 and 133 divisible by 3 = 26
The probability required[tex]=\frac{26}{77} = 0.337[/tex]
Using the hypergeometric distribution, it is found that there is a 0.0128 = 1.28% probability that if you randomly select five of the balls from the bag all of the numbers on the selected balls will be divisible by 3 or by 7.
The balls are chosen without replacement, hence, the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
x is the number of successes.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Multiples in the interval:
- From 57 to 132, there are (132 - 57)/3 + 1 = 26 multiples of 3.
- From 63 to 133, there are (133 - 63)/7 + 1 = 11 multiples of 7.
- 63, 84, 105, 126 are multiples of both.
Hence, [tex]k = 26 + 11 - 4 = 33[/tex]
As for the other parameters:
- 133 - 57 = 76 balls, hence [tex]N = 76[/tex]
- 5 are chosen, hence [tex]n = 5[/tex]
The probability is P(X = 5), hence:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]
[tex]P(X = 5) = h(5,76,5,33) = \frac{C_{33,5}C_{43,0}}{C_{76,5}} = 0.0128[/tex]
0.0128 = 1.28% probability that if you randomly select five of the balls from the bag all of the numbers on the selected balls will be divisible by 3 or by 7.
A similar problem is given at https://brainly.com/question/24826394
