Respuesta :
Answer:
a) 0.3429 = 34.29% probability there is 1 red ball out of the 2 balls drawn.
b) 0.3715 = 37.15% probability that there is at least 1 red ball out of 2 balls drawn
Step-by-step explanation:
The balls are drawn without replacement, which means that the hypergeometric distribution is used to solve this question.
Hypergeometric distribution:
The probability of x successes is given by the following formula:
[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]
In which:
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.
Combinations formula:
[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]
In this question:
3 + 5 + 7 = 15 balls, which means that [tex]N = 15[/tex]
3 red, which means that [tex]k = 3[/tex]
2 drawn, which means that [tex]n = 2[/tex]
a) Probability there is 1 red ball out of the 2 balls drawn
This is [tex]P(X = 1)[/tex]
So
[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 = 1) = h(1,15,2,3) = \frac{C_{3,1}*C_{12,1}}{C_{15,2}} = 0.3429[/tex]
0.3429 = 34.29% probability there is 1 red ball out of the 2 balls drawn.
b) Probability there is at least 1 red ball out of 2 balls drawn
This is:
[tex]P(X \geq 1) = P(X = 1) + P(X = 2)[/tex]
So
[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 = 1) = h(1,15,2,3) = \frac{C_{3,1}*C_{12,1}}{C_{15,2}} = 0.3429[/tex]
[tex]P(X = 2) = h(2,15,2,3) = \frac{C_{3,2}*C_{12,0}}{C_{15,2}} = 0.0286[/tex]
[tex]P(X \geq 1) = P(X = 1) + P(X = 2) = 0.3429 + 0.0286 = 0.3715[/tex]
0.3715 = 37.15% probability that there is at least 1 red ball out of 2 balls drawn
