Respuesta :
Answer:
The total number of samples that contain at least 3 red balls is 115.
Step-by-step explanation:
Total number of balls = 11
Total number of red balls = 6
Total number of white balls = 5
A sample of 4 balls is to be selected that contain at least 3 red. It means either 3 out of 4 balls are red or 4 out of 4 ball are red.
[tex]\text{Total ways}=\text{Three balls are red}+\text{Four balls are red}[/tex]
[tex]\text{Total ways}=^6C_3\times ^5C_1+^6C_4\times ^5C_0[/tex]
Combination formula:
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
Using this formula we get
[tex]\text{Total ways}=\frac{6!}{3!(6-3)!}\times \frac{5!}{1!(5-1)!}+\frac{6!}{4!(6-4)!}\times \frac{5!}{0!(5-0)!}[/tex]
[tex]\text{Total ways}=20\times 5+15\times 1[/tex]
[tex]\text{Total ways}=115[/tex]
Therefore the total number of samples that contain at least 3 red balls is 115.
Using the combination formula, it is found that 115 samples contain at least 3 red balls.
The balls are chosen without replacement, which is why the combination formula is used.
Combination formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem, the outcomes with at least 3 red balls are:
- 3 red from a set of 6 and 1 white from a set of 5.
- 4 red from a set of 6.
Hence:
[tex]T = C_{6,3}C_{5,1} + C_{6,4} = \frac{6!}{3!3!}\frac{5!}{1!4!} + \frac{6!}{4!2!} = 20(5) + 15 = 100 + 15 = 115[/tex]
115 samples contain at least 3 red balls.
A similar problem is given at https://brainly.com/question/24437717
