Respuesta :
Total number of children = 16 + 19 = 35
Total number of boys = 16
Total number of girls = 19
Favorable outcomes = BBBBB, BBBBG, BBBGB, BBGBB, BGBBB, GBBBB
P(at least 4 boys) = (16/35)(15/34)(14/33)(13/32)(12/31) + 4 (16/35)(15/34)(14/33)(13/32)(19/31) = 78/5797 + 494/5797 = 52/527
Answer: 52/527
Total number of boys = 16
Total number of girls = 19
Favorable outcomes = BBBBB, BBBBG, BBBGB, BBGBB, BGBBB, GBBBB
P(at least 4 boys) = (16/35)(15/34)(14/33)(13/32)(12/31) + 4 (16/35)(15/34)(14/33)(13/32)(19/31) = 78/5797 + 494/5797 = 52/527
Answer: 52/527
Using the hypergeometric distribution, it is found that there is a 0.1199 = 11.99% probability that at least four boys are chosen.
The students are chosen without replacement, which is the reason why the hypergeometric distribution is used.
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.
In this problem:
- 16 + 19 = 35 students, thus [tex]N = 35[/tex].
- 16 are boys, thus [tex]k = 16[/tex].
- Five are chosen, thus [tex]n = 5[/tex].
The probability that at least four are boys is:
[tex]P(X \geq 4) = P(X = 4) + P(X = 5)[/tex]
In which:
[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 = 4) = h(4,35,5,16) = \frac{C_{16,4}C_{19,1}}{C_{35,5}} = 0.1065[/tex]
[tex]P(X = 5) = h(5,35,5,16) = \frac{C_{16,5}C_{19,0}}{C_{35,5}} = 0.0134[/tex]
Then:
[tex]P(X \geq 4) = P(X = 4) + P(X = 5) = 0.1065 + 0.0134 = 0.1199[/tex]
0.1199 = 11.99% probability that at least four boys are chosen.
A similar problem is given at https://brainly.com/question/8174838
