Answer:
The required probability = 0.7143
Step-by-step explanation:
From the information given:
From a group of eight candidates
The no. of candidates that enrolled in internships = 5
The no. of candidates that enrolled in teaching = 3
Also, supposed all the eight candidates are equally qualified;
Then, Let assume that:
Y to represent the number of internship trainee candidates hired.
N to represent no. of candidates in a group = 8
r to represent those who enrolled in paid internship = 5
Now, N - r = 3 (for those who enrolled in traditional teaching program)
Suppose; n represent the positions for local teaching which is given as 3;
Then; selecting 3 from 8 whereby some enrolled in internships and some in traditional teaching programs;
Then, Â let assume Y is a random variable that follows a hypergeometric distribution; we have:
[tex]p(Y = y) = \left \{ {{\dfrac{ \bigg (^r_y \bigg)\bigg (^{N-r}_{n-y} \bigg) }{ \bigg ( ^N_n \bigg) } } _\atop { ^{0, otherwise} } } \right.[/tex]
[tex]p(Y = y) = \left \{ {{\dfrac{ \bigg (^5_y \bigg)\bigg (^{3}_{3-y} \bigg) }{ \bigg ( ^8_3 \bigg) } } } } \right, y= 0,1,2,3[/tex]
Thus, the probability that two or more internship trained candidates are hired can be computed as:
p(Y ≥ 2) = p(Y=2) + p(Y =3)
[tex]p(Y \geq 2) = \dfrac{ \bigg ( ^5_2\bigg) \bigg ( ^3_1 \bigg)}{\bigg (^8_3 \bigg)} + \dfrac{\bigg (^5_3 \bigg) \bigg (^3_0 \bigg)}{\bigg ( ^8_3\bigg)}[/tex]
[tex]p(Y \geq 2) = \dfrac{40}{56}[/tex]
[tex]\mathbf{p(Y \geq 2) = 0.7143}[/tex]
