Answer:
Step-by-step explanation:
Total number of workers = 20+ 15 + 10 = 45
Suppose 6 workers are selected from 45 workers, the number of ways they can be selected is [tex](^{45}_6)[/tex] ways
If 7 workers are selected from 20 workers on a day shift;
Then, the number of ways to select 7 members from a group of 20 is [tex](^{20}_7)[/tex] ways;
∴
P( all 7 selected workers will be from the day shift) is:
[tex]= \dfrac{(^{20}_7)}{(^{45}_6)}[/tex]
[tex]=\dfrac{ \dfrac{20!}{7!(20-7)!} } { \dfrac{45!}{6!(45-6)!}}[/tex]
= 0.0095
The possible number of ways to select all day shift workers is:
[tex]=(^{20}_{7})[/tex]
The possible number of ways to select all swing shift workers is:
[tex]=(^{15}_{7})[/tex]
The possible number of ways to select all graveyard shift workers is:
[tex]= ( ^{10}_{7})[/tex]
∴
The probability that all selected workers are from the same shift is:
[tex]= \Bigg [ \dfrac{ (^{20}_7) }{(^{45}_6) } + \dfrac{ (^{15}_7) }{(^{45}_6) } + \dfrac{ (^{10}_7) }{(^{45}_6) } \Bigg][/tex]
[tex]= \begin {bmatrix} \dfrac{ \dfrac{20!} {7!(20-7)!}}{ \dfrac{ 45! }{6!(45-6)!} } + \dfrac{ \dfrac{15!} {7!(15-7)!}}{ \dfrac{ 45! }{6!(45-6)!} } + \dfrac{ \dfrac{10!} {7!(10-7)!}}{ \dfrac{ 45! }{6!(45-6)!} } \end {bmatrix}[/tex]
= 0.0095 + 0.00079 + 0.000015
= 0.010305
P(selected workers are from at least two different shifts)
= 1 - P(selected workers are from 1 shift from any of the 3 shifts)
= 1 - 0.010305
= 0.989695
≅ 0.9897
We can determine the probability that at least one shift is unrepresented through the following ways:
Let assume that:
[tex]X_1 =[/tex] event that day shift is unrepresented
[tex]X_2 =[/tex] event that swing shift is unrepresented
[tex]X_3 =[/tex] event that graveyard shift is unrepresented
Thus; [tex](X_1 \cap X_2)[/tex] = event that all of them are drawn from graveyard shift.
[tex](X_1 \cap X_3) =[/tex] event that all are drawn from swing shift
[tex](X_2 \cap X_3) =[/tex] event that they are all drawn from day shift.
So;
[tex]P(X_1) = \dfrac{(^{25}_{7})} { (^{45}_{6}) } = 0.0590[/tex]
[tex]P(X_2) = \dfrac{(^{30}_{7})} { (^{45}_{6}) } = 0.2499[/tex]
[tex]P(X_3) = \dfrac{(^{35}_{7})} { (^{45}_{6}) } = 0.8256[/tex]
[tex]P(X_1 \cap X_2) = \dfrac{(^{10}_{7})} { (^{45}_{6}) } = 0.000015[/tex]
[tex]P(X_1 \cap X_3) = \dfrac{(^{15}_{7})} { (^{45}_{6}) } = 0.00079[/tex]
[tex]P(X_2 \cap X_3) = \dfrac{(^{20}_{7})} { (^{45}_{6}) } = 0.0095[/tex]
[tex]P(X_1 \cap X_2 \cap X_3) =0[/tex]
Probability (that at least one shift is unrepresented) is:
[tex]= P(X_1) + P(X_2) +P(X_3) - P(X_1 \cap X_2) - P(X_1 \cap X_3) - P(X_2 \cap X_3) + P(X_1 \cap X_2 \cap X_3)[/tex]
= 0.0590 + 0.2499 + 0.8256 - 0.000015 - 0.00079 - 0.0095 + 0
≅ 1.124
