1. ## workers probability

In a large factory with many workers , 30 % of the workers belong to union A , 60 % Union B and the remaining 10 % doesn't join any union . A sample of 5 workers is selected .

(1) Find the probability that the sample has less than 2 members of UNion A .

(2) It's found that exactly 4 of these workers are members of union . What is the probability that more of the members come from Union A than UNion B .

so For (1) ,

it could be 1 member from A , then 4 from A' OR 0 from A and 5 from A'

P(x<2)=(0.3)(0.7)^4+(0.7)^5=0.2401

(2) I did not consider the one member who has no union because it doesn't make a difference so Union A - 4 , B - 0 , OR A-3 , B-1

P(A>B)=(0.3)^4+(0.3)^3(0.6)=0.0243

But unfortunately both are wrong . The answer given is (1) 0.528 (2) 1/9

2. Originally Posted by thereddevils
In a large factory with many workers , 30 % of the workers belong to union A , 60 % Union B and the remaining 10 % doesn't join any union . A sample of 5 workers is selected .

(1) Find the probability that the sample has less than 2 members of UNion A .

(2) It's found that exactly 4 of these workers are members of union . What is the probability that more of the members come from Union A than UNion B .

so For (1) ,

it could be 1 member from A , then 4 from A' OR 0 from A and 5 from A'

P(x<2)=(0.3)(0.7)^4+(0.7)^5=0.2401

(2) I did not consider the one member who has no union because it doesn't make a difference so Union A - 4 , B - 0 , OR A-3 , B-1

P(A>B)=(0.3)^4+(0.3)^3(0.6)=0.0243

But unfortunately both are wrong . The answer given is (1) 0.528 (2) 1/9

(a) Denote the $\displaystyle X_i$ to equal 1 if the ith pick is from member A, and 0 otherwise. then we're interested in $\displaystyle P(\sum_{i=0}^5{X_i}\leq1)$. If you then note that the $\displaystyle X_i$ are (5,.3) binomial random variables, you compute
$\displaystyle P(\sum_{i=0}^5{X_i}\leq1) = \sum_{k=0}^{1}\binom{5}{k}(.3)^k(.7)^{5-k} = .52822$

(b)We're now interested in:
$\displaystyle P(\sum_{i=0}^4{X_i}=3|I) + P(\sum_{i=0}^4{X_i}=4|I)$

where I is the condition that all samples consist of union workers from A or B.
Again this can be solved using the binomial probability distribution, but with an adjusted probability of success equal to: 3/10*10/9=1/3. Using this, we get

$\displaystyle \Bigg{(}P(\sum_{i=0}^4{X_i}=3|I) + P(\sum_{i=0}^4{X_i}=4|I)\Bigg{)}$ = $\displaystyle \binom{4}{3}(1/3)^3(2/3)+\binom{4}{4}(1/3)^4 = \frac1{9}$

P.S. by the way, $\displaystyle P(A|I) = \frac{P(A\cap{I})}{P(I)} = \frac{P(A\cap({A\cup{B}}))}{1-P(I^c)} = \frac3{10}*\frac{10}{9}$

3. Hello, thereddevils!

Here's the first one.
(I'm still working on the second one.)

In a large factory with many workers, 30% of the workers belong to union A,
60% to Union B, and the remaining 10% didn't join any union.
A sample of 5 workers is selected.

(1) Find the probability that the sample has less than 2 members of Union A.
"Less than two A's" means "no A's or one A."

. . $\displaystyle P(\text{no A's}) \;=\;9.07)^5 \:=\:0.16807$

. . $\displaystyle P(\text{one A}) \:=\:{5\choose1}(0.3)^1(0.7)^4 \:=\:0.36015$

Therefore: .$\displaystyle P(\text{less than two A's}) \;=\;0.16807 + 0.36015 \;=\;0.52822$