Question

solve

Answer 1

http://mathworld.wolfram.com/HypergeometricDistribution.html

have a look

Answer 2

yes i have taken a look

Answer 3

2 groups... well trained in first aid, well trained in hospitality

sample size of 2, let x = number of scouts who are wll trained in the sample.

x = 0, 1 or 2 these are the only possible values of x.

Answer 4

i didnt get it

Answer 5

its a binomial distribution

p = 30/50 = 0.6

mean = n*p = 2*.6 = 1.2

Answer 6

no its not

Answer 7

can u plse explain me a little bit @pgpilot326

Answer 8

the probabilities change... so it's not a binomial

let F = those who are well trained in first aid
let H = those who are well trained in hospitality

P(1st selected is F) = 30/50
P(2nd selected is F| 1st selected is F) = 29/49 so these events are not independent which they would need to be in order to model using a binomial distribution.

however it can be modelled by the hypergeometric distribution

if f = # of well trained in first aid scouts
and h = # of well trained in first aid scouts

and we sample 2 then

\[P \left( X=k \right)= \frac{ \left(\begin{matrix}30 \\ k\end{matrix}\right) \left(\begin{matrix}20 \\ 2-k\end{matrix}\right) }{ \left(\begin{matrix}50 \\ 2\end{matrix}\right) } \text{, where }k=\left\{ 0, 1, 2 \right\}\]

X is hypergeometric.

the equation above is the probability distribution.
if you read the link than you should be able to identify which parameters are needed in order to find the expected value and variance.

I have to go...

good luck!

Answer 9

@Loser66
can u plse help me

Answer 10

@sirm3d

Answer 11

i would have said it like @pgpilot326 , but without mentioning the type of distribution

Answer 12

i know @pgpilot326
had done correct
but i m not getting it
i mean how i have to solve the following step

Answer 13

step to finding the mean?

Answer 14

plse see the attachment

Answer 15

\[\binom{n}{k}=\frac{n!}{k!(n-k)!}\]
for instance, \[\binom{30}{1}=\frac{30!}{1!29!}=30\]

Answer 16

okay
and i have to put k=0, 1, 2 ......
turn by turn , i have to solve for k

Answer 17

solve for P(x=k) for each instance of k

Answer 18

i didnt get this step

Answer 19

when k=0\[P(x=0)=\frac{\binom{30}{0}\binom{20}{2}}{\binom{50}{2}}=\frac{1\cdot190}{1225}=\frac{190}{1225}\]

Answer 20

okay
till P(x=2) ,i have to calculte the values

Answer 21

right.
as for the mean, \[mean=\sum_{\text{all }k}k\cdot P(x=k)\]

Answer 22

the mean is the same as @dumbcow gave, 1.2

Answer 23

@pgpilot326
@sirm3d
thank you so much

Answer 24

yw