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atjari
Pls help. A box contains 50 rose cuttings out of that 20 red, 15 yellow, 5 white and 10 purple. If 6 rose cuttings are selected randomly a.P(5 red rose cuttings) b.P(2 red, 2 yellow, 2 purple) c.(<3 purple)
\[P(x) = ^nC_x p^x (1 - p)^{n - x}\] where p is the success rate, n is number of trials and x is the amount wanted a.P(5 red rose cuttings) p=20/50=2/5 n=6 x=5 \[P(5 red) = ^6C_5 0.4^5 (1 - 0.4)^{6 - 5}\]\[= 6* 0.4^5 *0.6^1\]\[=0.036864\]
We use the same method to work out the separate probabilities for b and c. b. P(2 red, 2 yellow, 2 purple)=P(2red)*P(2yellow)*P(2purple) c. P(<3 purple)=P(0purple)+P(1purple)+P(2purple)
For the other two I get b. P=0.00992 and c. P=0.98415
Is this right|dw:1329097834854:dw|
I'm not sure, I have not seen it set out like that before, sorry.
It's actually this |dw:1329098081859:dw|
This is the rule to use \[P(x)=^nC_xp^x(1−p)^{n−x}\] ^----Binomial Probability Formula The information that has a box contains 50 rose cuttings out of that 20 red, 15 yellow, 5 white and 10 purple is used to work out the corresponding p values (rate of success). In your response I don't think the probabilities of picking out a certain colour was used.
I think for P(5 red rose cuttings) I would do 6* (20/50)*(19/49)...(16/46)*(44/45) or 6C5* P(20,5)/P(50,5) * 44/45 the logic is, 20/50 chances of picking the first red rose, 19/49 of getting the second, etc times 44/45 of not getting a rose on the last pick. then 6 different orderings
but I always find the problems unintuitive
This seems to be pertinent http://en.wikipedia.org/wiki/Hypergeometric_distribution If we believe it, then P(5 red rose cuttings) out of 6 picks with 20 red roses of 50 roses is C(20,5)*C(30,1)/C(50,6) = 0.029
Hi Phi. If u dnt mind cn u tel me the answer for the c part.
Can any1 tel me the answer for the c part with the method.
I see I had a small bug in my very first post the chance of not getting a red rose is 30/45 and not 44/45. So I would do 6* (20/50)*(19/49)...(16/46)*(30/45) or 6C5* P(20,5)/P(50,5) * 30/45 which gives 0.029, matching the hypergeometric formula. I'll think about (c)
How about calculating P(0 purple) , P(1 purple), P(2 purple) and adding them up?
That sounds good. Bt if we cn make up a hypergeometric formula it wud be better.
\[\frac{\left(\begin{matrix}10 \\ 0\end{matrix}\right)\left(\begin{matrix}40 \\ 6\end{matrix}\right)}{\left(\begin{matrix}50 \\ 6\end{matrix}\right)}+\frac{\left(\begin{matrix}10 \\ 1\end{matrix}\right)\left(\begin{matrix}40 \\ 5\end{matrix}\right)}{\left(\begin{matrix}50 \\ 6\end{matrix}\right)}+\frac{\left(\begin{matrix}10 \\ 2\end{matrix}\right)\left(\begin{matrix}40 \\ 4\end{matrix}\right)}{\left(\begin{matrix}50 \\ 6\end{matrix}\right)}\]
R u quite sure of this answer dr? Jst gonna copy dis and go cos so busy nw to think more.
Thanx a lot lol. Will never forget this great help of urs.