Probabilty (1 Viewer)

[Kutay]

i am having a problem with this question i was wondering if anyone could help me on the right track on how to solve it or provide solutions to it or something...

There is a row of five coloured lights, each with its own switch. when switched on, each light is equally likely to show green, red or amber.

(i) if all the lights are switched on:
1) HOW MANY different coloured patterns are possible?
2) what is the Probabilty
- the first three lights from the left are green, but not the fourth
- the first three lights from the left are the same colour, but not the fourth
- exactly three of the lights are green
(ii) if all the lights are switched on five times, find, as a decimal correct to three decimal places, the probabilty that exactly three lights will be the same colour on two or three occasions.

thankyou

 

[Sepulchres]

i) 1) 5 x 3C1
2) i) (1/3)^3 x (1/2) x 3C1
ii) 3 {(1/3)^3 x (1/2) x 3C1}
iii) (1/3)^3 x (1/2)^2 = 0.00926...
ii) For this you use binomial probability and the part above... I think.
{0.9907 + 0.0093 } ^5

Actually let me work on the last one...

 

[Sepulchres]

ii) For this you use binomial probability and the part above... I think.
{0.9907 + 0.0093 } ^5

P(X=2)+ P(X=3) = 5C2 x (0.9907)^3 x (0.0093)^2 + 5C3 x (0.9907)^2 x (0.0093)^3
= 0.00084099 + 0.0000078946
= 0.000849 (3sf)

Now I dont know if it says to give it in significant figures or decimal places because otherwise the answer is 0? I dont know. I hope you can follow the steps so that you can use it to work it out properly. Sorry meng.

 

[acmilan]

Itd be 0.001 to 3 decimal places.

 

[Kutay]

Thankyou guys very much is this question a hard one or is it just me?

 

[ephemeral]

well I found it pretty hard, not that that counts for much

 

[lum]

ok, how come i'm wrong?? for the answers, i got

1. 3^5
2.i) (1/3)^3 x 2/3
ii) 3 {(1/3)^3 x 2/3}
iii) {(1/3)^3 x (2/3)^2}x 5C3

let answer to part iii) be Y

ii) 2 occasions - Y^2 x (1-Y)^3 x 5C2
3 occasions - Y^3 x (1-Y)^2 x 5C3

hmm, i'm really bad at probability, but i dun get how sephlchures got his answers.

 

[Sepulchres]

1)Heh, I just realised. You;re right for the first. Its {3C1}^5, which is what you have.
2) i) Yea I misread the whole thing.
(1/3)^3 x 2 x 3 - the probability of having 3 greens is (1/3), the fourth can be 2 colours only, and the fifth is any of the three colours.
ii) multipy the abovy by 3.
iii) (1/3)^3 x 2 x 2. exactly three green lights and the fourth and fifth light which can be one of two lights.
iv) use lum's method.

Sorry, I knew something was wrong.

 

[lum]

oh good, so i wasn't wrong, no worries, we all make silly mistakes... costed me around 8% in 3u trials