Select the probability that you will encounter only one red light out of five traffic signal lights.?

Question

Assume red, green, and yellow are equally likely occurrences.

A - 80/243
B - 4/243
C - 1/243
D - 1/5

Please explain how you tackled this problem or just explain to me how to attempt it. Thank you.

Answer 1

5 * (1/3)^1 * (2/3)^4 = 80/243. this is a binomial distribution with "success" being hitting a red light and five trials. the probability of success is 1/3, or red out of red, yellow, green

Answer 2

This is how i went about doing this question:
First since there is five events I make 5 blanks
_ _ _ _ _
then lets say the red light happens on the first light so there is only one possiblity on the first light, next since red can only happen once, there is 2 outcomes remaining for the last 4 spots with repetition.
1 x 2 x 2 x 2 x 2 = 16
now you times this by 5 as the red light can be in any of the five spots. (12222 21222 22122 22212 22221)
which is 16 x 5 = 80
Finally calculate the total number of possiblities
3 x 3 x 3 x 3 x 3 = 243
therefore the total probability is answer A. 80/243.

Answer 3

Any time you have to figure out the probability of a number of success trials that are independent this is a binomial distribution (Discrete):

Binomial(n,y,p)=(n y)p^y(1-p)^(n-y)

Where (n y) is the binomial coefficient.

So, what are our parameters:
p - probability of success (what is the chance that at any given light you will encounter a red light as opposed to yellow or green?)

n - Number of trials. 5 in your case.

y - number of successes. 1 in your case.

So:

Binomial(5,1,p)=(5 1)p(1-p)^4

Now (5 1) = 5 (I encourage you to check this)

Binomial(5,1,p)=5p(1-p)^4

Now all you have to do is figure out p and put it in the equation to get the correct answer.

Examine the equation closely, if you play around a little you'll see that really the equation represents actual things:

Since p is the probability that an event occurs (1-p) is the probability it doesn't. If you wanted to know what the probability is that you get two red lights on the first two and then three yellow or green on the next three it would be:

p^2 (two red lights) * (1-p)^3 (three red lights)

Now, this is a specific order. The (n y) counts how many ways you can shuffle around the above order and still get the same value of the above equation.

If you still need help, ask additional questions.

Answer 4

(5c1)*(1/3)^1*(2/3)^4
Explanation: the 5c1 comes from the fact that there are essentially 5 ways in which you could encounter only 1 red. I hope the rest is already clear. Email me if you want a more thorough explanation.