need help to see if i did wrong

[jays2]

Hi, I have a probability question and i feel like i did it wrong.


A comparative psychologist wished to assess the spatial memory of rats. A rat was placed in a maze with 12 pathways, only one of which had some food at the end of it. The rat was removed from the maze after finding the food. 24 hours later the rat was placed in the same maze and the experimenter recorded whether it first selected the same pathway to get food as before. This procedure was repeated with 5 rats in total. Three of the five rats went down the correct pathway at the first attempt. What is the probability that this would happen by chance?

I simply did 1/12 x 1/12 x 1/12 x 11/12 x 11/12 = 0.00486 which i feel like is probably wrong, please help to see if i did wrong and give me some hints on where to start the questions with.

 

[Dr.Peterson]

Hi, I have a probability question and i feel like i did it wrong.

A comparative psychologist wished to assess the spatial memory of rats. A rat was placed in a maze with 12 pathways, only one of which had some food at the end of it. The rat was removed from the maze after finding the food. 24 hours later the rat was placed in the same maze and the experimenter recorded whether it first selected the same pathway to get food as before. This procedure was repeated with 5 rats in total. Three of the five rats went down the correct pathway at the first attempt. What is the probability that this would happen by chance?

I simply did 1/12 x 1/12 x 1/12 x 11/12 x 11/12 = 0.00486 which i feel like is probably wrong, please help to see if i did wrong and give me some hints on where to start the questions with.

You found the probability that the first 3 were right and the last 2 were wrong. They want the probability that some 3 were right and the other 2 were wrong.

How can you modify your result to get what you were asked for? (Alternatively, you could think about the binomial distribution, if you have learned it.)

 

[jays2]

You found the probability that the first 3 were right and the last 2 were wrong. They want the probability that some 3 were right and the other 2 were wrong.

How can you modify your result to get what you were asked for? (Alternatively, you could think about the binomial distribution, if you have learned it.)

so i've done the binomial test, is the probability that this would happen by chance 0.5%?

 

[JeffM]

It greatly helps if you ALWAYS show your work even if you are just asking us to confirm an answer (which is a quite acceptable thing to do). That way if you are wrong, we can show you where and why. I presume you did something like

[MATH]\dbinom{5}{3} * \left ( \dfrac{1}{12} \right )^3 * \left ( \dfrac{11}{12} \right )^2 = \dfrac{5 * 4}{2} * \dfrac{11^2}{12^5} = \dfrac{1210}{248832} \approx 0.49\%[/MATH]
That is correct.

 

[jays2]

i

It greatly helps if you ALWAYS show your work even if you are just asking us to confirm an answer (which is a quite acceptable thing to do). That way if you are wrong, we can show you where and why. I presume you did something like

[MATH]\dbinom{5}{3} * \left ( \dfrac{1}{12} \right )^3 * \left ( \dfrac{11}{12} \right )^2 = \dfrac{5 * 4}{2} * \dfrac{11^2}{12^5} = \dfrac{1210}{248832} \approx 0.49\%[/MATH]
That is correct.

i see, will keep that in mid, thanks for the help!

 

[Dr.Peterson]

I simply did 1/12 x 1/12 x 1/12 x 11/12 x 11/12 = 0.00486 which i feel like is probably wrong, please help to see if i did wrong and give me some hints on where to start the questions with.

so i've done the binomial test, is the probability that this would happen by chance 0.5%?

I notice now that your result (but not your work!) was actually correct (and equal to what you now say it is, after correctly multiplying by \({5\choose 3}=10\), which is what makes up for the different possible orders. Did you notice that?