Calculating Variance for Randomly Drawn Beads in a Necklace

[Jessica21]

1. A necklace consists of 5 beads on a string. The beads for making the necklace are drawn at random from a box containing a very large number of beads. 2/3 of the beads are pink and 1/3 are blue. find the mean and variance of the number of unlike pairs of adjacent beads in the necklace.

I am having trouble understanding what they mean by " the number of unlike pairs of adjacent beads"

2. well i know mean = E(x) = sum of all x of xf(x)
and Var (x)= E[(X-mean)^2]
3. attempt: so if i number the beads that go in the necklace from 1 to 5.
let X1 = 1 if beads 1 and 2 are different , and X1 = 0 if they are the same.
let X2 = 1 if beads 2 and 3 are different, and X2 = 0 if they are the same.
...
let X5 = 1 if beads 5 and 1 are different , and X5 = 0 if they are the same.

but i don't know how to find the probability function.

 

[Filip Larsen]

Assuming you have just drawn a single bead, can you then calculate the probability for the next bead to be different? If so, can you extend this calculation to a third, fourth and fifth bead? Perhaps you can recognize this probability distribution and use this to calculate the mean and variance for unlike pairs?

 

[Jessica21]

Um I am not sure if I get what you are trying to say but I have given it some thought and come up with the following!

if X1 =1then that means bead 1 = pink and bead 2 = blue , or bead 1 = blue and bead 2 = pink

so P (X1 =1 ) = (2/3)(1/3)+ (1/3)(2/3) = 4/9
and P (x1= 0) = 5/9

Do I just continue this for the rest of X2 through X5?

thank you!

 

[Filip Larsen]

if X1 =1then that means bead 1 = pink and bead 2 = blue , or bead 1 = blue and bead 2 = pink

so P (X1 =1 ) = (2/3)(1/3)+ (1/3)(2/3) = 4/9
and P (x1= 0) = 5/9

Correct. Notice, that the reason both first and second bead have same (and independent) probability for each color is that there are a very large number of beads.

Do I just continue this for the rest of X2 through X5?

Yes.

Some more hints if you need it:
If you draw a third bead and compare that to the second bead, what is the probability that the second and third bead are unlike. And again when you draw the fourth and fifth bead? After the fifth it seems you now have four pairs, each with the same constant probability of being unlike. Do you know of any probability distribution that describes a sequence of events where each event has two outcomes with constant probability p and 1-p? Can you use this to calculate the mean and variance of the number of unlike pairs out of 4 pairs in total?

 

[Jessica21]

I get it now!
thank you so much !