Suppose that we go sequentially through a group of people asking each in turn to

Question

Suppose that we go sequentially through a group of people asking each in turn to pick one of 5 options (e.g., "rock," "paper," "scissors," "fire," or "water") at random. Let Xi (for 0 ? i ? 4) be the random variable giving the number of additional people we have to ask before getting an option that has not been picked yet. (So X0 is the number of people we have to ask to get the first option, which will be 1 person; X1 the number of additional people we ask to get the second, etc.)

Find the following:

P {Xi = k}
E [Xi]

Additionally, if X is the number of people we will have to ask before we've heard all all of the options, what is E[X]?

Explanation / Answer

X0 is the number of people we have to ask to get the first option, which will be 1 person.

So, P{X0 = 1} = 1

X1 the number of additional people we ask to get the second option with p = 4/5 (there are 4 options remaining out of 5 options). By Geometric distribution,

P{X1 = k} = (1/5)k-1 (4/5)

X2 the number of additional people we ask to get the third option with p = 3/5 (there are 3 options remaining out of 5 options). By Geometric distribution,

P{X2 = k} = (2/5)k-1 (3/5)

X3 the number of additional people we ask to get the fourth option with p = 2/5 (there are 2 options remaining out of 5 options). By Geometric distribution,

P{X3 = k} = (3/5)k-1 (2/5)

X4 the number of additional people we ask to get the fifth option with p = 1/5 (there are 1 option remaining out of 5 options). By Geometric distribution,

P{X4 = k} = (4/5)k-1 (1/5)

Thus,

P{X0 = 1} = 1

P{X1 = k} = (1/5)k-1 (4/5)

P{X2 = k} = (2/5)k-1 (3/5)

P{X3 = k} = (3/5)k-1 (2/5)

P{X4 = k} = (4/5)k-1 (1/5)

By Geometric distribution, E[Xi] = 1/pi

Thus,

E[X0] = 1

E[X1] = 1/(4/5) = 5/4

E[X2] = 1/(3/5) = 5/3

E[X3] = 1/(2/5) = 5/2

E[X4] = 1/(1/5) = 5/1 = 5

X is the number of people we will have to ask before we've heard all all of the options, then

X = X0 + X1 + X2 + X3 + X4

E[X] = E[X0 + X1 + X2 + X3 + X4]

= E[X0] + E[X1] + E[X2] + E[X3] + E[X4]

= 1 + (5/4) + (5/3) + (5/2) + 5

= 137/12

= 11.4167

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