Question

Suppose that a drawer contains 8 marbles: 2 are red, 2 are blue, 2 are green, and 2 are yellow. The marbles are rolling around in a drawer, so that all possibilities are equally likely when they are drawn. Alice selects marbles (without replacement) until she gets a red marble, and then she stops afterwards. Let X denote the number of draws that are needed until the first red appears.

a. Find P(X=1).?

b. Find P(X=2).?

c. Find P(X=3).?

d. Find P(X=4).?

e. Find P(X=5).?

f. Find P(X=6).?

g. Find P(X=7).

Answer 1

Red = 2

Others = 6

Total = 8

(a)
First draw = Red:

P(X=1) = 2/8 = 0.25

(b)

P(First Draw = Others) = 6/8 = 0.75

P(Second Draw=Red) = 2/7 = 0.2857

So,

P(X=2) = 0.75 X 0.2857 = 0.2143

(c)

P(Second draw=Others) = 5/7 = 0.7143

P(Third Draw) = Red) = 2/6 = 0.3333

So,

P(X=3) = 0.75 X 0.7143 X 0.3333 = 0.1786

(d)

P(Third Draw=Others) = 4/6 = 0.6667

P(Fourth=Red) = 2/5=0.4

So,

P(X=4) =0.75 X 0.7143 X 0.6667 X 0.4=0.1429

(e)

P(Fourth Draw=Others) = 3/5 = 0.6

P(Fifth Draw=Red) =2/4 = 0.5

So,

P(X=5) = 0.75 X 0.7143 X 0.6667 X 0.6 X 0.5 = 0.1072

(f)

P(Fifth Draw=Others) = 2/4 = 0.5

P(Sixth draw=Red) = 2/3 = 0.6667

So,

P(X=6) = 0.75 X 0.7143 X 0.6667 X 0.6 X 0.5 X 0.6667 = 0.0714

(g)

P(Sixth Draw=Others) = 1/3 = 0.3333

P(Seventh=Red) = 2/2 =1

So,

P(X=7) = 0.75 X 0.7143 X 0.6667 X 0.6 X 0.5 X 0.3333 X 1 = 0.0357