Question

a hat contains 4 marbles, one blue, one green, one orange, and one red. two marbles are selected at random without replacement from the hat. complete parts a) through e). use the fundamental counting principle to determine the answer to part a). assume that each event is equally likely to occur.
list the sample space. choose the correct answer below.
a. bg, bo, br, gb, go, gr, ob, og, or, rb, rg, ro
b. bb, bg, bo, br, gb, gg, go, gr, ob, og, oo, or, rb, rg, ro, rr
c. bb, gg, oo, rr
d. b, g, o, r
c) determine the probability of selecting exactly 1 blue marble.
the probability is \\( \frac { 1 } { 2 } \\)
(simplify your answer.)
d) determine the probability of selecting at least 1 marble that is not blue.
the probability is \\( \square \\)
(simplify your answer.)

Explanation:

Step1: Determine the sample space size

From part a), we know the sample space has \(n(S)=12\) elements (from option A: BG, BO, BR, GB, GO, GR, OB, OG, OR, RB, RG, RO).

Step2: Find the complement event

The event of "selecting at least 1 marble that is not blue" has a complement event. The complement of "selecting at least 1 marble that is not blue" is "selecting 2 blue marbles". But since there is only 1 blue marble, the number of ways to select 2 blue marbles \(n(\text{2 blue}) = 0\).

Step3: Use the probability formula for the complement

We know that \(P(A)=1 - P(\text{complement of }A)\). Let \(A\) be the event of "selecting at least 1 marble that is not blue". Then \(P(A)=1-\frac{n(\text{2 blue})}{n(S)}\)

Since \(n(\text{2 blue}) = 0\) and \(n(S)=12\), we have \(P(A)=1-\frac{0}{12}\)

Answer:

\(1\)