Question #4665

A bowl contains 27 M&M candies; 9 are red, 6 are green, and 12 are blue. A piece of candy is chosen, and its color is noted. Suppose you took 3 pieces of candy without looking. Define the event E: Each color is represented. What is P(E)?
Round your answer to 3 decimal places.

Expert's answer

Let RGB& be the event that the first candy was red, the second was blue, and

the third was red.

Similarly, define the events RBG, GRB, RBG, BRG,

BGR

Then these events are inconistent each with other and& E& is their union,

so

P(E) = P(RGB) + P(RBG) + P(GRB) + P(RBG) + P(BRG) + P(BGR)

Notice

that

& P(RGB)= 9/27 * 6/26 * 12/25 = (9*6*12)/(27*26*25)

& P(RBG)=

9/27 * 12/26 * 6/25 = (9*6*12)/(27*26*25) = P(RGB)

Similarly, it can be

shown that each of the events RBG, RBG, GRB, RBG, BRG, BGR has the same

probability

(9*6*12)/(27*26*25)

Hence

P(E) = 6 *

(9*6*12)/(27*26*25)

& = 6*9*6*12/(27*26*25)

& =

6*6*12/(3*26*25)

& = 6*6*12/(3*2*13*25)

& = 6*12/(13*25)

& =

72/325

& = 0.2215385

& = 0.222

the third was red.

Similarly, define the events RBG, GRB, RBG, BRG,

BGR

Then these events are inconistent each with other and& E& is their union,

so

P(E) = P(RGB) + P(RBG) + P(GRB) + P(RBG) + P(BRG) + P(BGR)

Notice

that

& P(RGB)= 9/27 * 6/26 * 12/25 = (9*6*12)/(27*26*25)

& P(RBG)=

9/27 * 12/26 * 6/25 = (9*6*12)/(27*26*25) = P(RGB)

Similarly, it can be

shown that each of the events RBG, RBG, GRB, RBG, BRG, BGR has the same

probability

(9*6*12)/(27*26*25)

Hence

P(E) = 6 *

(9*6*12)/(27*26*25)

& = 6*9*6*12/(27*26*25)

& =

6*6*12/(3*26*25)

& = 6*6*12/(3*2*13*25)

& = 6*12/(13*25)

& =

72/325

& = 0.2215385

& = 0.222

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