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lesya [120]
4 years ago
9

An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of

the balls will be (a) of the same color? (b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted
Mathematics
1 answer:
Dovator [93]4 years ago
5 0

Answers:

without replacement: a) 86/969   b)80/323

with replacement: a)853/6859      b)1440/6859

Step-by-step explanation:

<h3>a) probability that all three are of same color:</h3>

if the balls are not replaced

the equal probable results of experiment

the outcome space :

S=[R_{1} R_{2} R_{3} ,R_{1} R_{2} G_{1} ,R_{1} R_{2} B_{1}..........], and there are 19.18.17 elements in S.

where

R- event that all three balls are red.

G- events that all three balls are green.

B- events that all three balls are blue.

and the number of possibilities from S that are in Red is 5.4.3 there are 6.5.4 events in blue, and 8.7.6 in green

P(R)+P(B)+P(G)=(5.4.3/19.18.17)+(6.5.4/19.18.17)+(8.7.6/19.18.17)

=86/969

if the balls are placed

the equal probable results of experiment

the outcome space :

S=[R_{1} R_{2} R_{3}, R_{1} R_{2} G_{1},R_{1} R_{2} B_{1} .........]

and there are 19.19.19 elements in S.

R- event that all three balls are red.

G- events that all three balls are green.

B- events that all three balls are blue.

and the number of possibilities from S are in red is  5.5.5, there are 6.6.6 events in blue  and 8.8.8 in green

thus the result is:

P(R)+P(B)+P(G)=(5^3/19^3)+(6^3/19^3)+(8^3/19^3)=853/6859

<h3>b) probability that all three are in different colors:</h3>

if the balls are not replaced

the equal probable results of experiment

the outcome space :

S=[R_{1} R_{2} R_{3}, R_{1} R_{2} B_{1} ,R_{1} R_{2} G_{1}..... ]

and there are 19.18.17 elements in S

and if all balls have to be in different colors.

choose one of the 5 red balls, 1 of the 6 green balls and 1 of the 8 green balls. and for every choice of red, a green and a blue ball they can be permuted 3! ways

P(R,B,G)=(5.6.8.3!/19.18.17)=80/323

if the balls are replaces then

P(R,B,G)=(5.6.8.3!/19^3)=1440/6859

<h3 />

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