Question

a bag has 12 marbles, all identical except for their color. there are 6 red marbles, 4 blue marbles, and 2 yellow marbles:
if you draw two marbles from the bag without replacement, find the following probabilities:
leave all answers as fractions
a.) ( p(rr) ) =
b.) ( p(rb) ) =
c.) ( p(ry) ) =
d.) ( p(br) ) =
e.) ( p(bb) ) =
f.) ( p(by) ) =
g.) ( p(yr) ) =
h.) ( p(yb) ) =
i.) ( p(yy) ) =

Explanation:

Step1: Calculate total number of marbles

Total marbles = 12

Step2: Calculate probabilities for each case

a. \(P(RR)\)

• First draw: Probability of red \(P(R_1)=\frac{6}{12}\)
• Second draw (without replacement): Probability of red \(P(R_2)=\frac{5}{11}\)
• \(P(RR)=\frac{6}{12}\times\frac{5}{11}=\frac{30}{132}=\frac{5}{22}\)

b. \(P(RB)\)

• First draw: Probability of red \(P(R_1)=\frac{6}{12}\)
• Second draw (without replacement): Probability of blue \(P(B_2)=\frac{4}{11}\)
• \(P(RB)=\frac{6}{12}\times\frac{4}{11}=\frac{24}{132}=\frac{2}{11}\)

c. \(P(RY)\)

• First draw: Probability of red \(P(R_1)=\frac{6}{12}\)
• Second draw (without replacement): Probability of yellow \(P(Y_2)=\frac{2}{11}\)
• \(P(RY)=\frac{6}{12}\times\frac{2}{11}=\frac{12}{132}=\frac{1}{11}\)

d. \(P(BR)\)

• First draw: Probability of blue \(P(B_1)=\frac{4}{12}\)
• Second draw (without replacement): Probability of red \(P(R_2)=\frac{6}{11}\)
• \(P(BR)=\frac{4}{12}\times\frac{6}{11}=\frac{24}{132}=\frac{2}{11}\)

e. \(P(BB)\)

• First draw: Probability of blue \(P(B_1)=\frac{4}{12}\)
• Second draw (without replacement): Probability of blue \(P(B_2)=\frac{3}{11}\)
• \(P(BB)=\frac{4}{12}\times\frac{3}{11}=\frac{12}{132}=\frac{1}{11}\)

f. \(P(BY)\)

• First draw: Probability of blue \(P(B_1)=\frac{4}{12}\)
• Second draw (without replacement): Probability of yellow \(P(Y_2)=\frac{2}{11}\)
• \(P(BY)=\frac{4}{12}\times\frac{2}{11}=\frac{8}{132}=\frac{2}{33}\)

g. \(P(YR)\)

• First draw: Probability of yellow \(P(Y_1)=\frac{2}{12}\)
• Second draw (without replacement): Probability of red \(P(R_2)=\frac{6}{11}\)
• \(P(YR)=\frac{2}{12}\times\frac{6}{11}=\frac{12}{132}=\frac{1}{11}\)

h. \(P(YB)\)

• First draw: Probability of yellow \(P(Y_1)=\frac{2}{12}\)
• Second draw (without replacement): Probability of blue \(P(B_2)=\frac{4}{11}\)
• \(P(YB)=\frac{2}{12}\times\frac{4}{11}=\frac{8}{132}=\frac{2}{33}\)

i. \(P(YY)\)

• First draw: Probability of yellow \(P(Y_1)=\frac{2}{12}\)
• Second draw (without replacement): Probability of yellow \(P(Y_2)=\frac{1}{11}\) (since only 1 yellow left after first draw)
• \(P(YY)=\frac{2}{12}\times\frac{1}{11}=\frac{2}{132}=\frac{1}{66}\)

Answer:

a. \(\frac{5}{22}\)
b. \(\frac{2}{11}\)
c. \(\frac{1}{11}\)
d. \(\frac{2}{11}\)
e. \(\frac{1}{11}\)
f. \(\frac{2}{33}\)
g. \(\frac{1}{11}\)
h. \(\frac{2}{33}\)
i. \(\frac{1}{66}\)