Question

1. a bag contains 5 red marbles and 10 blue marbles. sally selects two marbles without replacement. (a) what is the probability of selecting two red marbles? (b) what is the probability of selecting a red and a blue marble?

Explanation:

Step1: Calculate total number of marbles

Total marbles = \(5 + 10=15\)

Step2: Probability of first red marble

Probability of first red marble \(P(R_1)=\frac{5}{15}=\frac{1}{3}\)

Step3: Probability of second red marble (without replacement)

After first red is taken, there are \(4\) red and \(14\) total marbles. \(P(R_2)=\frac{4}{14}=\frac{2}{7}\)

Step4: Probability of two red marbles (a)

Using multiplication rule \(P(R_1\cap R_2)=P(R_1)\times P(R_2)=\frac{1}{3}\times\frac{2}{7}=\frac{2}{21}\)

Step5: Probability of red - blue (b)

Case 1: Red then Blue. \(P(RB)=\frac{5}{15}\times\frac{10}{14}=\frac{1}{3}\times\frac{5}{7}=\frac{5}{21}\)
Case 2: Blue then Red. \(P(BR)=\frac{10}{15}\times\frac{5}{14}=\frac{2}{3}\times\frac{5}{14}=\frac{5}{21}\)

Step6: Probability of red and blue (b)

Using addition rule \(P((RB)\cup(BR)) = P(RB)+P(BR)=\frac{5}{21}+\frac{5}{21}=\frac{10}{21}\)

Answer:

(a) \(\frac{2}{21}\)
(b) \(\frac{10}{21}\)