Question

a bag contains one red pen, four black pens, and three blue pens. two pens are randomly chosen from the bag and are not replaced. to the nearest hundredth, what is the probability that a black pen is chosen first and then another black pen is chosen? 0.02 0.19 0.21 0.25

Explanation:

Step1: Calculate total number of pens

Total pens = \(1 + 4+3=8\) pens.

Step2: Probability of first - black pen

Number of black pens = 4. Probability of choosing a black pen first, \(P(B_1)=\frac{4}{8}\)

Step3: Probability of second - black pen (no replacement)

After choosing one black pen, number of black pens left = 3 and total pens left = 7. Probability of choosing a black pen second, \(P(B_2|B_1)=\frac{3}{7}\)

Step4: Use multiplication rule for dependent events

\(P(B_1\cap B_2)=P(B_1)\times P(B_2|B_1)\)
\(P(B_1\cap B_2)=\frac{4}{8}\times\frac{3}{7}=\frac{12}{56}\approx0.19\)

Answer:

0.19