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? 0.25 0.21 0.19 0.02

Explanation:

Step1: Calculate total number of pens

Total pens \(=1 + 4+3=8\) pens.
Probability of choosing a black pen first: \(P(\text{first black})=\frac{4}{8}\)

Step2: Calculate probability after first - draw

After choosing one black pen, number of black pens left \(=4 - 1=3\), total pens left \(=8 - 1 = 7\)
Probability of choosing a black pen second (given first was black): \(P(\text{second black}|\text{first black})=\frac{3}{7}\)

Step3: Use multiplication rule for dependent events

By the formula \(P(A\cap B)=P(A)\times P(B|A)\), where \(A\) is "first pen is black" and \(B\) is "second pen is black"
\(P(\text{first black and second black})=\frac{4}{8}\times\frac{3}{7}=\frac{12}{56}\approx0.21\)

Answer:

\(0.21\)