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

First, find the total number of pens. There is 1 red, 4 black, and 3 blue pens. So total pens \( n = 1 + 4 + 3 = 8 \).

Step2: Probability of first black pen

The probability of choosing a black pen first: there are 4 black pens out of 8 total. So \( P(\text{first black}) = \frac{4}{8} = \frac{1}{2} \).

Step3: Probability of second black pen (without replacement)

After choosing one black pen, there are now 3 black pens left and total pens left is \( 8 - 1 = 7 \). So \( P(\text{second black}) = \frac{3}{7} \).

Step4: Multiply the two probabilities

Since these are dependent events, the combined probability is the product of the two probabilities: \( P = \frac{4}{8} \times \frac{3}{7} = \frac{12}{56} \approx 0.2143 \), which rounds to 0.21.

Answer:

0.21