Question

assume that you are drawing two balls without replacement from an urn that contains 14 green balls, 6 blue balls, and 4 red balls. what is the probability that you will draw a green ball and a red ball? (hint: there are two ways that this can happen.) the probability that you draw a green ball and a red ball is (round to three decimal places as needed.)

Explanation:

Step1: Calculate the total number of balls

The total number of balls is \(14 + 8+4=26\) balls.

Step2: Calculate the probability of the two - order cases

Case 1: Green first, then red

The probability of drawing a green ball first is \(P(G_1)=\frac{14}{26}\). After drawing a green ball, there are \(26 - 1=25\) balls left. The probability of drawing a red ball second is \(P(R_2|G_1)=\frac{4}{25}\). So the probability of this case is \(P_1=\frac{14}{26}\times\frac{4}{25}=\frac{56}{650}\)

Case 2: Red first, then green

The probability of drawing a red ball first is \(P(R_1)=\frac{4}{26}\). After drawing a red ball, there are \(26 - 1 = 25\) balls left. The probability of drawing a green ball second is \(P(G_2|R_1)=\frac{14}{25}\). So the probability of this case is \(P_2=\frac{4}{26}\times\frac{14}{25}=\frac{56}{650}\)

Step3: Calculate the total probability

The total probability \(P = P_1+P_2\). Since \(P_1 = P_2=\frac{56}{650}\), then \(P=\frac{56 + 56}{650}=\frac{112}{650}=\frac{56}{325}\approx0.172\)

Answer:

\(0.172\)