Question

Four green disks, five purple disks, six yellow disks, and seven white disks are placed into a basket and drawn one at a time, replacing the disk after each draw. Find the probability that a yellow disk is drawn at least once by the third draw. Round your answer to two decimal places.

Answer 1

Answer:

The probability that a yellow disk is drawn at least once by the third draw is P=0.62.

Step-by-step explanation:

We have a total of 4+5+6+7=22 disks, of which 6 are yellow.

There is a probability of 6/22 of picking a yellow disk in the first draw.

Because the disks are replaced after each draw, the probability remain constant for every draw.

The probability of drawing at least one yellow disk by the third draw is the sum of:

- The probability of drawing one yellow disk in the 3 draws. This have 3 possible combinations.

- The probability of drawing two yellow disks in the 3 draws. This have 3 possible combinations.

- The probability of drawing three yellow disks in the 3 draws. This have only one combination possible.

The probability of picking a yellow disk in one draw is p=3/11.

$P(Y\geq 1)=P(Y=1)+P(Y=2)+P(Y=3)\\\\P(Y\geq 1)=3*p*(1-p)^2+3*p^2(1-p)+p^3\\\\P(Y\geq 1)=3*(3/11)*(8/11)^2+3*(3/11)^2*(8/11)+(3/11)^3\\\\P(Y\geq 1)=\frac{576}{1331} +\frac{216}{1331} +\frac{27}{1331}=\frac{819}{1331} = 0.62$