Question

75 gumballs are in a bag. 18 of them are green. what is the probability of drawing 2 green gumballs if the first one is placed back in the bag before the second draw? round your answer to the nearest hundredth.

Explanation:

Step1: Calculate first - draw probability

The probability of drawing a green gumball on the first draw is the number of green gumballs divided by the total number of gumballs. So, $P_1=\frac{18}{75}$.

Step2: Calculate second - draw probability

Since the first gumball is not replaced, for the second draw, there are now 17 green gumballs and 74 total gumballs. So, $P_2 = \frac{17}{74}$.

Step3: Calculate the probability of both events

The probability of both events (drawing two green gumballs) is the product of the probabilities of each event. So, $P = P_1\times P_2=\frac{18}{75}\times\frac{17}{74}=\frac{306}{5550}\approx0.05$.

Answer:

$0.05$