Question

In a bag of 60 candies, 36 are green and 45 have caramel in them. If the events of picking a green candy and picking a candy with caramel are independent, what is the probability of selecting two candies where the first is green and the second has caramel? Express your answer as a fraction in simplest form.

Answer 1

The probability of choosing a green candy and then a candy with caramel from the bag, given the selections are independent, is 9/20.

Step-by-step explanation:

To calculate the probability of selecting a green candy first and then a candy with caramel from a bag of 60 candies where 36 are green and 45 have caramel, given the events are independent:

1. First, determine the probability of selecting a green candy: P(Green) = 36/60.
2. Then, determine the probability of selecting a candy with caramel: P(Caramel) = 45/60.
3. Since the events are independent, multiply these probabilities together: P(Green and Caramel) = P(Green) × P(Caramel).

In simplest form, P(Green) = ⅔ (3/5) and P(Caramel) = ¾ (3/4). Multiplying the two gives us P(Green and Caramel) = ⅔ × ¾ = ⅜.

Answer 2

9/20

Step-by-step explanation:

Total number of candies =60

Number of Green candies, n(G)=36

Number of candies with caramel, n(C)=45

Since the events are independent, the probability of selecting two candies where the first is green and the second has caramel is given by:

P(GC)=P(G) X P(C)

`=(n(G))/(n(S)) X (n(C))/(n(S)) \\=(36)/(60) X (45)/(60)\\ =(9)/(20)`

The probability in its lowest form is 9/20.