A jar contains three red marbles and five green marbles. A marble is drawn at random and not replaced. A second marble is then d

Question

kiruha [24]

3 years ago

7

A jar contains three red marbles and five green marbles. A marble is drawn at random and not replaced. A second marble is then d

rawn from the jar. Find the probability that the first marble is red and the second marble is green.

Mathematics

2 answers:

scoundrel [369] · Answer 1 · 2022-02-28T13:16:21+03:00

scoundrel [369]3 years ago

8 0

Answer:

3/28

Step-by-step explanation:

3 red+ 5 green = 8 marbles in the jar

1st marble:

P(red) = number of red/ total = 3/8

We keep the marble

2 red+ 5 green = 7 marbles in the jar

P(green) = number of green/ total = 2/7

P (red,keep,green) = P(red)* P(green) = 3/8*2/7 = 6/56 = 3/28

zvonat [6] · Answer 2 · 2022-02-28T12:09:57+03:00

zvonat [6]3 years ago

6 0

Answer:

P(red then green) = 5/8 * 3/7 = 15/56

Step-by-step explanation: Since this is a logical "and" case, we have to multiply the separate events' probabilities...thus P(red then green) = 5/8 * 3/7 = 15/56