132k views
4 votes
Part A

A bag contains 4 blue and 6 green marbles. Two marbles are selected at random from the bag. If the first marble is replaced before the second marble is drawn, what is P(blue second | green first)?
Part B
If the first marble is not replaced before the second marble is drawn, what is P(blue second | green first)?

User Hoots
by
4.5k points

1 Answer

3 votes

Answer:

If the first marble is replaced before the second marble is drawn, the probability of P is 24%; while if the first marble is not replaced before the second marble is drawn, the probability of P is 26.66%.

Explanation:

Since a bag contains 4 blue and 6 green marbles, and two marbles are selected at random from the bag, to determine, if the first marble is replaced before the second marble is drawn, what is P (blue second | green first) and , if the first marble is not replaced before the second marble is drawn, what is P (blue second | green first) the following calculations must be performed:

Blue = 4/10

Green = 6/10

A) 6/10 x 4/10 = 0.24

B) 6/10 x 4/9 = 0.2666

Therefore, if the first marble is replaced before the second marble is drawn, the probability of P is 24%; while if the first marble is not replaced before the second marble is drawn, the probability of P is 26.66%.

User Hitesh Kansagara
by
5.2k points