Question

A bag contains 8 blue coins and 6 red coins. a coin is removed at random and replaced by three of the other color.

Answer 1

Answer:

a)0.57

b)0.56

c)0.31

Step-by-step explanation:

a) There are a total of 14 coins in the bag initially, of which 8 are blue, therefore:

$P(blue)=\frac{8}{14}=\frac{4}{7}=0.57$

b) After removing one blue, there are 7 blue coins and 6 red coins, to which 3 more red coins are added, making it a total of 9 red coins. That is a total of 16 coins, therefore:

$P(red)=\frac{9}{16}=0.56$

c)After removing one red from the initial scenario, there are 8 blue coins and 5 red coins, to which 3 blue coins are added. That makes a total of 16 coins of which 11 are blue, and 5 are red, therefore:

$P(red)=\frac{5}{16}=0.31$

Answer 2

Probability is equal to the favorable outcome divided by the total outcomes. So, in this case, the favorable outcome is blue, which is 8. The total amount of outcomes is 14.
8/14 reduces to 4/7, which is the answer to the first part.

Now you have a total of 18 coins in the bag, because you added three and subtracted one. The favorable outcome is now red, and there were six originally. Adding the three, you now have 9.
9/18 reduces to 1/2, which is the answer to this part.

After you remove a red, you now have 8 reds in the bag. The total is now 20, because you added three and took away one. 
8/12 reduces to 2/3, which is your final answer.

Good luck, and I hope my answer was helpful!