Answer:
p = 6/11.
Step-by-step explanation:
So we have a bag that contains 6 red marbles and 6 green marbles.
Then the total number of marbles that are in that bag is:
6 + 6 = 12
There are 12 marbles in the bag, and we assume that all marbles have the same probability of being randomly drawn.
Now we draw two marbles, we want to find the probability that one is red and the other is green.
The first marble that we draw does not matter, as we just want the second marble to be of the other color.
So, suppose we draw a green one in the first attempt.
Then in the second draw, we need to get a red one.
The probability of drawing a red one will be equal to the quotient between the red marbles in the bag (6) and the total number of marbles in the bag (12 - 1 = 11, because one green marble was drawn already)
Then the probability is:
p = 6/11.
Notice that would be the exact same case if the first marble was red.
Then we can conclude that the probability of getting two marbles of different colors is:
p = 6/11.
89.890 Is what I would think it would be. Hope it helped.
<u>Answer:</u>
-16 cents
<u>Step-by-step explanation:</u>
We are given that on three rolls of a single die, you will lose $19 if a 3 turns up at least once, and you will win $5 otherwise.
We are to find the expected value of the game.
P (at least one 5 in three rolls) = 1 - P (no. of 3 in three) = 
= 0.875
P (other results) = 1 - 0.875 = 0.125
Random game value = -19, +5
Probabilities: 0.875, 0.125
Expected game value (X) = 0.875 × (-19) + 0.125 × (5) = -16 cents
Therefore, every time you play the game, you can expect to lose 16 cents
Answer:
After simplifying, the first step in solving an equation with a variable on both sides is to get the variable on one side. This is done by reversing the addition or subtraction of one of the terms with the variable
