if you flip three fair coins what is the probability that you"ll get heads on the first two flips and tails on the last flip

To check all the events (6), we label the chips. Suppose one chip with 1 is labeled R1 and the other B1 (as if they were red and blue). Now, lets take all combinations; for the first chip, we have 4 choices and for the 2nd chip we have 3 remaining choices. Thus there are 12 combinations. Since we don't care about the order, there are only 6 combinations since for example R1, 3 is the same as 3, R1 for us.

The combinations are: (R1, B1), (R1, 3), (R1, 5), (B1, 3), (B1, 5), (3,5)

We have that in 1 out of the 6 events, Miguel wins 2$ and in five out of the 6 events, he loses one. The expected value of this bet is: 1/6*2+5/6*(-1)=-3/6=-0.5$. In general, the expected value of the bet is the sum of taking the probabilities of the outcome multiplied by the outcome; here, there is a 1/6 probability of getting the same 2 chips and so on. On average, Miguel loses half a dollar every time he plays.

Step-by-step explanation:

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15%

step-by-step explanation:

use the table and count the number of guesses that has two correct answers.

the following are two correct answers

10010, 01100, 11000

a total of 3 numbers having 2 correct guesses out of 20 questions.

probability of getting 2 correct guesses = the number of questions getting 2 correct answers / total number of questions.

= 3/20

to convert it into percent multiply by 100, we get

= 3/20 * 100

= 15%

the answer is 15%

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