Expected Value in Probability?

Question

Someone is using a loaded die. The probability of a "1" is 1/2 and the rest of the values have equal probability. Find the expected value of this die.
Please use the formula and list the possible numbers on the die. Thank You.

Thank you both. The answer is correct and i understand it. Thanks once again

Answer 1

I am not sure If I am understanding this question but let me give it a try.
E(x) = nP .. where n is the expected pay off and P is the probability .. .E is the expected outcome.

If P(1) = .5
P(2,3,4,5,6) = .5
If each have the same probability then its going to be .5/5 = .1 or 1/10

E(x) = (1(.5)) + (2(.1)) + (3(.1)) + (4(.1)) + (5(.1)) + (6(.1))
E(x) = 2.5

Therefore, the expected value is 2.5

Now the question asks to list the values .. I am not sure what this means .. And it also says 'loaded' I guess that takes care of the .5 probability of getting a 1.

Let me know If I am completely off track or otherwise.

Answer 2

There are 8 words: 1 has 2 letters 5 have 3 letters 1 has 4 letters 1 has 9 letters P(2 letters) = 1/8 P(3 letters) = 5/8 P(4 letters) = 1/8 P(9 letters) = 1/8 E(X) = 2 * P(2) + 3 * P(3) + 4 * P(4) + 9 * P(9) E(X) = 2 * 1/8 + 3 * 5/8 + 4 * 1/8 + 9 * 1/8 E(X) = 2/8 + 15/8 + 4/8 + 9/8 E(X) = 30/8 E(X) = 3.75

Answer 3

(2+3+4+5+6)(1/10)+1/2 = 2.5

Answer 4

1/2 ÷ 5 = 1/10, so

E = (1/2)(1) + (1/10)(2) + ... + (1/10)(6)
E = 1/2 + (1/10)(2+3+4+5+6)
E = 1/2 + 2
E = 2.5