Question

Which of the following events has an expected value that is in the sample space? A. tossing a number cube once B. flipping a coin C. randomly picking a number between one and nine, inclusive D. randomly picking a number between one and ten, inclusive

Answer 1

The answer is D because i used common sense i guess

Answer 2

A) The event is "tossing a number cube once"

The sample space of this event is {1,2,3,4,5,6}.

The expected value of this event is  

$1\times \frac{1}{6}+2\times \frac{1}{6}+3\times \frac{1}{6}+4\times \frac{1}{6}+5\times \frac{1}{6}+6\times \frac{1}{6}\\ \\ =\frac{1}{6}+\frac{1}{3}+\frac{1}{2}+\frac{2}{3}+\frac{5}{6}+1\\ \\ =\frac{7}{2} = 3.5$

Since 3.5 is not in the sample space of the event. Therefore, option (A) is not correct.

(B) The event is "Flipping a coin"

Sample space of this event is {HH,TT,HT,TH}

Since sample space of this event is not numbers, therefore, this cannot be the correct option either.

(C) The event is, "Randomly picking a number between 1 and 9, inclusive."

The sample space of this event is {1,2,3,4,5,6,7,8,9}.

Expected value of this event is $\frac{1}{9}(1+2+3+4+5+6+7+8+9) = \frac{1}{9}(45) = 5$

Since 5 is the expected value and it is present in the sample space for this event. Therefore, option (C) is a correct choice.

(D) The sample is "Randomly picking a number between one and ten, inclusive".

The sample space of this event is {1,2,3,4,5,6,7,8,9,10}.

Therefore, expected value of the event is $\frac{1}{10}(1+2+3+4+5+6+7+8+9+10) = \frac{1}{10}(55)=5.5$

Since 5.5 is not present in the sample space of this event. Therefore, option (D) is not correct either.

Hence, the correct choice is option (C).