Which of the following demonstrates a fair choice of 1 from among 12 candidates?

assigning each candidate a sequence while rolling a die and flipping a coin (for example: 1-h).

assigning each candidate a 4-digit number made of the digits from 0-4. shuffling a bag of four balls numbered 0-4 and picking one at a time in order.

assigning each candidate a sequence of 4 heads or tails (for example: h-t-t-h) and flipping a coin four times to pick the winning sequence.

assigning each candidate a 2-digit sequence that is decided by throwing two dice where order does not matter (for example: 3-4, 2-6).

assigning each candidate a number that is decided by throwing two dice of different colors to distinguish the order of the outcomes (for example: 3-4, 4-3, 6-

I option is right

Step-by-step explanation:

Given that there are 12 candidates.  One has to be selected at random. FOr a fair choice of 1 among 12, we must have each one have the equal probability

i.e. prob of selecting any one = 1/12

This means sample space should consist of 12 events and one to be assigned for one.

In option i, we have coin toss has 2 outcomes, a die has 6 outcomes, together we have 12 equally likely outcomes thus making a fair choice.

Option II is incorrect because sample space consists of 5 numbers and 5 balls  hence sample space would be 25 cannot give equal chances to 12.ption 3 is incorrect since when we flip a coin 4 times we have 16 different outcomes hence 12 assigning cannot be equaly likely.

When two dice of different colours are thrown, sample space is 36 and hence 12 cannot be assigned order in a fair manner.

So Option I is right

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !

my answer is$ 12

step-by-step explanation:

because 15 -2=12