Combinations

[EvoN1020v]

Did you know that the more multiple choices (4 choices) questions that you have, the lower percentage you will get all the answers right?

 

1 question with 4 choices = 25% chance to get the answer right.

2 question with 8 choices = 6.25% chance to get both questions right.

 

Am I correct?

I think you do this by 1/(4^n) where n is the number of questions.

[ecoli]

Did you know that the more multiple choices (4 choices) questions that you have, the lower percentage you will get all the answers right?

 

Yeah that's pretty obvious, though harder to express mathematically, I suppose.

[Klaynos]

second situation: Each question has 4 options or each has 8 options?

 

If the former then you're correct. This is if the probability of each option being correct is the same.

[EvoN1020v]

1 single question have 4 multiple choices. Therefore 2 questions have 8 choices in total.

[Klaynos]

right yes, but individually they have 4 each if you think of it like that you can vary the system by putting in ones with 5 choices and 3 choices etc...

[starbug1]

1 question with 4 choices = 25% chance to get the answer right.

2 question with 8 choices = 6.25% chance to get both questions right.

 

 

this works only if you are calculating the probablity of getting all problems correct. But for each question the probablity is the same--25%. When you fill out a test at random, the probability of your final test score is still 25%. Logically, and by laws of probablity, a mean of students taking a test at random, the grade comes by 1/4. Same thing for 4 multiple choice questions with four choices each. one in four will be correct. Only when you combine the two categories do you need a different equation.

 

Saying that the more multiple choice question you have the lower the change of getting them right is utterly obvious. Were you unsure of the equation?

[BigMoosie]

Odds of getting first wrong and second right:

 

(3/4)*(1/4) = 18.75%

 

Odds of getting first right and second wrong:

 

(1/4)*(3/4) = 18.75%

 

Odds of getting both right:

 

(1/4)*(1/4) = 6.25%

 

Odds of getting both wrong:

 

(3/4)*(3/4) = 56.25%

 

The sum of each scenario multiplied by their respective score:

 

(3/4)*(1/4)*1 + (1/4)*(3/4)*1 + (1/4)*(1/4)*2 + (3/4)*(3/4)*0 = 0.5

 

That's 0.5 out of 2 which is 25%, the average score never changes.