Question

In Mariokart DS, there are eight start up characters and additional 4 more unlockable characters. A photo will be taken of the top 3 racers. What is the

probability that all top three racers will only be unlockable characters?

Answer 1

Assuming you had all of them (all 12) in a race like the Wii version it would be nearly 20%

If you had just 8 with the three unlock able characters in there like the DS it would be literally about 35% but of course its a game so its kind of configured on how to arrange everything lol

Answer 2

Using the hypergeometric distribution, it is found that there is a 0.0182 = 1.82% probability that all top three racers will only be unlockable characters.

The characters are chosen without replacement, which means that the hypergoemetric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• N is the size of the population.
• n is the size of the sample.
• k is the total number of desired outcomes.

In this problem:

• There are 8 + 4 = 12 characters, hence $N = 12$.
• 4 are unlockable, hence $k = 4$.
• 3 are going to be on the photo, hence $n = 3$.

The probability that all top three racers will only be unlockable characters is P(X = 3), hence:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,12,3,4) = \frac{C_{4,3}C_{8,0}}{C_{12,3}} = 0.0182$

0.0182 = 1.82% probability that all top three racers will only be unlockable characters.

A similar problem is given at https://brainly.com/question/24826394