A computer game shows two rows of eight cups. A player selects clicks on one cup from each row to reveal if they have a bean underneath the cup. The first row of cups has three cups with beans and the second row of cups had one cup with a bean.

The game creator notices that very few people play the game more than once. He decides to change it so that at least half of the cups in each row have beans.

Which of the following is the **best **action for the game creator to take in order to make the change but still give the game the greatest advantage possible over the player?

- The choice is add bean to one of the empy the cars in the first fow and three of the empty in the secound row
- Add bean to twoof the empty in the first row and three of the empty up in the secound row.
- Add bean totree of the emty cup in the secound row.

shaniab29544 Dec 18, 2014

#4**+10 **

Ah, shania.....I missed the "at least half" part.......but, "at least half" means "4 or more"....* NOT *that we have to have 5 cups filled....!!!

This eliminates the third option.....since only 3 cups in the first row are filled.

Notice that your answer - adding two in the first row and 4 in the second row - * isn"t* listed as one of the original options....!!!

Under the "at least" constraint......I'll amend my answer to the first option.....= (4/8)(4/8) = 16/64 = 25%

CPhill Dec 20, 2014

#1**+10 **

Originally we have (3/8) * (1/8) = 3/64

Let's look at the options

a) (4/8) * ( 4/8) = 16/64

b) (5/8) * (4/8) = 20/64

c) (3/8) * ( 4/8) = 12/64

The third option gives the most advantage.....

CPhill Dec 18, 2014

#3**+5 **

Since more than half of each row of cups must contain beans, each row will need at least five beans. This means that at least two of the empty cups in the first row and at least four of the empty cups in the second row must have beans added to them.

For the game operator to have the advantage over the player, the probability that the two cups selected will both contain beans must be less than 50%. The probability that the cups will both contain beans is the product of the probabilities of the cup from each row containing a bean. Adding beans to two of the empty cups in the first row and four of the empty cups in the second row will result in the probability that both cups selected contain beans being 58⋅58=2564≈39.06%, which is less than 50%**.**

Add beans to two of the empty cups in the first row and four of the empty cups in the second row.

shaniab29544 Dec 19, 2014

#4**+10 **

Best Answer

Ah, shania.....I missed the "at least half" part.......but, "at least half" means "4 or more"....* NOT *that we have to have 5 cups filled....!!!

This eliminates the third option.....since only 3 cups in the first row are filled.

Notice that your answer - adding two in the first row and 4 in the second row - * isn"t* listed as one of the original options....!!!

Under the "at least" constraint......I'll amend my answer to the first option.....= (4/8)(4/8) = 16/64 = 25%

CPhill Dec 20, 2014