A restaurant gives each customer a scratch off ticket with two sections. The first section has five covered squares, only two of which are red. The second section has nine covered squares, four of which are red. The player scratches off only one square in each section to reveal the colors, and wins a free meal if they reveal a red square in both sections.

The last week of the promotion the restaurant will increase the number of hidden red squares in each section by changing a total of three of the non-red squares to red, but in such a way as to minimize the number of winners.

Which of the following is the **best **action for the restaurant to take when changing the scratch off tickets for the last week of the promotion?

shaniab29544 Dec 18, 2014

#2**+5 **

Change three of the non-red squares in the first section to red and leave the second section as is.

Change two of the non-red squares in the first section to red and change one of the non-red squares in the second section to red.

Change one of the non-red squares in the first section to red and change two of the non-red squares in the second section to red.

Leave the first section as is and change three of the non-red squares in the second section to red.

shaniab29544 Dec 18, 2014

#3**+10 **

Best Answer

Originally, we have ....(2/5) (4/9)

Let's look at each option

a) (5/5) x (4/9) = 20/ 45

b) (4/5) x (5/9) = 20/45

c) (3/5) x (6/9) = 18 / 45

d) (2/5) x (7/9) = 14 / 45

Notice that the last option minimizes the chances of a winner....!!!!

CPhill Dec 18, 2014

#5**0 **

The lower the probability of both revealed squares being red, the fewer winners the contest will have. The probability of revealing a red square in a section is the number of red squares in the section divided by the total number of squares in the section. The probability that the both revealed squares will be red is the product of the probabilities of revealing a red square in each section.

Changing three of the non-red squares in the first section to red and leaving the second section as is will result in the probability that both revealed squares will be red being 55⋅49=2045=49≈44.4%. Changing two of the non-red squares in the first section to red and changing one of the non-red squares in the second section to red will result in the probability that both revealed squares will be red being 45⋅59=2045=49≈44.4%. Changing one of the non-red squares in the first section to red and changing two of the non-red squares in the second section to red will result in the probability that both revealed squares will be red being 35⋅69=1845=25=40%. Leaving the first section as is and changing three of the non-red squares in the second section to red will result in the probability that both revealed squares will be red being 25⋅79=1445≈31.1%.** The lowest probability of both revealed squares being red occurs when the first section is left as is and three of the non-red squares in the second section are changed to red, so this is the best option to add three red squares and still minimize the number of winners****.**

Leave the first section as is and change three of the non-red squares in the second section to red.

shaniab29544 Dec 19, 2014