Question

a store is giving every customer who enters the store a scratch - off card labeled with numbers from 1 to 10. it is equally likely that any of the numbers from 1 to 10 will be labeled on a given card. if the card is an even number, the customer gets a 15% discount on a purchase. if the card is an odd number greater than 4, the customer gets a 25% discount. otherwise, the discount is 30%. complete parts a and b.
click the icon to view scratch - off cards.

a. what is the probability for each discount?
15% discount: (\frac{1}{2})
25% discount: (\frac{3}{10})
30% discount: (\frac{1}{5})
(type integers or fractions.)

b. the store manager gives out 400 scratch - off cards. which discount would you expect the most customers to receive? explain.
most customers receive the (\boldsymbol{
abla}) discount because (\boldsymbol{square}) customers receive that discount.
(round to the nearest whole number as needed.)

Explanation:

Step1: List total numbers

Total numbers: $1,2,3,4,5,6,7,8,9,10$ (count = 10)

Step2: Calculate 15% discount prob

Even numbers: $2,4,6,8,10$ (count = 5)
Probability: $\frac{5}{10} = \frac{1}{2}$

Step3: Calculate 25% discount prob

Odd >4: $5,7,9$ (count = 3? Correction: Wait, 5,7,9 are 3? No, wait 5,7,9 is 3? Wait no, 1-10 odd numbers >4: 5,7,9 is 3? Wait no, wait 1,3 are odd ≤4, 5,7,9 are odd >4 (3 numbers? Wait no, 5,7,9 is 3, but wait 10 total, 5 even, 3 odd>4, so remaining is 2? No, 1,3 are odd ≤4, that's 2. So 2 numbers for 30% discount.
Wait correction Step3: Odd >4: $5,7,9$ (count=3)
Probability: $\frac{3}{10}$? No, wait 1-10: odd numbers are 1,3,5,7,9. Odd >4: 5,7,9 (3 numbers). Correct.

Step4: Calculate 30% discount prob

Remaining numbers: $1,3$ (count=2)
Probability: $\frac{2}{10} = \frac{1}{10}$

Step5: Calculate expected customers

For 15%: $400 \times \frac{1}{2} = 200$
For 25%: $400 \times \frac{3}{10} = 120$
For 30%: $400 \times \frac{1}{10} = 40$

Step6: Identify most common discount

200 is the largest number, so 15% discount is most common.

Wait correction: The original 15% was correct, 25% is $\frac{3}{10}$, 30% is $\frac{2}{10}=\frac{1}{5}$? Wait no, 10 total numbers: 5 even (15%), 3 odd>4 (25%), 2 odd≤4 (30%). So 30% is $\frac{2}{10}=\frac{1}{5}$. Oh right, I made a mistake earlier. So correcting Step4:

Step4: Calculate 30% discount prob

Numbers not even/odd>4: $1,3$ (count=2)
Probability: $\frac{2}{10} = \frac{1}{5}$

Step5: Calculate expected customers

15%: $400 \times \frac{1}{2} = 200$
25%: $400 \times \frac{3}{10} = 120$
30%: $400 \times \frac{1}{5} = 80$

Step6: Identify most common discount

200 is the highest, so 15% discount has the most customers.

Final correction:

Step1: Total possible numbers

Total: $n=10$ (1 to 10)

Step2: 15% discount (even numbers)

Even numbers: 5 values ($2,4,6,8,10$)
$P(15\%) = \frac{5}{10} = \frac{1}{2}$

Step3: 25% discount (odd >4)

Odd >4: 3 values ($5,7,9$)
$P(25\%) = \frac{3}{10}$

Step4: 30% discount (remaining)

Remaining: 2 values ($1,3$)
$P(30\%) = \frac{2}{10} = \frac{1}{5}$

Step5: Calculate expected customers

15%: $400 \times \frac{1}{2} = 200$
25%: $400 \times \frac{3}{10} = 120$
30%: $400 \times \frac{1}{5} = 80$

Step6: Find largest expected count

200 is the highest, so 15% is most common.

Answer:

a. 15% discount: $\frac{1}{2}$; 25% discount: $\frac{2}{5}$; 30% discount: $\frac{1}{10}$
b. Most customers receive the 15% discount because 200 customers receive that discount.