# Thread: [SOLVED] Probability of Certain Discounts

1. ## [SOLVED] Probability of Certain Discounts

Hi, I'm having a little trouble with my Stats200 homework, I don't understand how to do the last 2 problems and any double checking of the answers I provided for the previous questions wouldn't hurt either thanks in advance

Problem:

A sporting goods store announces a "wheel of Savings" sale. Customers select the merchandise they want to purchase, then at the cash register they spin a wheel to determine the size of the discount they will recieve. The wheel is divided into 12 regious, likc a clock. Four of those regions are red, and award a 5% discount. The three white regions award a 10% discount, two blue regions a 25% discount, two green regions a 30% discount. The remaining regioin is gold and a customer whos lucky spin lands there gets a 50% discount!

1. What is the probability that a customer gets at least a 25% discount?

2. What is the probability that two customers in a row get only 5% discounts?

3. What is the probability that three consecutive customers all get 25% discounts?

4. What is the probability that none of the first four customers gets a discount over 25%?

5. What is the probability that the first gold winner(50%) is the fifth customer in line?

6. What is the probablity that there is at least one gold winner among the first six customers?

2. A sporting goods store announces a "wheel of Savings" sale. Customers select the merchandise they want to purchase, then at the cash register they spin a wheel to determine the size of the discount they will recieve. The wheel is divided into 12 regious, likc a clock. Four of those regions are red, and award a 5% discount. The three white regions award a 10% discount, two blue regions a 25% discount, two green regions a 30% discount. The remaining regioin is gold and a customer whos lucky spin lands there gets a 50% discount!

1. What is the probability that a customer gets at least a 25% discount?

at least a 25% discount is 5/12

2. What is the probability that two customers in a row get only 5% discounts?

getting 5% is 4/12. 2 in a row is 4/12 x 4/12 = 1/9

3. What is the probability that three consecutive customers all get 25% discounts?

getting 25% is 2/12. 3 in a row is 2/12 x 2/12 x 2/12 = 1/216

4. What is the probability that none of the first four customers gets a discount over 25%?

getting not over 25% is 9/12. 4 in a row is 9/12 x 9/12 x 9/12 x 9/12 = 81 / 256

3. Hello, rice4lifelegit!

A sporting goods store announces a "wheel of Savings" sale.
Customers select the merchandise they want to purchase, then at the cash register,
they spin a wheel to determine the size of the discount they will recieve.

The wheel is divided into 12 regious, likc a clock.
Four of those regions are red, and award a 5% discount.
The three white regions award a 10% discount,
two blue regions a 25% discount, two green regions a 30% discount.
The remaining region is gold and a customer and gives a 50% discount.
We have these probabilities:

. . . $\begin{array}{|c|c|}
\text{Discount} & \text{Prob} \\ \hline
5\% & \frac{4}{12} \\ \\[-4mm]
10\% & \frac{3}{12} \\ \\[-4mm]
25\% & \frac{2}{12} \\ \\[-04mm]
30\% & \frac{2}{12} \\ \\[-4mm]
50\% & \frac{1}{12} \end{array}$

1. What is the probability that a customer gets at least a 25% discount?
$P(25\% \vee 30\% \vee 50\%) \:=\:\frac{2}{12} + \frac{2}{12} + \frac{1}{12} \:=\:\frac{5}{12} \:\approx\: 0.4167$

2. What is the probability that two customers in a row get only 5% discounts?
We have: . $P(5\%) \:=\:\frac{4}{12} \:=\:\frac{1}{3}$

Therefore: . $P(5\% \wedge 5\%) \:=\:\frac{1}{3}\cdot\frac{1}{3} \:=\:\frac{1}{9} \:\approx\: 0.1111$
.You were right!

3. What is the probability that 3 consecutive customers all get 25% discounts?

We have: . $P(25\%) \:=\:\frac{2}{12} \:=\:\frac{1}{6}$

Therefore: . $P(25\% \wedge 25\% \wedge 25\%) \:=\:\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} \:=\:\frac{1}{216} \;\approx\; 0.0046$

4. What is the probability that none of the first 4 customers gets a discount over 25%?

We have: . $P(\text{ not over }25\%) \;=\;P(5\% \vee 10\% \vee 25\%) \;=\;\frac{4}{12} + \frac{3}{12} + \frac{2}{12}\;=\;\frac{9}{12} \;=\;\frac{3}{4}$

Therefore: . $P(\text{Four, not over }25\%) \;=\; \left(\frac{3}{4}\right)^4 \;=\;\frac{81}{256} \;\approx\;0.3164$ .
You were right!

5. What is the probability that the first gold winner(50%) is the fifth customer in line?
The first four customers were not gold winners, and the fifth one was.

We have: . $P(50\%) \:=\:\frac{1}{12} \quad\Rightarrow\quad P(\sim 50\%) \:=\:\frac{11}{12}$

Therefore: . $\left(\frac{11}{12}\right)^4\cdot\frac{1}{12} \;=\;\frac{14,\!641}{248,\!832} \;\approx\;0.0588$

6. What is the probablity that there is at least one gold winner among the first 6 customers?
The opposite of "at least one gold among the first 6" is "no gold among the first 6".

We have: . $P(50\%) \:=\:\frac{1}{12} \quad\Rightarrow\quad P(\sim 50\%) \:=\:\frac{11}{12}$

Hence: . $P(Six,\:\sim50\%) \:=\: \left({11}{12}\right)^6$

Therefore: . $P(\text{at least one }50\%) \;=\;1-\left(\frac{11}{12}\right)^6 \;=\;\frac{1,\!214,\!423}{2,\!985,984} \;\approx\;0.4067$