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Math Help - [SOLVED] Probability of Certain Discounts

  1. #1
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    [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?
    My answer: .1667

    2. What is the probability that two customers in a row get only 5% discounts?
    My answer: .111

    3. What is the probability that three consecutive customers all get 25% discounts?
    My answer: .01389

    4. What is the probability that none of the first four customers gets a discount over 25%?
    My answer: .3164

    5. What is the probability that the first gold winner(50%) is the fifth customer in line?
    My answer: no clue

    6. What is the probablity that there is at least one gold winner among the first six customers?
    My answer: no clue


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  2. #2
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    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



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  3. #3
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    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|}<br />
\text{Discount} & \text{Prob} \\ \hline<br />
5\% & \frac{4}{12} \\ \\[-4mm]<br />
10\% & \frac{3}{12} \\ \\[-4mm]<br />
25\% & \frac{2}{12} \\ \\[-04mm]<br />
30\% & \frac{2}{12} \\ \\[-4mm]<br />
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

    Last edited by mr fantastic; July 26th 2009 at 08:15 PM. Reason: No edit - just flagging this reply as having been moved from another thread.
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