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IRISSAK [1]
2 years ago
3

Suppose two drugs are routinely used for treatment of a particular kidney disorder. Drug 1 is known to cure the disease 85% of t

he time and costs $90. Drug 2 is known to cure the disease 70% of the time and costs $65. The two drugs work independent of each other (that is, administration of one has no effect on the efficacy of the other). The two treatment plans are as follows:
Plan A: Treatment with Drug 1—if not effective, treatment with Drug 2.
Plan B: Treatment with Drug 2—if not effective, treatment with Drug 1.

Which statement is most correct in this situation?
\
Based on the overall probability of a cure, plan A should be selected over plan B.
Based on the overall probability of a cure, plan B should be selected over plan A.
Based on the overall cost of treatment, plan A should be selected over plan B.
Based on the overall cost of treatment, plan B should be selected over plan A.
Based on the probability of a cure and the cost of treatment, both plans are equivalent, so either can be selected.
Mathematics
1 answer:
3241004551 [841]2 years ago
4 0
<span>If we use Plan A, to start with drug 1, we spend $90. Then in the event that drug 1 did not work (1 - 85% = 15% chance), we spend $65 more for drug 2. This is an expected value of $90 + (0.15)($65) = $99.75.

If we use Plan B, to start with drug 2, we spend $65. And if it does not work (1 - 70% = 30% chance), we spend $90 more for drug 2. This is an expected cost of $65 + (0.3)($90) = $92

Therefore, </span><span>b</span><span>ased on the overall cost of treatment, plan B should be selected over plan A. (The probabilities will actually be the same after both drugs have been tried, in either order.)</span>
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The <em>probaility</em> of an event is defined as the number of favorable outcomes divided by the number of total possible events.

P (event E) = number of outcomes for event E / number of possible events

1. <u>As </u><u>first step</u><u>, you may draw a table to find the </u><u>sample space</u><u> (set of all possible outcomes)</u>.

<u>Sample space</u>

The results of the rolling two dice are summarized in this table:

                    Second roll    1       2       3       4        5      6

First roll

  1                                       (1,1)  (1,2)    (1,3)   (1,4)   (1,5)  (1,6)

  2                                     (2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

  3                                     (3,1)  (3,2)  <u>(3,3)</u>   (3,4)  (3,5)  (3,6)

  4                                     (4,1)   (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

  5                                     (5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

  6                                     (6,1)  (6,2)  (6,3)  (6,4)  (6,5)  <u>(6,6)</u>

<u></u>

2.<u>Now, in that table, you can observe</u>:

The results (3,3) and (6,6) are highlited.

a) Total number of events: 6 × 6 = 36

b) Nnmber of outcomes for the event rolling two 6s (6,6): 1

c) Number of outcomes for the event rolling two 3s (3,3): 1

3) <u>Next, you can calculate the probabilities:</u>

a) Probability rolling two 6s = 1 / 36

b) Probability of rolling two 3s: 1 / 36

4. <u>Conclusion</u>: the probabilities prove that rolling two 6s is just as likely as rolling two 3s.

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