If a student attends every management science class, the probability of passing the course is 0.80; but if the student only atte
nds randomly, then the probability of passing the course is 0.50. If a student fails, they can take a makeup test where the probability of passing is 0.60 if the student has attended every class. This probability of passing the makeup test drops to 0.10 if the student has attended at random. Passing the course is worth 5 credits. Full time attendance "costs" 3 credits in terms of energy and time whereas random attendance "costs" only 1 credit.Use a decision tree to decide which is the best attendance pattern to adopt.
1 answer:
Answer:
EV of node 4 = 0.6×5 + 0.4×0 = 3
EV of node 5 = 0.1×5 + 0.9×0 = 0.5
EV of node 2 = (0.8×5 + 0.2×EV of node 4) - 3
EV of node 2 = (0.8×5 + 0.2×3 ) - 3
EV of node 2 = 4.6 - 3
EV of node 2 = 1.6
EV of node 3 = (0.5×5 + 0.5×EV of node 5) - 1
EV of node 3 = (0.5×5 + 0.5×0.5 ) - 1
EV of node 3 = 2.75 - 1
EV of node 3 = 1.75
EV of node 3 is higher, therefore, best attend pattern is to attend randomly.
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The number of bananas that Joey purchased would be represented on a number line with digits 0 to 10 by marking point 9 with a dot (circle) as shown in the image attached.
- Let the number of bananas be B.
<u>Given the following data:</u>
- A dozen oranges = $3. 50
- Bananas = $0. 25
- Total cost = $5.75
<u>Note:</u> A dozen oranges is equal to 12 oranges.
To determine the number line that best represents the number of bananas Joey purchased:
First of all, we would calculate the number of bananas that Joey purchased:
B = 9 bananas.
<u>For the </u><u>number line</u><u>:</u>
The number of bananas that Joey purchased would be represented on a number line with digits 0 to 10 by marking point 9 with a dot (circle) as shown in the image attached.
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