BATTERY LIFE

Part one:

1. Probability tree

60%

40%

Good battery life

Bad battery life

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*Battery Life*

Low bulb failure

Medium bulb failure

High bulb failure

Low bulb failure

Medium bulb failure

High bulb failure

70%

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Battery Life

50%

20%

10%

20%

30%

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Battery Life

Probability table

Let A – be good *battery life*

B – Be bad battery life

L – Be low bulb life

M – Be medium bulb life

H – Be high bulb life

L

M

H

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Battery Life

total

A

0.42

0.12

0.06

0.6

B

0.08

0.12

0.2

0.4

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Battery Life

total

0.5

0.24

0.26

1

2. A test was undertaken on a random bicycle *light* and it was found to have low bulb failure what is the probability that this bicycle light has a good battery life?

0.5 x 0.42 = 0.21

3. Memo

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Battery Life

The probability of getting a good battery outcome is 0.6, while the probability of getting a bad battery outcome is 0.4 and the probability of getting either a good or bad battery can be calculated by adding 0.4 + 0.6 = 1.

For the conditional __probability__ we will have good battery life then the probability that the bulb will have low failure is 0.7, the probability that we will have medium bulb failure is 0.2 and finally the probability that the bulb will have high failure is 0.1.

For the condition that the battery has bad battery life then the probability that we will have low bulb failure is 0.2, the probability that we will have medium bulb failure is 0.3 while the probability that there will be high bulb failure is 0.5.

From the above conditions it is __evident__ that the probability of bulb failure depends on the nature of the bulb whether the battery has good battery life or bad battery life. When we have good battery life then the lower is the probability that the bulb will fail and on the other hand if we __have__ bad battery life the higher the probability that we will have bulb failure. Another observation is that if we randomly choose a battery we have greater chances of choosing a battery with good battery life.

Part two:

Expansion or acquisition

1. Decision tree

70%

7/14

Battery Life

30%

Expansion

Success buy out

Increase revenue by 100million

Increase revenue by 30million

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Battery Life

60%

40%

China manufacturer

Glass competitor

90 million increase

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Battery Life

70million increase

20%

80%

86million increase

72 million increase

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__Battery Life__

50%

50%

50%

50%

2. Probabilities and expected values:

Glass competitor

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Battery Life

China manufacturer

expansion

total

amount

72 million

86 million

90 million

70 million

30 million

100 million

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*Battery Life*

probability

0.175

0.175

0.28

0.07

0.12

0.18

1

3. Optimum strategy

We get the above probabilities by multiplying the values of each probability, the probability of a 72 million increase in revenue through merging with a glass competitor is calculated by multiplying 0.7 x 0.5 x 0.5 = 0.175, and this probability is the same as the one that will give us an increase of 86 million through buying the glass competitor.

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**Battery Life**

By buying the china manufacturer the probability that there will be an increase in revenue by 90 million is got by multiplying 0.7 x 0.5 x 0.2 = 0.28, while an increase of 70 million in revenue probability is got by multiplying 0.7 x 0.5 x 0.2 = 0.07.

Through an expansion of the existing facility there is a 0.3 x 0.4 = 0.12 probability that there will be a 30 million increase in revenue whereas there is a 0.3 x 0.6 = 0.18 probability that there will be an increase in the level of revenue by 100 million.

The optimum strategy is that which has the highest probability, in our case by merging with the china manufacturer and expecting 90 million has an outcome of 0.28 which is the highest of all the others, this would be the best strategy because it has a high increase of 90 million on revenue and also highest probability to occur.

REFERENCE:

Ray R and C Mc Keenan (2001) Mathematics for Economics, MIT Publishers , USA

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