Chance of B and C After Accident: 2/5, 3/5

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In summary, before the race, the chances of winning for runners A, B, and C were proportional to 5, 3, and 2, respectively. After A's accident, his chance of winning decreased to one-third. This caused the chances of B and C to change as well. The new chances of winning for B and C are 2/5 and 4/15, respectively.
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utkarshakash

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Homework Statement


Before a race the chances of three runners, A,B,C were estimated to be proportional to 5,3,2; but during the race A meets with an accident which reduces his chance to one-third. What are now the respective chances of B and C?

Homework Equations



The Attempt at a Solution


The ratio of chance of A to B is 5/3. But the chance of A after accident is 1/3. Let chance of B be x. Since the ratio is same ∴ x= 1/5. But the correct answer is just the double of this.
 
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utkarshakash said:

Homework Statement


Before a race the chances of three runners, A,B,C were estimated to be proportional to 5,3,2;
5+3+ 2= 10 so their probabiliities of winning are 5/10, 3/10, and 2/10, respectively.

but during the race A meets with an accident which reduces his chance to one-third. What are now the respective chances of B and C?

Homework Equations



The Attempt at a Solution


The ratio of chance of A to B is 5/3. But the chance of A after accident is 1/3. Let chance of B be x. Since the ratio is same ∴ x= 1/5. But the correct answer is just the double of this.
Your error is assuming A's and B's chances of winning still have the same ratio. That can't be true since A's accident changes A's chance of winning while B's has not changed. It is the ratio of B and C that does not change.

If A's chance after the accident is 1/3, the sum of the other two probabilites must be 2/3. Assuming the relative chances of B and C have not changed, since, originally, B's chance of winning was 3/2 C's, letting "x" be C's chance of winning, B's chance is (3/2)x so that (3/2)x+ x= (5/2)x= 2/3. Dividing both sides by 5/2, x= (2/3)(2/5)= 4/15. C's chance of winning now is 4/15 and B's is (3/2)(4/15)= 2/5.
 
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1. What does the fraction "2/5" represent in the context of this accident?

The fraction "2/5" represents the probability or chance of event B occurring after the accident. In other words, there is a 2 out of 5 chance that event B will happen.

2. What is the significance of the fraction "3/5" in this scenario?

The fraction "3/5" represents the probability or chance of event C occurring after the accident. This means there is a 3 out of 5 chance that event C will happen.

3. How are the chances of events B and C related in this accident?

The chances of events B and C are independent in this scenario, meaning that one event does not affect the probability of the other event occurring. They each have their own separate probability of happening.

4. Is a 2/5 chance of event B and a 3/5 chance of event C considered high or low?

The interpretation of a 2/5 and 3/5 chance being high or low is subjective and can vary depending on the context of the accident. It is important to consult with a statistician or expert to determine the significance of these probabilities in relation to the specific accident.

5. How can the probabilities of events B and C be calculated or determined?

The probabilities of events B and C can be calculated using mathematical formulas, such as the multiplication rule for independent events. This involves multiplying the individual probabilities of each event occurring. However, in the case of an accident, these probabilities may also be estimated based on data and statistics from similar accidents or expert analysis.

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