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

[utkarshakash]

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.

 

[HallsofIvy]

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.