# OLd STEP question. Am I correct?

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#1
STEP 1991 Paper 1 Question16.
I posted a solution on the 1991 Solutions thread a long time ago but am not at all sure about it. Has anyone else got a solution to this question?
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6 years ago
#2
(Original post by brianeverit)
STEP 1991 Paper 1 Question16.
I posted a solution on the 1991 Solutions thread a long time ago but am not at all sure about it. Has anyone else got a solution to this question?
Agree on P(toad crosses safely), obviously.

There's a typo in the second term of your P(frog crosses safely). It's missing a factor of 1/3.

I get a small difference in the actual value too. P(Frog crosses safely) = 0.40816.

Were you assuming that they were crossing at the same time?
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6 years ago
#3
For the probability that the frog beats the toad (and survives) I get 0.323005896.

I used a calculator. Was that allowed?
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#4
(Original post by BabyMaths)
For the probability that the frog beats the toad (and survives) I get 0.323005896.

I used a calculator. Was that allowed?
Thanks. I'll have another look. Yes calculators were allowed.
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#5
(Original post by BabyMaths)
Agree on P(toad crosses safely), obviously.

There's a typo in the second term of your P(frog crosses safely). It's missing a factor of 1/3.

I get a small difference in the actual value too. P(Frog crosses safely) = 0.40816.

Were you assuming that they were crossing at the same time?
Thanks for pointing out the typo. I agree with 0.41816 but didn't think it necessary to give 5 decimal places.
How could I be assuming that they cross together when the question specifies that the frog crosses before the toad. Where do you think I am wrong?
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6 years ago
#6
(Original post by brianeverit)
Thanks for pointing out the typo. I agree with 0.41816 but didn't think it necessary to give 5 decimal places.
How could I be assuming that they cross together when the question specifies that the frog crosses before the toad. Where do you think I am wrong?
I'll assume that 0.41816 is a typo.

I gave 5 decimal places to assist in making a comparison of our results. I also mentioned it as your solution says 0.4081 when it should be 0.4082 if you're rounding to 4 decimal places.

I don't have your solution to hand just now so I'll say what I did, which may of course be wrong.

P(Frog crosses faster than the toad) = P(frog at n=0 and toad at n>0) + P(frog at n=1 and toad at n>1) etc. = 0.1 x 0.9 + 0.144 x 0.81 + 0.1188 x 0.729 + 0.04536 x 0.6561=0.323005896

For the final part I have 0.9288.
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6 years ago
#7

I think the problem is in your claim that:

P(both cross safely with frog before the toad) = P(frog crosses safely before it is safe to do so).

There can be times when it is safe for the frog but not safe for the toad, after the same wait.

In the final part we have the same method but you missed out one term in the denominator in the penultimate line, namely which you have correct in your first line.
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#8
(Original post by BabyMaths)

I think the problem is in your claim that:

P(both cross safely with frog before the toad) = P(frog crosses safely before it is safe to do so).

There can be times when it is safe for the frog but not safe for the toad, after the same wait.

In the final part we have the same method but you missed out one term in the denominator in the penultimate line, namely which you have correct in your first line.
How can there be times when it is safe for the frog but tut not the toad?. Either it is safe to cross, with prob 0.1 or it isn't with prob 0.9. t It's just that the frog will sometimes attempt to cross when it is not safe,but the toad never will. So, since the toad will cross as soon as it is safe to do so, in order for the frog to cross first he must do so before it is safe.
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6 years ago
#9
(Original post by brianeverit)
How can there be times when it is safe for the frog but tut not the toad?
This can happen because they are not crossing at the same time.

For example the frog may be able to cross safely at n=0 when the toad may have had to wait until n=1.
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#10
(Original post by BabyMaths)
This can happen because they are not crossing at the same time.

For example the frog may be able to cross safely at n=0 when the toad may have had to wait until n=1.
I still maintain that the frog must cross safely before it is safe to do so, because as soon as it is safe to cross then the toad will do so, hence for the frog to have crossed first then it must have been while it was not safe to do so.
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6 years ago
#11
(Original post by brianeverit)
I still maintain that the frog must cross safely before it is safe to do so, because as soon as it is safe to cross then the toad will do so, hence for the frog to have crossed first then it must have been while it was not safe to do so.

When the frog attempts to cross, the road may, for example, be

SAFE so the frog crosses.
UNSAFE (but this does not matter)
UNSAFE
SAFE
.
.

When the toad attempts to cross it could be

UNSAFE
UNSAFE
UNSAFE
.
.

The frog crossed first, when it was safe to do so.
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#12
(Original post by BabyMaths)

When the frog attempts to cross, the road may, for example, be

SAFE so the frog crosses.
UNSAFE (but this does not matter)
UNSAFE
SAFE
.
.

When the toad attempts to cross it could be

UNSAFE
UNSAFE
UNSAFE
.
.

The frog crossed first, when it was safe to do so.
During first minute
If safe to cross then they both do so frog does not cross BEFORE toad
If unsafe to cross both wait

After 1 minute
If safe to cross, again both do so but
If unsafe Toad waits whilst Frog attempts to cross with probability 1/3 and gets across successfully with further probability 0.2

After 2 minutes
If safe to cross both do so
If unsafe, as above but the 1/3 becomes 2/3

After 3 minutes
If safe to cross, as before
If unsafe, Toad waits but Frog attempts to cross and gets acroiss with probability 0.2

After 4 or more minutes, Toad will still only attempt to cross if safe to do so but Frog will attempt to cross.
So the only way the Frog can cross before the Toad is if he successfully attempts to when unsafe to do so.
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6 years ago
#13
(Original post by brianeverit)
During first minute
If safe to cross then they both do so frog does not cross BEFORE toad
If unsafe to cross both wait

After 1 minute
If safe to cross, again both do so but
If unsafe Toad waits whilst Frog attempts to cross with probability 1/3 and gets across successfully with further probability 0.2

After 2 minutes
If safe to cross both do so
If unsafe, as above but the 1/3 becomes 2/3

After 3 minutes
If safe to cross, as before
If unsafe, Toad waits but Frog attempts to cross and gets acroiss with probability 0.2

After 4 or more minutes, Toad will still only attempt to cross if safe to do so but Frog will attempt to cross.
So the only way the Frog can cross before the Toad is if he successfully attempts to when unsafe to do so.
The question says "Later on, a frog is also trying to cross the road".

Perhaps someone else would like to comment..

(Original post by DFranklin)
...
(Original post by ghostwalker)
...
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6 years ago
#14
(Original post by BabyMaths)
...
I'd interpret it as:

Some time after that frog makes the attempt.

There's no overlap between the two.

Nice to finally see the question.

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#15
(Original post by ghostwalker)
I'd interpret it as:

Frog crosses or gets squished.

Some time after that toad makes the attempt.

There's no overlap between the two.

Nice to finally see the question.
Crossing at different times with the frog taking less time than the toad is surely no different from them starting together and the frog crossing before the toad.
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6 years ago
#16
(Original post by brianeverit)
Crossing at different times with the frog taking less time than the toad is surely no different from them starting together and the frog crossing before the toad.
Just looking at when they first arrive at the crossing:

If they are together:

There are only two options, it's safe or it's not safe.

If frog crosses later:

We have four options, safe for frog, safe for toad,
and safe for frog, unsafe for toad,
and unsafe for frog, safe for toad,
and unsafe for frog, unsafe for toad.
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6 years ago
#17
(Original post by BabyMaths)
P(Frog crosses faster than the toad) = P(frog at n=0 and toad at n>0) + P(frog at n=1 and toad at n>1) etc. = 0.1 x 0.9 + 0.144 x 0.81 + 0.1188 x 0.729 + 0.04536 x 0.6561=0.323005896
I agree with this, and with 0.40816 for the first part.

Edit:. I also agree with 0.9288 as the final answer.
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6 years ago
#18
(Original post by brianeverit)
Crossing at different times with the frog taking less time than the toad is surely no different from them starting together and the frog crossing before the toad.
If they start together, the road conditions each of them face must be the same. (i.e. if it's safe for one, it's safe for the other).

As the question is written, I think the expectation is that the road conditions faced are independent.
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#19
(Original post by DFranklin)
If they start together, the road conditions each of them face must be the same. (i.e. if it's safe for one, it's safe for the other).

As the question is written, I think the expectation is that the road conditions faced are independent.
I now have what I believe is a definitive solution and has been confimed by Peter Mitchell who runs the Meiklerigg site
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6 years ago
#20
(Original post by brianeverit)
has been confimed by Peter Mitchell who runs the Meiklerigg site
What is the basis for Peter Mitchell being able to confirm it? Did he set the STEP question?
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