# OLd STEP question. Am I correct?

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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?

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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#2

(Original post by

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?

**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?

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.

I don't get your answers for the rest.

Were you assuming that they were crossing at the same time?

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#3

For the probability that the frog beats the toad (and survives) I get 0.323005896.

I used a calculator. Was that allowed?

I used a calculator. Was that allowed?

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(Original post by

For the probability that the frog beats the toad (and survives) I get 0.323005896.

I used a calculator. Was that allowed?

**BabyMaths**)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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(Original post by

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.

I don't get your answers for the rest.

Were you assuming that they were crossing at the same time?

**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.

I don't get your answers for the rest.

Were you assuming that they were crossing at the same time?

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

(Original post by

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?

**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 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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#7

I found your solution again.

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.

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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(Original post by

I found your solution again.

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.

**BabyMaths**)I found your solution again.

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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#9

(Original post by

How can there be times when it is safe for the frog but tut not the toad?

**brianeverit**)How can there be times when it is safe for the frog but tut not the toad?

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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(Original post by

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.

**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.

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#11

(Original post by

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.

**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 so the toad waits

UNSAFE

UNSAFE

UNSAFE

SAFE so the toad crosses.

.

.

The frog crossed first, when it was safe to do so.

The toad had to wait until n=4.

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(Original post by

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 so the toad waits

UNSAFE

UNSAFE

UNSAFE

SAFE so the toad crosses.

.

.

The frog crossed first, when it was safe to do so.

The toad had to wait until n=4.

**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 so the toad waits

UNSAFE

UNSAFE

UNSAFE

SAFE so the toad crosses.

.

.

The frog crossed first, when it was safe to do so.

The toad had to wait until n=4.

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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#13

(Original post by

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.

**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.

Perhaps someone else would like to comment..

(Original post by

...

**DFranklin**)...

(Original post by

...

**ghostwalker**)...

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#14

(Original post by

...

**BabyMaths**)...

Toad crosses.

Some time after that frog makes the attempt.

There's no overlap between the two.

Nice to finally see the question.

**Edit:**Had toad and frog the wrong way around.

And toad can't get squished!

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(Original post by

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.

**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.

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#16

(Original post by

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.

**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 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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#17

(Original post by

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

**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

**Edit:**. I also agree with 0.9288 as the final answer.

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#18

**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.

As the question is written, I think the expectation is that the road conditions faced are independent.

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(Original post by

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.

**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.

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#20

(Original post by

has been confimed by Peter Mitchell who runs the Meiklerigg site

**brianeverit**)has been confimed by Peter Mitchell who runs the Meiklerigg site

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