OLd STEP question. Am I correct?

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?

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

I used a calculator. Was that allowed?

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?

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 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 $0.9^2 \times \frac{2}{3} \times 0.1$ 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?

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.

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.

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.

...

...

...

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.

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

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.

has been confimed by Peter Mitchell who runs the Meiklerigg site