Probability Question - Bear Catching Fish

Question

A bear wants to catch $3$ fish from a river. When he has caught $3$ fish, he'll leave. When a fish comes, there is a $50\%$ chance he'll catch it. What's the probability that the $5^{th}$ fish will not be caught?

My attempt at a solution is below and in the answers section, could someone please verify that the logic and calculation checks out.

My solution:

The probability that the 5th fish will not be caught is equal to the probability that the bear catches 3 fish before he gets to the 5th fish, plus the probability that the bear doesn't catch 3 fish before he gets to the 5th fish, and he then doesn't catch the 5th fish.

P(Bear catches 3 fish before getting to fish 5) = P(Bear catches first 3 fish) + P(Bear catches 3 of first 4 fish)

Furthermore, if 1 represents that a fish has been caught, and 0 represents that a fish hasn't been caught, then we get 111 as the sequence that corresponds to the bear catching the first 3 fish, where the first fish is represented by the leftmost 1, and the last fish is represented by rightmost 1.

Then,

P(Bear catches first 3 fish) = (1/2)^3 = 1/8

For P(Bear catches 3 of first 4 fish), our sequences are as follows: 1101, 1011, and 0111. We exclude 1110, as the bear will not try to catch the 4th fish in this case, because it has already caught 3 fish.

Then,

P(Bear catches 3 of first 4 fish) = 3(1/2)^4 = 3/16

Thus,

P(Bear catches 3 fish before getting to fish 5) = P(Bear catches first 3 fish) + P(Bear catches 3 of first 4 fish) = 1/8 + 3/16 = 5/16

And,

P(Bear doesn't catch 3 fish before he gets to the 5th fish, and he then doesn't catch the 5th fish) = (1-P(Bear catches 3 fish before getting to fish 5))*P(Bear doesn't catch fish 5) = (1 - 5/16)(1/2) = 11/32

Finally,

P(5th Fish will not be caught) = P(Bear catches 3 fish before getting to fish 5) + P(Bear doesn't catch 3 fish before he gets to the 5th fish, and he then doesn't catch the 5th fish) = 5/16 + 11/32 = 21/32 = 0.65625

Can someone please verify whether my logic and calculations are indeed correct, and if not, identify my mistake?

Answer 1

My solution:

The probability that the 5th fish will not be caught is equal to the probability that the bear catches 3 fish before he gets to the 5th fish, plus the probability that the bear doesn't catch 3 fish before he gets to the 5th fish, and he then doesn't catch the 5th fish.

P(Bear catches 3 fish before getting to fish 5) = P(Bear catches first 3 fish) + P(Bear catches 3 of first 4 fish)

Furthermore, if 1 represents that a fish has been caught, and 0 represents that a fish hasn't been caught, then we get 111 as the sequence that corresponds to the bear catching the first 3 fish, where the first fish is represented by the leftmost 1, and the last fish is represented by rightmost 1.

Then,

P(Bear catches first 3 fish) = (1/2)^3 = 1/8

For P(Bear catches 3 of first 4 fish), our sequences are as follows: 1101, 1011, and 0111. We exclude 1110, as the bear will not try to catch the 4th fish in this case, because it has already caught 3 fish.

Then,

P(Bear catches 3 of first 4 fish) = 3(1/2)^4 = 3/16

Thus,

P(Bear catches 3 fish before getting to fish 5) = P(Bear catches first 3 fish) + P(Bear catches 3 of first 4 fish) = 1/8 + 3/16 = 5/16

And,

P(Bear doesn't catch 3 fish before he gets to the 5th fish, and he then doesn't catch the 5th fish) = (1-P(Bear catches 3 fish before getting to fish 5))*P(Bear doesn't catch fish 5) = (1 - 5/16)(1/2) = 11/32

Finally,

P(5th Fish will not be caught) = P(Bear catches 3 fish before getting to fish 5) + P(Bear doesn't catch 3 fish before he gets to the 5th fish, and he then doesn't catch the 5th fish) = 5/16 + 11/32 = 21/32 = 0.65625

Can someone please verify whether my logic and calculations are indeed correct, and if not, identify my mistake?

Answer 2

If we assume the bear does not leave, there are $2^5=32$ possible outcomes, which fish five would expect to survive 16. Because the bear leaves after three fish, this will increase the number of ways it can survive for sure. This is the number of ways trimmed from the tree (C = fish caught, S = fish survives, X = bear left):

CCCXX = $2^2$ = 4

SCCCX = $2^1$ = 2

CSCCX = $2^1$ = 2

CCSCX = $2^1$ = 2

4+2+2+2 = 10 ways.

This means 10 of the 32 tree branches the fish survives for sure, and this leaves 32-10 = 22 ways with a 50% chance of survival, therefore the fish would survive 11. This gives 10+11=21 possible ways for the fish to survive.

Thus the probability that fish five survives is 21/32.