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?
Best Answer
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?