Question
Brian has 2 apples and 3 bananas. Every morning for 5 days, he randomly chooses and packs his lunch with an apple or with a banana. What is the probability that Brian will get a banana at least two days in a row?
Answers
GPT-4o mini
To solve the problem, we first calculate the total number of ways in which Brian can choose his lunch over the 5 days. Since he can choose either an apple (A) or a banana (B) each day, the total number of arrangements of his choices is given by:
\[
2^5 = 32
\]
Next, we need to determine the number of arrangements that do not have a banana on two consecutive days. We can use a combinatorial approach to count these arrangements.
Let's define:
- \( a_n \): the number of valid arrangements of lunches over \( n \) days that do not have two consecutive bananas.
To construct these arrangements:
1. If the first day is an apple (A), the next \( n-1 \) days can be any valid arrangement of \( n-1 \) days.
2. If the first day is a banana (B), the second day must be an apple (A) to avoid consecutive bananas. The next \( n-2 \) days can be any valid arrangement of \( n-2 \) days.
This gives us the recurrence relation:
\[
a_n = a_{n-1} + a_{n-2}
\]
Now, we need base cases:
- \( a_1 \): For one day, the choices are either A or B, hence \( a_1 = 2 \).
- \( a_2 \): The valid arrangements are AA, AB, and BA, so \( a_2 = 3 \).
Now we can compute \( a_n \) for \( n = 3, 4, 5 \):
\[
\begin{align*}
a_3 & = a_2 + a_1 = 3 + 2 = 5 \\
a_4 & = a_3 + a_2 = 5 + 3 = 8 \\
a_5 & = a_4 + a_3 = 8 + 5 = 13 \\
\end{align*}
\]
Thus, \( a_5 = 13 \), meaning there are 13 valid arrangements of the lunches over 5 days without two consecutive bananas.
Now, we calculate the number of arrangements where there are at least two consecutive bananas by subtracting the number of valid arrangements from the total arrangements:
\[
\text{Number of arrangements with at least 2 consecutive bananas} = 32 - 13 = 19
\]
Finally, to find the probability that he will get a banana at least two days in a row, we take the ratio of the number of favorable arrangements to the total arrangements:
\[
P(\text{at least 2 consecutive bananas}) = \frac{19}{32}
\]
Therefore, the final answer is
\[
\boxed{\frac{19}{32}}
\]
\[
2^5 = 32
\]
Next, we need to determine the number of arrangements that do not have a banana on two consecutive days. We can use a combinatorial approach to count these arrangements.
Let's define:
- \( a_n \): the number of valid arrangements of lunches over \( n \) days that do not have two consecutive bananas.
To construct these arrangements:
1. If the first day is an apple (A), the next \( n-1 \) days can be any valid arrangement of \( n-1 \) days.
2. If the first day is a banana (B), the second day must be an apple (A) to avoid consecutive bananas. The next \( n-2 \) days can be any valid arrangement of \( n-2 \) days.
This gives us the recurrence relation:
\[
a_n = a_{n-1} + a_{n-2}
\]
Now, we need base cases:
- \( a_1 \): For one day, the choices are either A or B, hence \( a_1 = 2 \).
- \( a_2 \): The valid arrangements are AA, AB, and BA, so \( a_2 = 3 \).
Now we can compute \( a_n \) for \( n = 3, 4, 5 \):
\[
\begin{align*}
a_3 & = a_2 + a_1 = 3 + 2 = 5 \\
a_4 & = a_3 + a_2 = 5 + 3 = 8 \\
a_5 & = a_4 + a_3 = 8 + 5 = 13 \\
\end{align*}
\]
Thus, \( a_5 = 13 \), meaning there are 13 valid arrangements of the lunches over 5 days without two consecutive bananas.
Now, we calculate the number of arrangements where there are at least two consecutive bananas by subtracting the number of valid arrangements from the total arrangements:
\[
\text{Number of arrangements with at least 2 consecutive bananas} = 32 - 13 = 19
\]
Finally, to find the probability that he will get a banana at least two days in a row, we take the ratio of the number of favorable arrangements to the total arrangements:
\[
P(\text{at least 2 consecutive bananas}) = \frac{19}{32}
\]
Therefore, the final answer is
\[
\boxed{\frac{19}{32}}
\]