Mom has 2 apples and 3 bananas. Every morning for 5 days she is going to give her son an apple or a banana.
What is the probability that the boy will get a banana at least two days in a row during these five days?
number of ways to give the fruits
= 5!/(3!2!) = 10
(you could list them)
Bananas at least two days in a row:
BABAB is the only case where two or more B could not be together. Look at your list from above
prog(at least two B's in a row) = 9/10
9/10
To find the probability that the boy will get a banana at least two days in a row during these five days, we can use a combination of counting the possible scenarios and applying probability principles.
Step 1: Determine the total number of possible outcomes.
Since the boy can either get an apple or a banana each morning for 5 days, there are 2 possible outcomes for each day. Thus, the total number of possible outcomes is 2^5 = 32.
Step 2: Determine the number of desired outcomes.
To get a banana at least two days in a row, we need to consider the following scenarios:
1) Getting a banana for 2 days in a row: BBAAA, ABBAA, AAABB (3 scenarios).
2) Getting a banana for 3 days in a row: BBBAA, ABBBA (2 scenarios).
3) Getting a banana for 4 days in a row: BBBBA (1 scenario).
4) Getting a banana for 5 days in a row: BBBBB (1 scenario).
The total number of desired outcomes is 3 + 2 + 1 + 1 = 7.
Step 3: Calculate the probability.
The probability is equal to the number of desired outcomes divided by the total number of possible outcomes.
Therefore, the probability is 7/32 = 0.21875, or approximately 21.9%.
To find the probability that the boy will get a banana at least two days in a row during these five days, we need to calculate the probability for each possible outcome where the boy gets a banana at least two days in a row.
Let's analyze this step by step:
1. First, we need to calculate the total number of possible ways the boy can receive an apple or a banana each morning for five days.
The boy can receive either an apple or a banana each morning, and since there are five mornings, the total number of possible outcomes is:
2 (choices for each morning) * 2 (choices for the second morning) * 2 (choices for the third morning) * 2 (choices for the fourth morning) * 2 (choices for the fifth morning) = 2^5 = 32
The total number of possible outcomes is 32.
2. Now, we need to calculate the number of favorable outcomes where the boy receives a banana at least two days in a row.
To do this, we can consider two scenarios:
a) The boy receives a banana for two consecutive days, followed by any choice on the remaining three days.
b) The boy receives a banana for three consecutive days, followed by any choice on the remaining two days.
For Scenario a):
In this case, the first two days must be bananas, and for the remaining three days, the boy can choose either an apple or a banana.
The total number of outcomes for this scenario is:
1 (banana on day 1) * 1 (banana on day 2) * 2^3 (choices for day 3, 4, and 5) = 8
For Scenario b):
In this case, the first three days must be bananas, and for the remaining two days, the boy can choose either an apple or a banana.
The total number of outcomes for this scenario is:
1 (banana on day 1) * 1 (banana on day 2) * 1 (banana on day 3) * 2^2 (choices for day 4 and 5) = 4
The total number of favorable outcomes for both scenarios is: 8 + 4 = 12
3. Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
Probability = 12 / 32
Therefore, the probability that the boy will get a banana at least two days in a row during these five days is 3/8 or 0.375 (37.5%).