During a single day at radio station WMZH, the probability that a particular song is played is 3/4. What is the probability that this song will be played on at most 2 days out of 5 days? Round your answer to the nearest thousandth.
LOL --- Do they stop playing if it is played the first two? I assume not
Anyway this can be done by calculating the probability of 5 days played + prob of 4 days played + prob of 3 days played.
subtract that sum from 1.00 to get the probability that if it was ever played it was less than 3 days
To find the probability that the song will be played on at most 2 days out of 5, we can calculate the probability that it will be played on 0 days, 1 day, or 2 days and then add them together.
The probability that the song is played on any given day is 3/4. Therefore, the probability that the song is not played on any given day is 1 - 3/4 = 1/4.
The probability of the song being played on 0 days out of 5 is (1/4)^5 = 1/1024.
The probability of the song being played on 1 day out of 5 is (1/4)^4 * (3/4) * 5C1 = 5/256.
The probability of the song being played on 2 days out of 5 is (1/4)^3 * (3/4)^2 * 5C2 = 15/256.
Adding these probabilities together, we get:
1/1024 + 5/256 + 15/256 ≈ 0.067.
Therefore, the probability that the song will be played on at most 2 days out of 5 is approximately 0.067 (rounded to the nearest thousandth).
To find the probability that the song will be played on at most 2 days out of 5, we can use the binomial probability formula.
The probability of success (playing the song) on any given day is 3/4, which means the probability of failure (not playing the song) is 1/4.
Let's calculate the individual probabilities for each scenario:
1. The song is played on 0 days out of 5:
- Probability: (1/4)^5 (since the song is not played on any of the 5 days)
- Simplifying: 1/1024
2. The song is played on 1 day out of 5:
- Probability: (3/4)^1 * (1/4)^4 (since the song is played once and not played the remaining 4 days)
- Simplifying: 3/1024
3. The song is played on 2 days out of 5:
- Probability: (3/4)^2 * (1/4)^3 (since the song is played twice and not played the other 3 days)
- Simplifying: 9/1024
To find the probability of at most 2 days, we sum up the individual probabilities:
1/1024 + 3/1024 + 9/1024 = 13/1024 ≈ 0.0127
Therefore, the probability that the song will be played on at most 2 days out of 5 is approximately 0.0127, rounded to the nearest thousandth.