Perfect Solo Problem: Clara Nett plays a musical
solo. She is quite good and guesses that her
probability of playing any one note right is 99%.
The solo has 60 notes.
a. Find the probability that
i. She plays every note right
ii. She makes exactly one mistake
iii. She makes exactly two mistakes
iv. She makes at least two mistakes
v. She makes more than two mistakes
b. What must be Clara’s probability of playing
any one note right if she wants to have a 95%
probability of playing all 60 notes right?
To solve the given problem, we need to understand the concept of probability and apply it to the given scenario.
a)
i. The probability that Clara plays every note right can be found by multiplying the probability of playing each note right together. Since her probability of playing any one note right is 99% (or 0.99), the probability of playing all 60 notes right is:
P(play every note right) = 0.99^60 = 0.5905 (rounded to 4 decimal places)
ii. The probability that Clara makes exactly one mistake can be calculated using the binomial probability formula. The binomial probability is given by:
P(k successes in n trials) = (n C k) * p^k * (1-p)^(n-k)
where n is the total number of trials, k is the number of successful outcomes, and p is the probability of success.
In this case, n = 60, k = 1, and p = 0.99. Plugging these values into the formula, we get:
P(make exactly one mistake) = (60 C 1) * (0.99)^1 * (1-0.99)^(60-1)
= 60 * 0.99 * 0.01^59
≈ 0.0060 (rounded to 4 decimal places)
iii. The probability that Clara makes exactly two mistakes can be calculated in a similar way. Using the same formula as above, but with k = 2, we get:
P(make exactly two mistakes) = (60 C 2) * (0.99)^2 * (1-0.99)^(60-2)
= 1770 * (0.99)^2 * (0.01)^58
≈ 0.050 (rounded to 3 decimal places)
iv. The probability that Clara makes at least two mistakes can be found by subtracting the probabilities of making zero and one mistake from 1. So:
P(make at least two mistakes) = 1 - P(play every note right) - P(make exactly one mistake)
≈ 1 - 0.5905 - 0.0060
≈ 0.4035 (rounded to 4 decimal places)
v. The probability that Clara makes more than two mistakes is the same as the probability of making at least three mistakes. We can calculate this by subtracting the probabilities of making zero, one, and two mistakes from 1. So:
P(make more than two mistakes) = 1 - P(play every note right) - P(make exactly one mistake) - P(make exactly two mistakes)
≈ 1 - 0.5905 - 0.0060 - 0.050
≈ 0.3545 (rounded to 4 decimal places)
b) To find the probability of playing all 60 notes right with a 95% probability, we need to find the probability of making zero mistakes. Let's denote the probability of playing any one note right as x.
Using the formula from part a) (i), we can set up the following equation:
0.99^60 = x^60 = 0.95
Solving this equation for x, we find:
x ≈ 0.9938 (rounded to 4 decimal places)
Therefore, Clara must have a probability of playing any one note right approximately 99.38% (or 0.9938) to have a 95% probability of playing all 60 notes right.
R -- right = .99
W -- wrong = .01
prob(all correct) = C(60,60) (.99^60) (.01)^0)
= .00^60 = ...
prob(59right,1wrong)
= C(60,59)(.99^59)(.01^1) = ....
Prob(58right,2wrong) = ....... ** (see above and adjust)
prob(at least 2 wrong) = 1 - **
you try the others.