Suppose that 50 identical batteries are being tested. After 8 hours of continuous use, assume that a given battery is still operating with a probability of .70 and has failed with a probability of .30. What is the probability that between 25 and 30 batteries (inclusive) will last at least 8 hours?
Recall that if the following conditions are met, we can use a binomial distribution to model the situation:
1. there are only two possible outcomes (operating or not operating, i.e. a bernoulli experiment)
2. probability applies to all units observed, and does not change.
3. random events
If we use the binomial distribution, where p=0.7 (success) and q=0.3 (failure) then the probability of success of i batteries out of n=50 are given by:
nCi*p^i*q^(n-i)
where nCi = n!/((n-i)!i!)
So calculate the probabilities for i=20,21,22,23,24,25 and sum them to get the probability that 20-25 batteries will remain operational.
I get about 0.0024
Sorry, it should be for 25-30 operating batteries, I get about P(25-30)=0.084
To solve this problem, we will use the binomial distribution formula. Let's break down the problem step by step:
Step 1: Identify the values given in the problem:
- Number of batteries tested (n) = 50
- Probability of a battery operating after 8 hours (p) = 0.70
- Probability of a battery failing after 8 hours (q) = 0.30
- Range of batteries that will last at least 8 hours (between 25 and 30, inclusive)
Step 2: Calculate the probability of each number of batteries that will last at least 8 hours:
We want to find the probability that the number of batteries (X) that will last at least 8 hours is between 25 and 30 (inclusive), so we need to calculate the probability of X = 25, X = 26, X = 27, X = 28, X = 29, and X = 30.
Use the binomial distribution formula: P(X = k) = C(n, k) * p^k * q^(n-k), where C(n, k) is the number of combinations.
For each value, plug in the given values and calculate the individual probabilities (taking into account the range):
P(X = 25) = C(50, 25) * (0.70)^25 * (0.30)^(50-25)
P(X = 26) = C(50, 26) * (0.70)^26 * (0.30)^(50-26)
P(X = 27) = C(50, 27) * (0.70)^27 * (0.30)^(50-27)
P(X = 28) = C(50, 28) * (0.70)^28 * (0.30)^(50-28)
P(X = 29) = C(50, 29) * (0.70)^29 * (0.30)^(50-29)
P(X = 30) = C(50, 30) * (0.70)^30 * (0.30)^(50-30)
Step 3: Calculate the total probability by summing up all the individual probabilities:
P(25 <= X <= 30) = P(X = 25) + P(X = 26) + P(X = 27) + P(X = 28) + P(X = 29) + P(X = 30)
Add up the probabilities calculated in Step 2 to find the final solution.
To solve this problem, we can use the binomial probability formula. The formula is:
P(X = k) = C(n, k) * p^k * q^(n-k)
Where:
- P(X = k) is the probability of getting exactly k successes,
- C(n, k) is the binomial coefficient, which represents the number of ways to choose k successes from n trials,
- p is the probability of success on a single trial,
- q is the probability of failure on a single trial,
- n is the total number of trials.
In this case, we want to find the probability that between 25 and 30 batteries (inclusive) will last at least 8 hours, so we need to calculate the probabilities for k = 25, 26, 27, 28, 29, and 30, and then sum them up.
Let's calculate the probability for k = 25 as an example:
P(X = 25) = C(50, 25) * (0.70)^25 * (0.30)^(50-25)
To calculate C(50, 25), we can use the formula:
C(n, k) = n! / (k! * (n-k)!)
where "!" denotes a factorial.
Therefore, C(50, 25) = 50! / (25! * (50-25)!).
For convenience, we can use a scientific calculator or a software like Excel to calculate this value.
Repeat the same process to calculate the probabilities for k = 26, 27, 28, 29, and 30, and then add them up to find the final probability.
Keep in mind that this calculation assumes that the batteries are operating independently of each other, and the probabilities of success (battery lasting at least 8 hours) and failure (battery failing before 8 hours) remain constant throughout the 8-hour period.