The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 10%. If 15 calculators are selected at random, what is the probability that 4 or more of the calculators will be defective
prob of defect = .1
prob of NOT defect = .9
4 or more defective means
exclude cases of 0, 1, 2, or 3 defective
prob of none defective = C(15,0) (.1^0)( .9^15) = .20589
prob of one defective = C(15,1) (.1)^1 (.9)^14 =.34315
prob of two defective = C(15,2) (.1^2)(.9^13) = .266896
prob of three defective = C(15,3) (.1^3)(.9^12) = .128505
So prob of 4 or more defective
= 1 - sum of the above cases
= .944444
prob of three defective = C(15,3) (.1^12)(.9^3) =
To find the probability that 4 or more of the calculators will be defective, we can use the binomial probability formula. The binomial formula is given by:
P(X = k) = nCk * p^k * q^(n-k)
Where:
P(X = k) is the probability of getting exactly k successes,
n is the total number of trials,
k is the number of successes,
p is the probability of success on a single trial,
q is the probability of failure.
In this case, n = 15, k = 4, p = 0.10 (probability of a defective calculator), and q = 0.90 (probability of a non-defective calculator).
Now, we need to calculate the probability of getting 4, 5, 6, 7, ..., 15 defective calculators and then sum them together.
P(4 or more defective calculators) = P(X = 4) + P(X = 5) + P(X = 6) + ... + P(X = 15)
P(X = k) = nCk * p^k * q^(n-k)
So, let's calculate the probability for each value of k and then sum them up:
P(X = 4) = 15C4 * (0.10)^4 * (0.90)^(15-4)
P(X = 5) = 15C5 * (0.10)^5 * (0.90)^(15-5)
P(X = 6) = 15C6 * (0.10)^6 * (0.90)^(15-6)
...
P(X = 15) = 15C15 * (0.10)^15 * (0.90)^(15-15)
After calculating each of these probabilities, we can sum them together to find the probability that 4 or more of the calculators will be defective.
To find the probability that 4 or more of the calculators will be defective, we can use the binomial probability formula.
The binomial probability formula is given by:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of having exactly k successful outcomes (i.e., having exactly k defective calculators)
- n is the total number of trials (i.e., the number of calculators selected)
- p is the probability of a successful outcome (i.e., the probability of a defective calculator)
- k is the number of successful outcomes (i.e., the number of defective calculators)
In this case, n = 15 (since 15 calculators are selected) and p = 0.10 (since the probability of a defective calculator is 10%).
Now, let's calculate the probability of having 4 or more defective calculators by summing up the probabilities for k = 4, 5, 6, ..., 15 (since there are 15 calculators in total).
P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) + ... + P(X = 15)
Using the binomial probability formula, the probability for each value of k can be calculated. Finally, sum up all these probabilities to get the final answer.