A company produces light bulbs, and finds that approximately 2% of the light bulbs they produce are defective. The company wants to make a money-back guarantee that in every box of 10 light bulbs, no more than 1 will be defective. What is the probability that a random box of 10 bulbs will fail to satisfy the guarantee?
My Approach:
If 100 are produced 2 light bulbs will be defective.
In 10 cases (10 lightbulbs each); we have a 50% chance that the 2 will be in 1 box and a 50% chance that 2 defectives will be in 2 boxes.
Therefore the probability of it failing its guarantee is (1/10)(1/2)= 5%
Is this correct? I am not sure how else to approach this problem.
Your approach to the problem is not correct. Let's break down the problem and explain how to solve it correctly.
To solve this problem, we can use the binomial probability formula. In this case, we have 10 trials (representing the 10 light bulbs in a box) with a probability of success (a light bulb being defective) of 2% (or 0.02).
The probability of having exactly "k" defective light bulbs out of 10 can be calculated using the formula:
P(k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(k) is the probability of having exactly "k" defective light bulbs out of 10
- n is the number of trials (10 in this case)
- k is the number of successful trials (defective light bulbs)
- p is the probability of success (defective light bulb)
- (nCk) represents the binomial coefficient, which can be calculated as n! / (k! * (n-k)!)
In this case, we want the probability that no more than 1 light bulb is defective, so we need to find P(0) + P(1) (P(0) represents no defective light bulbs, and P(1) represents exactly 1 defective light bulb).
Calculating P(0):
P(0) = (10C0) * (0.02^0) * (1-0.02)^(10-0) = 1 * 1 * 0.98^10 = 0.817
Calculating P(1):
P(1) = (10C1) * (0.02^1) * (1-0.02)^(10-1) = 10 * 0.02 * 0.98^9 = 0.163
Now we can find the probability of failing to satisfy the guarantee by adding the probabilities:
P(failure) = P(0) + P(1) = 0.817 + 0.163 = 0.98
Therefore, the probability that a random box of 10 light bulbs will fail to satisfy the money-back guarantee is approximately 0.98, or 98%.