A company manufactures washers, about 5% of which are defective. If a random sample
of 100 washers are inspected, what is the probability that fewer than 4 are defective?
fewer than 4 defective ---> 0 ,1,2, or 3 are defective
prob(defective) = 5/100 = 1/20
prob(not defective) = 19/20
prob
= C(100,0) (1/20)^0 (19/20)^100 + C(100,1)(1/20)(19/20^99 + C(100,2)(1/20)^2 (19/20)^98 + C(100,3)(1/20)^3 (19/20)^97
= .00592.. + .03116.. + .08118...+ .13957..
= .2578...
check my arithmetic
To find the probability that fewer than 4 washers are defective, we need to calculate the probability for each possible number of defective washers (0, 1, 2, and 3) and then sum them up.
Given that 5% of washers are defective, the probability of a washer being defective is 0.05.
To calculate the probability of getting k defective washers in a sample of size n, we can use the binomial distribution formula:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of getting exactly k defective washers,
- (n C k) is the number of ways to choose k defective washers out of n washers,
- p is the probability of a washer being defective,
- (1-p) is the probability of a washer not being defective,
- n is the total number of washers in the sample, and
- k is the number of defective washers.
Now, let's calculate the probability for each possible number of defective washers.
For k = 0:
P(X = 0) = (100 C 0) * 0.05^0 * (1-0.05)^(100-0)
For k = 1:
P(X = 1) = (100 C 1) * 0.05^1 * (1-0.05)^(100-1)
For k = 2:
P(X = 2) = (100 C 2) * 0.05^2 * (1-0.05)^(100-2)
For k = 3:
P(X = 3) = (100 C 3) * 0.05^3 * (1-0.05)^(100-3)
Finally, sum up the probabilities for all the possible numbers of defective washers:
P(fewer than 4 defective washers) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Calculating these probabilities will give you the answer.