A factory finds out that on the average 20% of bolts produced by the given machine will be defective.if 10 bolts are selected at random from the day's production of this machine,find the probability that 1.exactly 2 will be defective.
2. 2 or more will be defective
3. More than 5 will be defective
(base on probability and combinatorial analysis)
prob(fail) = .2
prob(not fail) = .8
prob(exactly 2 of 10 will fail)
= C(10,2) .2^2 .8^8
= appr .302
2 or more defective ---> exclude 0 not fail and 1 not fail
= 1 - C(10,0) .8^0 .2^10 - C(10,1) (.8)( .2^9)
= ....
more than 5
= prob(exactly 6) + prob(exactly 7) + ... + prob(exactly 10)
= ...
just grind that one out
To find the probability in each of these cases, we need to use combinatorial analysis and the concept of probability. In combinatorial analysis, we use combinations to calculate the number of ways we can select items without regard to their order.
Before we calculate the probability, let's understand the problem. We know that the average percentage of defective bolts produced by the machine is 20%. This means that out of every 100 bolts produced, 20 of them are defective.
Now, let's move on to each question and calculate the probability.
1. Exactly 2 will be defective:
To find the probability that exactly 2 bolts will be defective, we need to calculate the probability of selecting 2 defective bolts and 8 non-defective bolts from the total of 10 bolts.
The formula for the probability of exactly "k" successes in "n" trials is given by the binomial probability formula:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
n = total number of trials (in this case, 10 bolts)
k = number of successful trials (in this case, 2 defective bolts)
p = probability of success in a single trial (in this case, 20% or 0.2)
(1 - p) = probability of failure in a single trial
Using the formula, we can plug in the values:
P(X = 2) = (10 C 2) * (0.2)^2 * (1 - 0.2)^(10 - 2)
Calculating this will give us the probability that exactly 2 bolts will be defective.
2. 2 or more will be defective:
To find the probability that 2 or more bolts will be defective, we need to calculate the probability of selecting 2, 3, 4,..., 10 defective bolts out of 10 bolts. We can then sum up these probabilities to get the total probability.
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + ... + P(X = 10)
We can use the same binomial probability formula as in case 1 for each value of k from 2 to 10 and sum them up to get the final probability.
3. More than 5 will be defective:
To find the probability that more than 5 bolts will be defective, we need to calculate the probability of selecting 6, 7, 8, 9, and 10 defective bolts out of 10 bolts. We can then sum up these probabilities to get the total probability.
P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
Again, we can use the binomial probability formula for each value of k from 6 to 10 and sum them up.
Now that we have explained the approach to solving each question, you can apply the binomial probability formula and the concept of combinations to calculate the probabilities in each case.