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

Prob(defect) = .2

prob(not defective = .8

prob(exactly 2 of 10 are defective_
= C(10,2) (.2)^2 (.8)^8
= ....

2. Use the "back-door" approach

if 2 or more are defective, then
we can't have:
none defective or only 1 defective.
= 1 - C(10,0) (.2)^0 (.8)^10 - C(10,1) (.2)(.8^9
= ...

3. For this you will just have to grind out
C(10,5) (.2)^5 (.8)^5 + C(10,6) (.2)^6 (.8)^4 + ... + C(10,10) .2^10 .8^0

Well, well, well, looks like we have some defective bolts here! Let's calculate the probability for each scenario:

1. To find the probability that exactly 2 bolts will be defective, we can use the binomial probability formula. The probability of 2 defective bolts out of 10 can be calculated as:
P(exactly 2 defective bolts) = C(10, 2) * (0.2)^2 * (0.8)^8, where C(10, 2) is the combination formula for selecting 2 items out of 10. I'll let you plug in the numbers and do the math!

2. To find the probability that 2 or more bolts will be defective, we need to find the probability of having exactly 2, 3, 4, 5, 6, 7, 8, 9, or 10 defective bolts and add them up. You can use the same formula as above for each case and sum them up.

3. To find the probability that more than 5 bolts will be defective, we need to find the probability of having 6, 7, 8, 9, or 10 defective bolts and add them up. Again, same formula for each case and sum them up.

Remember, always keep a sense of humor when dealing with probabilities, because they can sometimes drive you a little nuts! Good luck with your calculations!

To find the probabilities, we will use the binomial probability formula:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k)

Where:
- n is the number of trials (number of bolts selected)
- k is the number of successful outcomes (number of defective bolts)
- p is the probability of success (probability of a bolt being defective)

1. To find the probability that exactly 2 bolts will be defective:
n = 10 (10 bolts selected)
k = 2 (exactly 2 defective bolts)
p = 0.20 (20% defective)
Using the formula:

P(X = 2) = (10 C 2) * 0.20^2 * (1 - 0.20)^(10 - 2)

Calculating this, we get:
P(X = 2) = 45 * 0.04 * 0.8^8 ≈ 0.302

Therefore, the probability that exactly 2 bolts will be defective is approximately 0.302.

2. To find the probability that 2 or more bolts will be defective:
We need to find the probabilities of having exactly 2, 3, 4, 5, 6, 7, 8, 9, and 10 defective bolts, and then sum them up.
Using the formula for each case:

P(X = 2) = (10 C 2) * 0.20^2 * (1 - 0.20)^(10 - 2)
P(X = 3) = (10 C 3) * 0.20^3 * (1 - 0.20)^(10 - 3)
P(X = 4) = (10 C 4) * 0.20^4 * (1 - 0.20)^(10 - 4)
P(X = 5) = (10 C 5) * 0.20^5 * (1 - 0.20)^(10 - 5)
P(X = 6) = (10 C 6) * 0.20^6 * (1 - 0.20)^(10 - 6)
P(X = 7) = (10 C 7) * 0.20^7 * (1 - 0.20)^(10 - 7)
P(X = 8) = (10 C 8) * 0.20^8 * (1 - 0.20)^(10 - 8)
P(X = 9) = (10 C 9) * 0.20^9 * (1 - 0.20)^(10 - 9)
P(X = 10) = (10 C 10) * 0.20^10 * (1 - 0.20)^(10 - 10)

Calculating each probability and summing them up, we get:
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

P(X ≥ 2) ≈ 0.302 + 0.201 + 0.088 + 0.026 + 0.005 + 0.001 + 0 + 0 + 0
P(X ≥ 2) ≈ 0.623

Therefore, the probability that 2 or more bolts will be defective is approximately 0.623.

3. To find the probability that more than 5 bolts will be defective:
We need to find the probabilities of having 6, 7, 8, 9, and 10 defective bolts, and then sum them up.

P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

Using the same formula as before and calculating each probability, we get:
P(X > 5) ≈ 0.001 + 0 + 0 + 0 + 0
P(X > 5) ≈ 0.001

Therefore, the probability that more than 5 bolts will be defective is approximately 0.001.

To find the probabilities in this scenario, we will use the binomial probability formula. The formula is given as:

P(x) = (nCx) * p^x * (1-p)^(n-x)

Where:
P(x) is the probability of x successes,
n is the total number of trials,
x is the number of successful trials,
p is the probability of success in a single trial, and
(1-p) is the probability of failure in a single trial.

Let's calculate the probabilities step by step using the given information:

1. Probability of exactly 2 defective bolts:
In this case, we want to calculate P(x=2), where x represents the number of defective bolts. So, we have:
n = 10 (total number of bolts selected)
x = 2 (number of defective bolts)
p = 0.20 (probability of a bolt being defective)

Using the formula:
P(x=2) = (10C2) * (0.20)^2 * (1-0.20)^(10-2)

Calculating the values:
(10C2) = 10! / (2! * (10-2)!)
= 10! / (2! * 8!)
= (10 * 9) / (2 * 1)
= 45

P(x=2) = 45 * (0.20)^2 * (0.80)^8

Now, we can calculate the value of P(x=2).

2. Probability of 2 or more defective bolts:
In this case, we want to calculate P(x>=2). It means we need to find the probability of having 2, 3, 4, 5, 6, 7, 8, 9, or 10 defective bolts.

To calculate this, we can find the probability of the complement event (having less than 2 defective bolts) and subtract it from 1.

P(x>=2) = 1 - P(x<2)

P(x<2) = P(x=0) + P(x=1)

P(x=0) = (10C0) * (0.20)^0 * (0.80)^10
P(x=1) = (10C1) * (0.20)^1 * (0.80)^9

Calculate P(x<2) and then find P(x>=2).

3. Probability of more than 5 defective bolts:
In this case, we want to calculate P(x>5). It means we need to find the probability of having 6, 7, 8, 9, or 10 defective bolts.

To calculate this, we can find the probability of the complement event (having less than or equal to 5 defective bolts) and subtract it from 1.

P(x>5) = 1 - P(x<=5)

P(x<=5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5)

Calculate P(x<=5) and then find P(x>5).

By following these steps, you can calculate the probabilities for each scenario using the binomial probability formula.

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