In the mass production of bolts it is found that 5% are defective. Bolts selected at random and put into
packets of ten. A packet is selected at random. Find the probability that it contains
(a) three defective bolts.
(b) less than three are defective.
Two packets are selected at random.
(c) Find the probability that there are no defective bolts in either packet.
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To find the probabilities, we will use the binomial probability formula. The formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
where:
- P(X = k) represents the probability of getting exactly k successes (in this case, defective bolts).
- C(n, k) is the binomial coefficient, which represents the number of ways to choose k successes from n trials.
- p is the probability of success (defective bolt) on a single trial.
- n is the number of trials (bolts in a packet).
(a) To find the probability of getting three defective bolts in a packet of ten, we substitute the values into the formula:
P(X = 3) = C(10, 3) * 0.05^3 * (1 - 0.05)^(10 - 3)
Using a calculator, we can evaluate this expression. The binomial coefficient C(10, 3) is calculated as 120, and the probability is 0.05^3 ≈ 0.000125. The complement of the probability (1 - 0.05)^(10 - 3) is approximately 0.9515. Multiplying these values together gives:
P(X = 3) ≈ 120 * 0.000125 * 0.9515 ≈ 0.0143
So, the probability that a randomly selected packet contains three defective bolts is approximately 0.0143.
(b) To find the probability of having less than three defective bolts, we need to calculate the probabilities of having zero, one, or two defective bolts and sum them up:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
P(X = 0) = C(10, 0) * 0.05^0 * (1 - 0.05)^(10 - 0)
P(X = 1) = C(10, 1) * 0.05^1 * (1 - 0.05)^(10 - 1)
P(X = 2) = C(10, 2) * 0.05^2 * (1 - 0.05)^(10 - 2)
Calculating each of these probabilities will give us:
P(X = 0) ≈ 0.5987
P(X = 1) ≈ 0.3152
P(X = 2) ≈ 0.0746
Summing these probabilities together:
P(X < 3) ≈ 0.5987 + 0.3152 + 0.0746 ≈ 0.9885
So, the probability that a randomly selected packet contains less than three defective bolts is approximately 0.9885.
(c) To find the probability that there are no defective bolts in either packet, we assume independence between the two packets. This means that the probability of no defective bolts in the first packet is the same as the second packet. Therefore, we square the probability of having no defective bolts in a single packet:
P(no defective bolts in one packet) = (1 - 0.05)^10
Calculating this probability gives:
P(no defective bolts in one packet) ≈ (0.95)^10 ≈ 0.5987
Since we assume independence, the probability of both packets having no defective bolts is the product of the individual probabilities:
P(both packets have no defective bolts) ≈ 0.5987 * 0.5987 ≈ 0.3581
So, the probability that there are no defective bolts in either packet is approximately 0.3581.
To solve these probability problems, we need to apply the concepts of probability and combinatorics.
Let's start with part (a): finding the probability that a packet contains three defective bolts.
Step 1: Determine the probability of selecting a defective bolt
Since 5% of the bolts are defective, the probability of selecting a defective bolt is 0.05.
Step 2: Determine the probability of selecting three defective bolts
If we assume that each bolt in the packet is selected independently, we need to calculate the probability of selecting a defective bolt three times in a row. Since there are ten bolts in a packet, we use the binomial distribution formula:
P(X=r) = C(n,r) * p^r * (1-p)^(n-r)
where:
P(X=r) represents the probability of observing r successes (defective bolts)
C(n,r) is the binomial coefficient that calculates the number of combinations
p is the probability of a single success (selection of a defective bolt)
n is the number of trials (selection of bolts in the packet)
Using these values:
P(X=3) = C(10,3) * 0.05^3 * (1-0.05)^(10-3)
Step 3: Calculate the result
Evaluate the formula using the values from step 2.
For part (b): finding the probability that less than three bolts are defective, we need to find the probability of having 0, 1, or 2 defective bolts.
P(X<3) = P(X=0) + P(X=1) + P(X=2)
Again, use the binomial distribution formula for each value of X.
For part (c): finding the probability that there are no defective bolts in either packet, we need to determine the probability for each packet individually and multiply them together. Since each packet is selected independently, the probability of no defective bolts in either packet is the product of the probabilities.
I hope this helps! Let me know if you need any further clarification or assistance.