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The probability of a resistor being defective is 0.02. There are 50 such resistors in a circuit. The probability of two or more defective resistors in the circuit (round off to two decimal places) is __________.
Correct answer is '0.26'. Can you explain this answer?
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Verified Answer
The probability of a resistor being defective is 0.02. There are 50 su...
P = 0.02
n = 50
λ = np = 50 (0.02) = 1
P (x ≥ 2) = 1 – P (x < 2)
= 1 – [P(x =0) + P (x = 1)]
P (x ≥ 2) = 1 – e–1 (1 + 1) = 0.26
n = 50
λ = np = 50 (0.02) = 1
P (x ≥ 2) = 1 – P (x < 2)
= 1 – [P(x =0) + P (x = 1)]
P (x ≥ 2) = 1 – e–1 (1 + 1) = 0.26Most Upvoted Answer
The probability of a resistor being defective is 0.02. There are 50 su...
Probability of a resistor being defective
The probability of a resistor being defective is given as 0.02. This means that out of every 100 resistors produced, 2 of them are expected to be defective.
Number of resistors in the circuit
In the given circuit, there are 50 resistors. Let's denote this as n = 50.
Probability of two or more defective resistors
To find the probability of having two or more defective resistors in the circuit, we need to calculate the probability of having exactly two, exactly three, and so on, up to 50 defective resistors, and then sum them up.
Calculating the probability of having exactly two defective resistors
To calculate the probability of having exactly two defective resistors, we use the binomial probability formula:
P(X=k) = C(n,k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of having exactly k defective resistors,
C(n,k) is the number of combinations of n items taken k at a time,
p is the probability of a resistor being defective (0.02),
n is the total number of resistors in the circuit (50), and
k is the number of defective resistors (2).
Plugging in the values, we get:
P(X=2) = C(50,2) * (0.02)^2 * (1-0.02)^(50-2)
Calculating this value gives us the probability of having exactly two defective resistors in the circuit.
Calculating the probability of having exactly three, four, and so on, defective resistors
Similarly, we can calculate the probabilities of having exactly three, four, and so on, defective resistors using the same formula.
Calculating the probability of two or more defective resistors
To find the probability of having two or more defective resistors, we need to sum up the probabilities of having exactly two, exactly three, and so on, up to 50 defective resistors.
P(X>=2) = P(X=2) + P(X=3) + ... + P(X=50)
Calculating this sum will give us the probability of having two or more defective resistors in the circuit.
Final answer
After performing the calculations, the final answer is obtained as 0.26 (rounded off to two decimal places). This means that there is a 26% chance of having two or more defective resistors in the circuit.
The probability of a resistor being defective is given as 0.02. This means that out of every 100 resistors produced, 2 of them are expected to be defective.
Number of resistors in the circuit
In the given circuit, there are 50 resistors. Let's denote this as n = 50.
Probability of two or more defective resistors
To find the probability of having two or more defective resistors in the circuit, we need to calculate the probability of having exactly two, exactly three, and so on, up to 50 defective resistors, and then sum them up.
Calculating the probability of having exactly two defective resistors
To calculate the probability of having exactly two defective resistors, we use the binomial probability formula:
P(X=k) = C(n,k) * p^k * (1-p)^(n-k)
Where:
P(X=k) is the probability of having exactly k defective resistors,
C(n,k) is the number of combinations of n items taken k at a time,
p is the probability of a resistor being defective (0.02),
n is the total number of resistors in the circuit (50), and
k is the number of defective resistors (2).
Plugging in the values, we get:
P(X=2) = C(50,2) * (0.02)^2 * (1-0.02)^(50-2)
Calculating this value gives us the probability of having exactly two defective resistors in the circuit.
Calculating the probability of having exactly three, four, and so on, defective resistors
Similarly, we can calculate the probabilities of having exactly three, four, and so on, defective resistors using the same formula.
Calculating the probability of two or more defective resistors
To find the probability of having two or more defective resistors, we need to sum up the probabilities of having exactly two, exactly three, and so on, up to 50 defective resistors.
P(X>=2) = P(X=2) + P(X=3) + ... + P(X=50)
Calculating this sum will give us the probability of having two or more defective resistors in the circuit.
Final answer
After performing the calculations, the final answer is obtained as 0.26 (rounded off to two decimal places). This means that there is a 26% chance of having two or more defective resistors in the circuit.
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