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A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is
- a)0.0036
- b)0.1937
- c)0.2234
- d)0.3874
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A lot has 10% defective items. Ten items are chosen randomly from this...
Let A be the event that items are defective and B be the event that items are non- defective
∴ P( A )= 0.1 and P(B) = 0.9
∴ Probability that exactly two of those items are defective
Most Upvoted Answer
A lot has 10% defective items. Ten items are chosen randomly from this...
To solve this problem, we can use the concept of binomial probability.
Understanding the Problem:
We have a lot of items in which 10% are defective. We randomly choose ten items from this lot. We need to find the probability that exactly 2 of the chosen items are defective.
Solution:
Step 1: Defining the Events:
Let's define the events:
A: The chosen item is defective
B: The chosen item is not defective
Step 2: Calculating Probability of Defective Item:
We are given that 10% of the items are defective. Therefore, the probability of a randomly chosen item being defective is:
P(A) = 0.10
Step 3: Calculating Probability of Non-defective Item:
Since the probability of an item being defective is 0.10, the probability of an item not being defective is:
P(B) = 1 - P(A) = 1 - 0.10 = 0.90
Step 4: Calculating the Probability of Exactly 2 Defective Items:
To find the probability of exactly 2 defective items, we can use the binomial probability formula:
P(X=k) = nCk * p^k * q^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes (in our case, 2 defective items).
n is the number of trials (in our case, 10 items).
k is the number of successes (in our case, 2 defective items).
p is the probability of success (in our case, P(A)).
q is the probability of failure (in our case, P(B)).
Using these values, the probability of exactly 2 defective items is:
P(X=2) = 10C2 * (0.10)^2 * (0.90)^(10-2)
Calculating the values:
10C2 = (10!)/(2!(10-2)!) = (10*9)/(2*1) = 45
(0.10)^2 = 0.01
(0.90)^(10-2) = 0.90^8 ≈ 0.43046721
P(X=2) = 45 * 0.01 * 0.43046721 ≈ 0.1937
Therefore, the probability that exactly 2 of the chosen items are defective is approximately 0.1937, which corresponds to option B.
Understanding the Problem:
We have a lot of items in which 10% are defective. We randomly choose ten items from this lot. We need to find the probability that exactly 2 of the chosen items are defective.
Solution:
Step 1: Defining the Events:
Let's define the events:
A: The chosen item is defective
B: The chosen item is not defective
Step 2: Calculating Probability of Defective Item:
We are given that 10% of the items are defective. Therefore, the probability of a randomly chosen item being defective is:
P(A) = 0.10
Step 3: Calculating Probability of Non-defective Item:
Since the probability of an item being defective is 0.10, the probability of an item not being defective is:
P(B) = 1 - P(A) = 1 - 0.10 = 0.90
Step 4: Calculating the Probability of Exactly 2 Defective Items:
To find the probability of exactly 2 defective items, we can use the binomial probability formula:
P(X=k) = nCk * p^k * q^(n-k)
Where:
P(X=k) is the probability of getting exactly k successes (in our case, 2 defective items).
n is the number of trials (in our case, 10 items).
k is the number of successes (in our case, 2 defective items).
p is the probability of success (in our case, P(A)).
q is the probability of failure (in our case, P(B)).
Using these values, the probability of exactly 2 defective items is:
P(X=2) = 10C2 * (0.10)^2 * (0.90)^(10-2)
Calculating the values:
10C2 = (10!)/(2!(10-2)!) = (10*9)/(2*1) = 45
(0.10)^2 = 0.01
(0.90)^(10-2) = 0.90^8 ≈ 0.43046721
P(X=2) = 45 * 0.01 * 0.43046721 ≈ 0.1937
Therefore, the probability that exactly 2 of the chosen items are defective is approximately 0.1937, which corresponds to option B.
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