Question:
It is known that \(10\%\) of certain articles manufactured are defective. What is the probability that in a random sample of \(12\) such articles, \(9\) are defective?
It is known that \(10\%\) of certain articles manufactured are defective. What is the probability that in a random sample of \(12\) such articles, \(9\) are defective?
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Use the binomial probability formula with n equal to 12, r equal to 9 and p equal to one tenth.
Updated On: Nov 12, 2026
- \(22\left(\frac{9^{3}}{10^{12}}\right)\)
- \(22\left(\frac{9^{12}}{10^{12}}\right)\)
- \(22\left(\frac{9^{3}}{10^{9}}\right)\)
- \(22\left(\frac{9^{3}}{10^{11}}\right)\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
The number of defective articles follows a binomial distribution with \(n=12\), \(p=\dfrac1{10}\), \(q=\dfrac9{10}\).
Step 2: Formula
\(P(X=9)={}^{12}C_9\,p^{9}q^{3}\).
Step 3: Substitute
\({}^{12}C_9=220\).
\(P=220\left(\dfrac1{10}\right)^{9}\left(\dfrac9{10}\right)^{3}=\dfrac{220\cdot9^{3}}{10^{12}}\).
Step 4: Simplify
\(\dfrac{220\cdot9^{3}}{10^{12}}=22\left(\dfrac{9^{3}}{10^{11}}\right)\).
Final Answer:
The probability is 22 times 9 cubed over 10 to the power 11. \[ \boxed{\text{(D) }22\left(\dfrac{9^{3}}{10^{11}}\right)} \]
The number of defective articles follows a binomial distribution with \(n=12\), \(p=\dfrac1{10}\), \(q=\dfrac9{10}\).
Step 2: Formula
\(P(X=9)={}^{12}C_9\,p^{9}q^{3}\).
Step 3: Substitute
\({}^{12}C_9=220\).
\(P=220\left(\dfrac1{10}\right)^{9}\left(\dfrac9{10}\right)^{3}=\dfrac{220\cdot9^{3}}{10^{12}}\).
Step 4: Simplify
\(\dfrac{220\cdot9^{3}}{10^{12}}=22\left(\dfrac{9^{3}}{10^{11}}\right)\).
Final Answer:
The probability is 22 times 9 cubed over 10 to the power 11. \[ \boxed{\text{(D) }22\left(\dfrac{9^{3}}{10^{11}}\right)} \]
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