WBJEE · Maths · Probability
Suppose a machine produces metal parts that contain some defective parts witl probability \(0.05 .\) How many parts should be produced in order that the probability 0 atleast one part being defective is \(1 / 20\) more? (Given that, \(\log _{10} 95=1.977\) an \(\log _{10} 2=0.3\) )
- A 11
- B 12
- C 15
- D 14
Answer & Solution
Correct Answer
(D) 14
Step-by-step Solution
Detailed explanation
Given probability of delective pat \(=0.05=\frac{1}{20}\) Probability of non-defective part \(=1-0.05=0.95=\frac{19}{20}\) We know that, \(P(X=r)={ }^{n} Cr p^{r} q^{n-r}\) where, \(\quad p=\frac{1}{20}, q=\frac{19}{20}\) \(r \geq 1\) and \(n=?\) Also,…
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