AP EAMCET · Maths · Probability
A box contains 20% defective bulbs. Five bulbs are chosen randomly from this box. The probability that exactly 3 of the chosen bulbs are defective is
- A \(\frac{32}{625}\)
- B \(\frac{32}{125}\)
- C \(\frac{16}{625}\)
- D \(\frac{16}{125}\)
Answer & Solution
Correct Answer
(A) \(\frac{32}{625}\)
Step-by-step Solution
Detailed explanation
Given, \(P(\) defective bulbs \()=20 \%=\frac{1}{5}\) Let, \(p=\frac{1}{5} \Rightarrow q=\frac{4}{5}\) Now, P (exactly 3 defective bulbs)…
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