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MHT CET · Maths · Probability

It is known that a box of 8 batteries contains 3 defective pieces and a person randomly selects two batteries from the box. If \(X\) is the number of defective batteries selected,
then \(P(X \leq 1)=\)

  1. A \(\frac{25}{28}\)
  2. B \(\frac{14}{28}\)
  3. C \(\frac{10}{28}\)
  4. D \(\frac{13}{28}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{25}{28}\)

Step-by-step Solution

Detailed explanation

We have probability of selecting defective batteries \(=\frac{3}{8}\)
\(\therefore p=\frac{3}{8} \text { and } q=\frac{5}{8} \)
\( \therefore P(X=0)+P(X=1) \)
\( =\left[{ }^{2} C_{0}\left(\frac{3}{8}\right)^{0} \times\left(\frac{5}{8}\right)^{2}\right]+\left[{ }^{2} C_{1}\left(\frac{3}{8}\right)^{1} \times\left(\frac{5}{8}\right)^{1}\right] \)
\( =\frac{25}{64}+\frac{2 \times 3 \times 5}{64}=\frac{55}{64} \)