TS EAMCET · Maths · Probability
A person fails 4 times in a game when he plays 9 times. If he plays 15 times, the probability of having success at most one is
- A \(\frac{65}{9}\left(\frac{5}{9}\right)^{14}\)
- B \(\frac{65}{9}\left(\frac{5}{9}\right)^{15}\)
- C \(\frac{79}{9}\left(\frac{4}{9}\right)^{14}\)
- D \(\frac{79}{9}\left(\frac{4}{9}\right)^{15}\)
Answer & Solution
Correct Answer
(C) \(\frac{79}{9}\left(\frac{4}{9}\right)^{14}\)
Step-by-step Solution
Detailed explanation
We have, \(P=\frac{5}{9}\) \[ \therefore \quad q=1-p=1-\frac{5}{9}=\frac{4}{9} \] and \[ n=15, r \leq 1 \]…
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