AP EAMCET · Maths · Probability
Each of the two boxes \(A\) and \(B\) contain 10 chits numbered 1 to 10 . If one chit is drawn at random from each of \(A\) and \(B\), then the probability that the number on the chit drawn from \(A\) is smaller than the number on the chit drawn from \(B\), is
- A \(\frac{9}{10}\)
- B \(\frac{9}{20}\)
- C \(\frac{19}{20}\)
- D \(\frac{17}{20}\)
Answer & Solution
Correct Answer
(B) \(\frac{9}{20}\)
Step-by-step Solution
Detailed explanation
According to the given information, If drawn number from \(A\) is 1 , then the favourable drawn number from \(B\) are \(2,3,4, \ldots \ldots 10\), are total 9 cases. Similarly, for 2 , there are 8 cases and so on.…
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