AP EAMCET · Maths · Probability
A bag contains \(2 n+1\) coins. It is known that \(n\) of these coins have a head on both sides, whereas the remaining \(n+1\) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is \(\frac{31}{42}\), then \(n\) is equal to
- A \(10\)
- B \(11\)
- C \(12\)
- D \(13\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
The probability that the toss results is a tail \(=\frac{(n+1)}{2(2 n+1)}\) \(\therefore 1-\frac{(n+1)}{2(2 n+1)}\) is the probability that the toss result is a head.…
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