AP EAMCET · Maths · Probability

A box contains \(n\) coins, \(m\) of which are fair and the rest are biased. When a biased coin is tossed, the probability of getting a head is twice as likely as tail. A coin is drawn from the box at random and is tossed twice. It is found that first time it shows head and the second time it shows tail. Then, the probability that the coin drawn is fair is

A \(\frac{7 m}{8 n+m}\)
B \(\frac{9 m}{8 n+m}\)
C \(\frac{7 m}{8 m+n}\)
D \(\frac{9 m}{8 m+n}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{9 m}{8 n+m}\)

Step-by-step Solution

Detailed explanation

Let the event \(A\) and \(B\) \(A=\) fair coin tossed \(B=\) biased coin tossed \(E=\) first time head second time tail \(P(A)=\frac{m}{n}, P(B)=\frac{n-m}{n}\) \(P(E / A)=\frac{1}{4}, P(E / B)=\frac{2}{9}\) Required probability…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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