JEE Mains · Maths · STD 11 - 14. probability
In a game two players \(A\) and \(B\) take turns in throwing a pair of fair dice starting with player \(A\) and total of scores on the two dice, in each throw is noted. \(A\) wins the game if he throws a total of \(6\) before \(B\) throws a total of \(7\) and \(B\) wins the game if he throws a total of \(7\) before \(A\) throws a total of six The game stops as soon as either of the players wins. The probability of \(A\) winning the game is
- A \(\frac{31}{61}\)
- B \(\frac{5}{6}\)
- C \(\frac{5}{31}\)
- D \(\frac{30}{61}\)
Answer & Solution
Correct Answer
(D) \(\frac{30}{61}\)
Step-by-step Solution
Detailed explanation
\(P(6)=\frac{1}{6}, P(7)=\frac{5}{36}\) \(P(A)=W+F F W+F F F F W+\ldots . .\) \(=\frac{1}{6}+\frac{5}{6} \times \frac{31}{36} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{31}{36}\right)^{2} \frac{1}{6}+\ldots\)…
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