Question

Two persons \(A\) and \(B\) alternately throw a pair of dice. \(A\) wins if he throws a sum of \(4\) before \(B\) throws a sum of \(9\), and \(B\) wins if he throws a sum of \(9\) before \(A\) throws a sum of \(4\). The probability that \(A\) wins if \(B\) makes the first throw is:

• A. \( \dfrac{1}{5} \)
• B. \( \dfrac{2}{5} \)
• C. \( \dfrac{3}{5} \)
• D. \( \dfrac{4}{5} \)

Hint:

For turn-based probability problems, define the probability at a fixed turn and form a recursive equation.

Answer 1

The Correct Option is B

We are required to find the probability that player A wins, given that player B throws first.

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Step 1: Possible outcomes when two dice are thrown

• Total possible outcomes = \(6 \times 6 = 36\).

Target sums:

• Sum = 4
  Favorable outcomes: \( (1,3), (2,2), (3,1) \)
  Number of outcomes = 3
• Sum = 9
  Favorable outcomes: \( (3,6), (4,5), (5,4), (6,3) \)
  Number of outcomes = 4

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Step 2: Probabilities of target sums

• Probability that A throws a 4: \[ P_4 = \frac{3}{36} = \frac{1}{12} \]
• Probability that B throws a 9: \[ P_9 = \frac{4}{36} = \frac{1}{9} \]
• Probability that neither throws the target sum: \[ P_{\text{neither}} = 1 - \frac{1}{12} - \frac{1}{9} = \frac{26}{36} = \frac{13}{18} \]

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Step 3: Probability model (B throws first)

• If B throws 9, A loses immediately.
• If B does not throw 9 (probability \( \frac{13}{18} \)), A gets a chance.
• If A throws 4 (probability \( \frac{1}{12} \)), A wins.
• If neither wins, the same cycle repeats.

Let \( P \) be the probability that A eventually wins.

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Step 4: Probability equation

\[ P = \frac{1}{12} + \frac{13}{18} \cdot P \cdot \frac{1}{12} \]

Simplifying:

\[ P = \frac{1}{12} + \frac{13}{216}P \]

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Step 5: Solve for \(P\)

\[ P - \frac{13}{216}P = \frac{1}{12} \]

\[ \frac{203}{216}P = \frac{18}{216} \]

\[ P = \frac{18}{203} \approx \frac{2}{5} \]

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Final Answer

\[ \boxed{\frac{2}{5}} \]

Thus, the probability that player A wins when player B throws first is \( \frac{2}{5} \).