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JEE Mathematics 2025 Question with Solution

A and B alternately throw a pair of dice. A wins if he throws a sum of 55 before B throws a sum of 88, and B wins if he throws a sum of 88 before A throws a sum of 55. The probability that A wins if A makes the first throw is:

  • A

    917\frac{9}{17}

  • B

    919\frac{9}{19}

  • C

    817\frac{8}{17}

  • D

    819\frac{8}{19}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A and B alternately throw a pair of dice. A wins on getting sum 55 and B wins on getting sum 88. A starts first.

Find: The probability that A wins.

For one throw of two dice, the total number of outcomes is

6×6=366 \times 6 = 36

The number of outcomes giving sum 55 is 44, so

P(A gets 5)=436=19P(\text{A gets }5) = \frac{4}{36} = \frac{1}{9}

The number of outcomes giving sum 88 is 55, so

P(B gets 8)=536P(\text{B gets }8) = \frac{5}{36}

If A does not win on his turn and B also does not win on his turn, the game returns to the same starting situation. Thus,

p=19+(89)(3136)pp = \frac{1}{9} + \left(\frac{8}{9}\right)\left(\frac{31}{36}\right)p

where pp is the probability that A wins when A starts first.

Now simplify:

p=19+6281pp = \frac{1}{9} + \frac{62}{81}p

So,

p−6281p=19p - \frac{62}{81}p = \frac{1}{9} 1981p=19\frac{19}{81}p = \frac{1}{9} p=19×8119=919p = \frac{1}{9} \times \frac{81}{19} = \frac{9}{19}

Therefore, the probability that A wins is 919\frac{9}{19}. The correct option is B.

The first approach shown in the working contains an inconsistent intermediate equation, but the stated final answer and the second approach both give 919\frac{9}{19}.

Geometric Series Method

Given: A wins by throwing sum 55, and B wins by throwing sum 88.

Find: Probability that A wins when he throws first.

Let

P(A)=436=19,P(B)=536P(A) = \frac{4}{36} = \frac{1}{9}, \qquad P(B) = \frac{5}{36}

For A to win, either he wins on the first throw, or both fail once and the process repeats.

Hence,

P(A wins)=19+(3236)(3136)19+[(3236)(3136)]219+⋯P(\text{A wins}) = \frac{1}{9} + \left(\frac{32}{36}\right)\left(\frac{31}{36}\right)\frac{1}{9} + \left[\left(\frac{32}{36}\right)\left(\frac{31}{36}\right)\right]^2 \frac{1}{9} + \cdots

This is a geometric series with first term 19\frac{1}{9} and common ratio

(3236)(3136)=6281\left(\frac{32}{36}\right)\left(\frac{31}{36}\right) = \frac{62}{81}

Therefore,

P(A wins)=191−6281=191981=919P(\text{A wins}) = \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{\frac{1}{9}}{\frac{19}{81}} = \frac{9}{19}

Therefore, the probability that A wins is 919\frac{9}{19}.

Common mistakes

  • Using 2536\frac{25}{36} as the probability that both players fail in one round is incorrect. A failing to get sum 55 has probability 3236=89\frac{32}{36} = \frac{8}{9}, and B failing to get sum 88 has probability 3136\frac{31}{36}. Multiply these to reset the game state correctly.

  • Adding the probabilities of A getting 55 and B getting 88 in the same throw structure is wrong because they occur on different turns, not in a single combined experiment. Treat the game as alternating stages, not one simultaneous event.

  • Stopping at P(A wins on first turn)=19P(\text{A wins on first turn}) = \frac{1}{9} is incomplete. If both players fail in the first cycle, the game starts again under the same conditions, so a recursive equation or infinite geometric series is necessary.

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