Question

Two persons A and B are playing a game of rolling 2 dice alternately. The person who first gets the sum of the numbers appearing on the dice as a prime number will win the game. If A starts the game, then the probability of B winning the game is

• A. \(\dfrac{12}{19}\)
• B. \(\dfrac{5}{19}\)
• C. \(\dfrac{7}{19}\)
• D. \(\dfrac{14}{19}\)

Hint:

When players take alternate independent turns, the probability usually forms an infinite geometric series. Identify the common ratio and apply \[ \boxed{1+r+r^2+\cdots=\frac1{1-r}}. \]

Answer 1

The Correct Option is C

Step 1: Find the probability of getting a prime sum.

The possible prime sums are& nbsp;

\[ 2,\;3,\;5,\;7,\;11. \]

Their corresponding frequencies are

\[ 1,\;2,\;4,\;6,\;2, \]

respectively. Therefore,

\[ P(\text{prime sum}) = \frac{1+2+4+6+2}{36} = \frac{15}{36} = \frac{5}{12}. \]

Hence,

\[ P(\text{not prime}) = 1-\frac{5}{12} = \frac{7}{12}. \]

Step 2: Write the probability that B wins.

B wins if:

• A fails and B succeeds, or
• Both A and B fail once, then A fails again and B succeeds, and so on.

Thus,

\[ P(B) = \frac{7}{12}\cdot\frac{5}{12} + \left(\frac{7}{12}\right)^3\cdot\frac{5}{12} + \left(\frac{7}{12}\right)^5\cdot\frac{5}{12} +\cdots. \]

Step 3: Evaluate the geometric series.

Taking common terms,

\[ P(B) = \frac{35}{144} \left[ 1+ \left(\frac{49}{144}\right)+ \left(\frac{49}{144}\right)^2+\cdots \right]. \]

Using the formula

\[ 1+r+r^2+\cdots=\frac{1}{1-r}, \]

where

\[ r=\frac{49}{144}, \]

we obtain

\[ P(B) = \frac{35}{144} \times \frac{144}{95} = \frac{35}{95} = \boxed{\frac{7}{19}}. \]

Hence, the correct option is \[ \boxed{(C)}. \]