CAMANISHPARMAR wrote:

Romeo and Juliet play a dice game in which the two participants take turns rolling a single fair six-sided die. The first player to roll a 6 wins. If Romeo rolls first, what is the probability that Juliet will win?

A) \(\frac{1}{4}\)

B) \(\frac{1}{3}\)

C) \(\frac{4}{9}\)

D) \(\frac{5}{11}\)

E) \(\frac{1}{2}\)

(Lets consider the SIMPLE CASE first) the probability of Juliet winning the game considering Romeo looses in first turn and Juliet wins in second turn was \(\frac{5}{6}*\frac{1}{6}\) = \(\frac{5}{36}\)

But what if Juliet does not win in the her first turn but wins in her second turn. Then we have a scenerio where Romeo looses, Juliet also looses, again Romeo looses & finally Juliet wins which translates into \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\)

The total probability that juliet wins in either of her first two chances will be sum of the individual probabilities of she either winning in her first or second chance.

= \(\frac{5}{6}*\frac{1}{6}\)+ \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\)

Similarly:-

The total probability that juliet wins in either of the first THREE chances is

= \(\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6} *\frac{5}{6}*\frac{1}{6}\)

We can continue like this towards infinity & if we take out \(\frac{5}{6}*\frac{1}{6}\) from this series, we realise that this is a geometric progression with common ratio, r = \(\frac{5}{6}*\frac{5}{6}\)

If we add the probabilities of all individual possible events... we have....

\(\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\) + \(\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6} *\frac{5}{6}*\frac{1}{6}\) + ..... so on

Note the formula for a sum of GP is:-

Attachment:

GP sum.jpg [ 2.28 KiB | Viewed 915 times ]
Since common ratio of this GP is less than one then \(r^n\) (where n is infinite) means \(r^n\) will become very minute and we can ignore it.

Attachment:

Expo.jpg [ 19.49 KiB | Viewed 912 times ]
Therefore substituting the values of a = \(\frac{5}{6}*\frac{1}{6}\) and \(r^n\) = 0 and r = \(\frac{5}{6}*\frac{5}{6}\) we get:-

\(\frac{5}{6}*\frac{1}{6}\)*\(\frac{36}{11}\) = \(\frac{5}{11}\) (Ans)

Hence option D is the correct option.

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Manish

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