Two fair dice, each with faces numbered $$1$$,$$2$$,$$3$$,$$4$$,$$5$$ and $$6$$, are rolled together and the sum of thenumbers on the faces is observed. This process is repeated till the sum is either a prime number or aperfect square. Suppose the sum turns out to be a perfect square before it turns out to be a primenumber. If 𝑝 is the probability that this perfect square is an odd number, then the value of $$14$$𝑝 is

Question

Two fair dice, each with faces numbered $$1$$,$$2$$,$$3$$,$$4$$,$$5$$ and $$6$$, are rolled together and the sum of thenumbers on the faces is observed. This process is repeated till the sum is either a prime number or aperfect square. Suppose the sum turns out to be a perfect square before it turns out to be a primenumber. If 𝑝 is the probability that this perfect square is an odd number, then the value of $$14$$𝑝 is

Infinity Learn · Accepted Answer

$$Prime/$$ Perfect square  Prime →$$2$$,$$3$$,$$5$$,$$7$$,$$11(1$$,$$1)$$,(1$$,$$2),(2$$,$$1),(1$$,$$4),(4$$,$$1),(2$$,$$3),(3$$,$$2)(6$$,$$1),(5$$,$$2),(4$$,$$3),(1$$,$$6),(2$$,$$5),(3$$,$$4),(5$$,$$6),(6$$,$$5) Square $$=4$$,$$9$$→(1$$,$$3),(3$$,$$1),(2$$,$$2),(6$$,$$3),(3$$,$$6),(5$$,$$4),(4$$,$$5) (Perfect$$ square before prime $$)=736+1536736+15362736+$$…… P $$(Perfect$$ square is $$odd) $$=436+1536436+15362436+$$…….$$=4/361$$−$$15367/361$$−$$1536=47$$⇒$$14p=47$$×$$14=8$$

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