A person throws a dice while he gets a number greater than $$2$$. The probability that he gets a $$6$$ in the last throw is

Question

Infinity Learn · Accepted Answer

Let $$E1$$, be the event that six shows when a dice is thrown. Let $$E2$$ be the event that number less than or equal to $$2$$ shows when a dice is thrown.$$PE1=1/6$$ and $$PE2=2/6=1/3E$$ $$=$$ the event that six turns up in the last throw $$=$$ the event that $$E1$$ happens in the all previous throws and $$E2$$ happens in the last throw $$P(E)=PE1E2$$ or $$E1E1E2$$ or $$E1E1E1E2$$…..$$=PE1$$⋅$$PE2+PE1$$⋅$$PE1$$⋅$$PE2+PE1$$⋅$$PE1$$⋅$$PE1$$⋅$$PE2+$$…$$=13$$⋅$$16+132$$⋅$$16+133$$⋅$$16+$$…$$P(E)=13$$⋅$$16$$×$$11$$−$$1/3$$⇒$$P(E)=112$$

A person throws a dice while he gets a number greater than 2. The probability that he gets a 6 in the last throw is

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