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Question

Mathematics Question on Probability

A and B alternately throw a pair of dice. A wins if he throws 6 before B throws 7; and B wins if he throws 7 before Athrows 6. What are their respective chances of winning, if A throws the dice first?

A

1316,3116\frac{13}{16},\frac{31}{16}

B

3061,3161\frac{30}{61},\frac{31}{61}

C

3161,4161\frac{31}{61},\frac{41}{61}

D

3861,2861\frac{38}{61},\frac{28}{61}

Answer

3061,3161\frac{30}{61},\frac{31}{61}

Explanation

Solution

The probability that A throws ′6′'6' =(536)=(\frac{5}{36}) and

The probability of A not getting ′6′'6' =1−536=3136=1-\frac{5}{36}=\frac{31}{36}

The probability that B throws ′7′'7' =636=\frac{6}{36}

The probability of A winning the game is given by,

= (In the first throw if A wins) or (In the 1st throw A looses, B also looses and in the 2nd throw A wins) or ( A and B looses in the 1st and 2nd throw's and A wins in the 3rd throw .....it goes on like that)

=\frac{5}{36}$$+\frac{31}{36}.\frac{30}{36}.\frac{5}{36}$$+\frac{31}{36}.\frac{30}{36}.\frac{31}{36}.\frac{30}{36}$$\frac{5}{36}$$+.......

It is in arithmetic progression,

=(\frac{5}{36})$$(1-\frac{\frac{1}{31}}{36}$$.\frac{30}{36}$$)

=(\frac{5}{36})$$(\frac{1296}{1296-930})

=(\frac{5}{36})$$(\frac{1296}{366})

=\frac{5×6}{61}$$=\frac{30}{61}

Probability of B winning the game is given by,

=1−3061=1-\frac{30}{61}

=3161=\frac{31}{61}

Hence, option B is the correct answer.The correct option is (B): 3061,3161\frac{30}{61},\frac{31}{61}