Question

A has 3 shares in a lottery in which there are 3 prizes and 6 blanks; B has 1 share in a lottery in which there is 1 prize and 2 blanks: show that A's chance of success is to B's as 16 to 7.

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Solution

A may draw 3 prizes in 1 way;

He may draw 2 prizes and 1 blank in 3.21.2×6 ways ;

He may draw 1 prize and 2 blanks in 3×6.51.2 ways;

The sum of these numbers is 64, which is the number of ways in which A can win a prize. Also he can draw 3 tickets in 9.8.71.2.3, or 84 ways;

Therefore A's chance of success =6484=1621.

B's chance of success is clearly =13;

Therefore A's chance : B's chance =1621:13

=16:7.

He may draw 2 prizes and 1 blank in 3.21.2×6 ways ;

He may draw 1 prize and 2 blanks in 3×6.51.2 ways;

The sum of these numbers is 64, which is the number of ways in which A can win a prize. Also he can draw 3 tickets in 9.8.71.2.3, or 84 ways;

Therefore A's chance of success =6484=1621.

B's chance of success is clearly =13;

Therefore A's chance : B's chance =1621:13

=16:7.

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