List down numbers when the player loses.(lesser cases)

1 with any number - he wins

(2-3) and (2-5) - 4 cases he loses (2*2 = the 2 numbers can appear in either 2 die)

(3-4) (3-5) - 2*2 = 4 cases (numbers can appear on either die)

(4-5) and (4-6) = 4 cases (numbers can appear on either die)

(5-6) = 2 cases. (numbers can appear on either die)

Total cases = 36

Losing cases = 14

Winning cases = 22

P(winning) = 22/36 = 11/18

there are total 36 different ways two dices can be thrown
when 1 is thrown on one dice, the other dice can be any number, hence 6 ways
when 2 is thrown on a dice, the other can be only even numbers (2,4,6) and reversed dices (can be 1) 4 ways
with 3 is thrown on a dice, the other can be 3,6 and rev.dices (can be 1,6) 4 ways
with 4 on a dice, the other can be 4 and rev.dices (can be 1,4) 3 ways
with 5 only 5 and rev.dices (can be 1,5) 3 ways
with 6 only 6 and rev.dices (can be 1,6) 3 ways
subtotal 6+4+4+3+3+3=22

SO 22/36 is 11/18

venmic wrote:A certain dice game can only be won if, when a player throws
two fair six-sided dice, the number showing on one of the dice
is a multiple of the number showing on the other. What is the
probability that a player wins this game?

[spoiler]11/18[/spoiler]

If 1st throw comes - 1 2nd throw 1-6 all are multiple of 1
1st throw - 2 2nd throw 2,4,6 multiple of 2
1st throw - 3 2nd throw 3,6 multiple of 3
and reverse result as 1st throw 1-6 and 2nd throw 1 is also possible.
For 4,5,6 there are no muliples.

Sol: (1/6*1+ 1/6*3/6 + 1/6* 2/6)+ (1*1/6+ 3/6*1/6+ 2/6*1/6)

Ans: 11/18

Just doing it by finding the number of co-primes

if 2 - 3,5
if 3 - 2,4,5
if 4 - 3,5,6
if 5 - 2,3,4,6
if 6 - 4,5

total = 36 - 14 = 22
=> propability = 22/36 = 11/18

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