Meyer rolls two fair, ordinary dice with the numbers on their sides. What is the probability that AT LEAST one of the dice shows a square number?

SmartMathMan Jan 18, 2018

#1**+2 **

on each dice there are 2 square numbers (1 and 4). so for each dice there is 1 in three chance that there will be a square number if rolled, or a two in three chance it will not be rolled. rolling two dice at once is the same as rolling one dice twice so the probability the probabilty that she DOESN'T roll a square number is:

\(\frac{2}{3}\times\frac{2}{3}\)

which is, rounded off, 0.4445 or 44.45%

now the chances that she DOES roll a square number is 1 - 0.4445 which is 0.5556 or 55.56%

so the chances that she does roll a square number is 55.56% , which means that her rolling a square number is slightly more likely than not!

Scruffy23. Jan 18, 2018

#1**+2 **

Best Answer

on each dice there are 2 square numbers (1 and 4). so for each dice there is 1 in three chance that there will be a square number if rolled, or a two in three chance it will not be rolled. rolling two dice at once is the same as rolling one dice twice so the probability the probabilty that she DOESN'T roll a square number is:

\(\frac{2}{3}\times\frac{2}{3}\)

which is, rounded off, 0.4445 or 44.45%

now the chances that she DOES roll a square number is 1 - 0.4445 which is 0.5556 or 55.56%

so the chances that she does roll a square number is 55.56% , which means that her rolling a square number is slightly more likely than not!

Scruffy23. Jan 18, 2018