I think that the equation would certainly be \$frac16 + frac16 + frac16 = frac12\$, bereason each dice can roll a 4, and also if I sum them up, I"ll acquire 4 on each dice.

You are watching: What is the probability of rolling two six-sided dice and obtaining two 4s?

(The dice is fair, in case you were wondering) Required probcapability of at least two 4"s,

is the probcapability of obtaining all 3 4"s, i.e. \$frac16^3=frac1216\$, added to the probcapability of getting precisely 2 4"s, which is:\$frac15216\$

(Tright here are fifteenager methods to obtain 2 fours: \$441,414,144,442,424,244..\$ and also so on until \$446,464,644\$; another means to obtain the solution is making use of permutations and also combinations: \$frac(^1C_1)(^1C_1)(^5C_1)+(^1C_1)(^5C_1)(^1C_1)+(^5C_1)(^1C_1)(^1C_1)(^6C_1)^3\$) First point you need to take cases with 2 time 4"s out of 3 rolls.

Probability = \$frac 16 imes frac 16 imes frac 56 + frac 16 imes frac 56 imes frac 16 + frac 56 imes frac 16 imes frac 16\$

=\$frac 5216 + frac 5216 + frac 5216\$

=\$frac 15216\$

Second below at leastern is used. So might be tbelow is 4"s on all dice. So you need to take instances of 3 time 4"s.

Probcapacity = \$frac 16 imes frac 16 imes frac 16\$

=\$frac 1216\$

Total probability = \$frac 15216 + frac 1216\$ Total number of possibilities are \$6^3\$ considering that there are three dices.

Now, there is one way of gaining all \$4\$ in all dices \$geq\$ at least 2 4"s and \$15=35 select 1\$ methods of gaining two fours \$geq\$ at leastern 2 4"s. So...

\$\$ extProbability of event = frac16216=frac 227\$\$  Thanks for contributing an answer to nlinux.orgematics Stack Exchange!

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