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: $frac(^1C_1)(^1C_1)(^5C_1)+(^1C_1)(^5C_1)(^1C_1)+(^5C_1)(^1C_1)(^1C_1)(^6C_1)^3$)

As such your answer is simply: $frac1+15216=frac227$


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...

$$ extProbability of event = frac16216=frac 227$$



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