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 = $frac 16 imes frac 16 imes frac 56 + frac 16 imes frac 56 imes frac 16 + frac 56 imes frac 16 imes frac 16$
=$frac 5216 + frac 5216 + frac 5216$
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 = $frac 16 imes frac 16 imes frac 16$
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!Please be certain to answer the question. Provide details and share your research!
But avoid …Asking for help, clarification, or responding to other answers.Making statements based upon opinion; ago them up through references or personal experience.
Use nlinux.orgJax to format equations. nlinux.orgJax reference.
See more: Why Is Much Ado About Nothing A Comedy Much Ado About Nothing
To learn even more, view our tips on writing excellent answers.
Article Your Answer Discard
Not the answer you're looking for? Browse various other concerns tagged probcapability combinatorics or ask your very own question.
website design / logo © 2021 Stack Exreadjust Inc; user contributions licensed under cc by-sa. rev2021.9.2.40133