From my course i understand that i have actually N^n Possibilities = 6^2 = 36.6 = sides of dice and 2 = number of dice rolls

Kcurrently i calculated the possibilities for all also values. P = Probability and x = value:

P(x=2)=1/36

P(x=4)=3/36

P(x=6)=5/36

P(x=8)=5/36

P(x=10)=3/36

P(x=12)=1/36

Sums of all P should be the probcapacity i am looking for:

P(x=2)+...+P(x=12)=18/36=1/2=50%

Is this correct?Is tbelow any kind of way to fix this "easier" or by making use of a formula. Since the next job is for 3 dice rolls :-)

Thanks a lot.

You are watching: What is the probability that the sum of the numbers on two dice is even when they are rolled?

probcapability
Share
Cite
Follow
edited Oct 22 "17 at 12:51 N. F. Taussig
asked Oct 22 "17 at 12:48 hukachakahukachaka
\$endgroup\$
1

6

See more: When Is An Atom Unlikely To React And Form Chemical Bonds? ?

\$egingroup\$
Yes, there is a less complicated means. It is referred to as principle of deferred decisions. It goes like this. Condition on the parity of the first die throw. It is either 0 or 1. Now, the parity determines uniquely what the parity of the second die throw should be to have actually the resultant amount have parity 0. Therefore,

eginalignP( exteven sum) &= P( extalso amount, first was even + P( exteven amount, initially was odd) \&= P( exteven)P( extfirst was even) + P( exteven)P( extinitially was odd)endalign

Now, note that \$P( extfirst was even) = P( extsecond is even) = 1/2\$. Similarly, \$P( exteven) = P( extsecond is odd) = 1/2\$. Hence, we have \$P( extalso sum) = 1/2left(P( extfirst was even) + P( extinitially was odd) ight) = 1/2left(1 ight) = 1/2\$.

ADDENDUM: The principle of deferred decisions helps deal with a lot harder comparable problems. Consider this problem - you throw a die 10 times. What is the probcapability that the sum is a multiple of 6?

eginalignP( extsum \$equiv 0 mod 6\$) &= sum_i=0^5 P( extamount \$equiv 0 mod 6\$| extprevious amount was \$equiv i mod 6\$) cdot P( extprevious amount was \$equiv i mod 6\$)endalign

Keep in mind that \$P( extsum \$equiv 0 mod 6\$| extprevious sum was \$equiv i mod 6\$) = 1/6\$ bereason tright here is just one particular die value that will certainly give us \$(6-i) mod 6\$. Thus, \$P( extsum \$equiv 0 mod 6\$) = 1/6left(sum_i=0^5 P( extprevious amount was \$equiv i mod 6\$) ight) = 1/6left(1 ight) = 1/6.\$