Question

Question #9c722

Answer 1

First we will solve for the probability of solving for getting the sum of six with a pair of fair dice, we would need to specify that the dice are fair, as you can not always assume such.

we thus know that throwing either number on one die has equal probability. now if we throw 2 die, lets look at all the results we get:

`1&1 , 1&2, 1&3, 1&4, color(blue)(1&5), 1&6`
`2&1, 2&2, 2&3, color(blue)(2&4), 2&5, 2&6`
`3&1, 3&2, color(blue)(3&3), 3&4, 3&5, 3&6`
`4&1, color(blue)(4&2), 4&3, 4&4, 4&5, 4&6`
`color(blue)(5&1), 5&2, 5&3, 5&4, 5&5, 5&6`
`6&1, 6&2, 6&3, 6&4, 6&5, 6&6`

Now we add up the total amount of results:

Total results `= 36`

then add up the amount that sum to `6` when you add their vaules:

sum to 6 `= 5`

Knowing that all results are equally likely, we can conclude that to roll two fair dice, and get a sum of 6 from the two values is:

p(sum to 6) `=5/36`

Now to answer the question of having this result happen in two consecutive attempts.

we know that the throwing of each trial is independent, and thus:

`P(A nn B) = P(A).P(B)`

Thus we can set `P(A) =` Throwing a sum of six
and `P(B) =` throwing a sum of six

Which now we can calculate to fin that having both happen is:

`P(A).P(B) = (5/36).(5/36) = (5/36)^2 ~~ 0.0193`

or can say `~~2%`