Question

can someone explain?

two fair die are tossed 5 times, find probablity that sum of 7 will show on the first three tosses and will not show on the other two?

Answer 1

There are 6 ways for a seven to show on one roll of two dice.
2,5
5,2
3,4
4,3
6,1
1,6
There are 36 ways for the two dice to show a total of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
There are 36-6 ways to roll not 7.

\[\left(\frac{6}{36}\right)^3 \left(\frac{30}{36}\right)^2=\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2=\frac{25}{7776}=0.00321502 \]

Answer 2

oh, so 6! for total outcome

Answer 3

i was thinking this way
there are 6^5 outcomes rolling two dice 5 times
6 of those sum up to 7 in one roll

Answer 4

couldn't complete my thinking here!

Answer 5

so i need some combinatorics to finish up this
seems that way with your result

Answer 6

@lordofpens total outcomes is not 6!
rather it is 6^5
because there are 6 possibilities for one roll
and another 6 for the second roll
and third and so one
is total 6x6x6x6x6=6^5

Answer 7

@robtobey has the same thing
although went a different approach, but this probabilities are like that
can be done differently :)

Answer 8

When I was much younger I calculated the player's winning percent playing craps in a Nevada gambling hall.

Answer 9

what was the the percentage of winning the gamble?

Answer 10

i need to work on this probabilities more! sometimes i get them sometimes i'm off the picture. lack of practice

Answer 11

Close to even, but the house is still expected to win. Don't recall the exact number.
I believe the players winning percentage is near 48% overall.

Answer 12

i see, that is not a good percentage to play the game hehehe

Answer 13

If you never play, you will never lose.
Someone is paying for the Vegas hotels, shows and so forth.
I am no longer one of them.