A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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?
anonymous
 one year ago
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?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There 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 366 ways to roll not 7. \[\left(\frac{6}{36}\right)^3 \left(\frac{30}{36}\right)^2=\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2=\frac{25}{7776}=0.00321502 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, so 6! for total outcome

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i was thinking this way there are 6^5 outcomes rolling two dice 5 times 6 of those sum up to 7 in one roll

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0couldn't complete my thinking here!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0so i need some combinatorics to finish up this seems that way with your result

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0@robtobey has the same thing although went a different approach, but this probabilities are like that can be done differently :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When I was much younger I calculated the player's winning percent playing craps in a Nevada gambling hall.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0what was the the percentage of winning the gamble?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i need to work on this probabilities more! sometimes i get them sometimes i'm off the picture. lack of practice

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Close 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.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i see, that is not a good percentage to play the game hehehe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.