Question:
Morgan is playing a board game that requires three standard dice to be thrown at one time. Each die has six sides, with one of the numbers 1 through 6 on each side. She has one throw of the dice left, and she needs a 17 to win the game. What is the probability that Morgan wins the game (order matters)?
Answer:
1/72
Step-by-step explanation:
<em>Morgan can roll a 17 in 3 different ways. The first way is if the first die comes up 5, the second die comes up 6, and the third die comes up 6. The second way is if the first die comes up 6, the second die comes up 5, and the third die comes up 6. The third way is if the first die comes up 6, the second die comes up 6, and the third die comes up 5. For each way, the probability of it occurring is 1/6 x 1/6 x 1/6 = 1/216. Therefore, since there are 3 different ways to roll a 17, the probability that Morgan rolls a 17 and wins the game is 1/216 + 1/216 + 1/216 = 3/216 = 1/72</em>
<em>I had this same question on my test!</em>
<em>Hope this helped! Good Luck! ~LILZ</em>
Yk I really don’t know what to do but I’m just saying that I’m going to be gone for a while
There are various ways in which to do this problem. I'd suggest
converting 5 3/4 into an improper fraction and then dividing that improper fraction by 4:
20+3 23 1 23
------- divided by 4 is --------- * ----- = ------ (answer)
4 4 4 16
I can not see your ferris wheel and so I can not answer it.
Heya user
we have
7x - 6y and x + y
according to the question
( 7x - 6y ) + ( x + y )
→ 7x - 6 y +x + y
→ 8x - 5y
Hope This helps you
