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>
Y = 30 - 2(2)
y= 30 - 4
y = 26
7033/10000 is the answer you're looking for
___________________________
◆ AREA RELATED TO CIRCLES ◆ ___________________________
As shown in the figure ,
Radius of circle = 5 cm
Side of square = 2 × (Radius of inscribed circle)
Side of square = 2 × 5 cm
Side of square = 10 cm = a
___________________________
Now ,
Area of shaded region = (Area of square) - ( Area of inscribed circle )
Area of shaded region =

Area of shaded region =

Area of shaded region = 100 - 78.571