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>
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3+x
This is a trick question. Unless we know how much she’s eaten for lunch, we can’t figure out how much she ate.
Answer:
47
Step-by-step explanation:
There are total 52 cards.
Certain number of cards are lost.
When divided by 4, leaves 3 cards remain ; 51, 47, 43, 39, . . .
When divided by 5, leaves 2 cards remain : 47, 42, 37, 32, . . .
When divided by 3, leaves 2 cars remain : 50, 47, 44, 41, . . .
The common number among the three is 47.
I can’t see it it too blurry