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ozzi
2 years ago
13

Hey there! how are you? have you ate? make sure you grab something to eat. have you been sleeping well? try to get loots of slee

p cause you need it!
you're braver than you believe, and stronger than you seem, and smarter than you think. dont forget that YOU are so mf beautiful/handsome!! you are perfect exactly how you are. dont ever listen to someone who tells you otherwise. cause we are all different. love yourself because you are an amazing person and If you need to appreciate that. please remember you are loved by so much, and you matter. i might not know what your going through, or if your even going through something. but just know I'm so proud of you for continuing. you are so strong and i love you so do other people! whenever you find yourself doubting how far you can go, just remember how far you have come. remember everything you have faced..
love you have a great day/night.!
Mathematics
2 answers:
nikklg [1K]2 years ago
7 0
I really needed this. lol
andriy [413]2 years ago
6 0
Thanks, I didn’t realize how much I needed this. Hope you have a good day!
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If you roll two six-faced dice together, you will get 36 possible outcomes. 4 pts 1. List all possible outcomes of the experimen
bezimeni [28]

Given:

Two dice are rolled together.

Total number of possible outcomes.

To find:

The list of total possible outcomes.

The probability of getting a sum of 11 in these outcomes.

The probability of getting a sum less than or equal to 4.

The probability of getting a sum of 13 or more.

Solution:

If two dice are rolled together, then the total number of possible outcomes is 36 and list of total possible outcomes is

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sum of 11 in these outcomes = {(5,6),(6,5),(6,6)} = 3

The probability of getting a sum of 11 in these outcomes is

P(\text{sum=11})=\dfrac{3}{36}

P(\text{sum=11})=\dfrac{1}{12}

Therefore, the probability of getting a sum of 11 in these outcomes is \dfrac{1}{12}.

Sum less than or equal to 4 = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)} = 6

The probability of getting a sum less than or equal to 4 is

P({sum\leq 4})=\dfrac{6}{36}

P({sum\leq 4})=\dfrac{1}{6}

Therefore, the probability of getting a sum less than or equal to 4 is \dfrac{1}{6}.

Sum of 13 or more = empty set because maximum sum is 12.

The probability of getting a sum of 13 or more is

P(sum\geq 13)=\dfrac{0}{36}

P({sum\geq 13})=0

Therefore, the probability of getting a sum of 13 or more is 0.

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