The probability of having a sum of either 6 or 10 is [tex]\frac{2}{9}[/tex]
The sample space has the size [tex]|S|=36[/tex]
Let
[tex]A=\text{the event of having a sum of 6}\\B=\text{the event of having a sum of 10}[/tex]
then
[tex]A=\{(1,5),(2,4),(3,3),(4,2),(5,1)\}\\|A|=5[/tex]
and
[tex]B=\{(4,6),(5,5),(6,4)\}\\|B|=3[/tex]
then, the probability of having a sum of either 6 or 10 will be
[tex]P(A\cup B)=P(A)+P(B)[/tex]
since [tex]A[/tex] and [tex]B[/tex] are mutually exclusive events. Substituting
[tex]P(A\cup B)=P(A)+P(B)\\=\frac{|A|}{|S|}+\frac{|B|}{|S|}\\=\frac{2}{9}[/tex]
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When two balanced dice are rolled, there are 36 possible outcomes. To find the probability that the sum of the numbers on the dice is 6 or 10, count the favorable outcomes and divide it by the total outcomes.
Explanation:When two balanced dice are rolled, there are 36 possible outcomes. To find the probability that the sum of the numbers on the dice is 6 or 10, we need to count the favorable outcomes and divide it by the total outcomes.
For the sum of 6, there are 5 possible outcomes: (1,5), (2,4), (3,3), (4,2), (5,1). For the sum of 10, there are 3 possible outcomes: (4,6), (5,5), (6,4). Therefore, there are 8 favorable outcomes out of 36 total outcomes.
The probability is calculated as:
Probability = Favorable outcomes / Total outcomes = 8 / 36 = 2 / 9
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WILL SOMEONE HELP ME ANSWER THIS PLEASE!!!!
Luca and William are playing a video game and they have scored a total of 20,000 points. Luca scored 4,000 more points than William. If you let l= the number of points that Luca scored, and w= the number of points that William scored, then the problem can be represented by the system:
l+w=20,000 and l=w+4,000
Graph the system. How many points did each boy score?
My possible answers are:
a. William = 8,000 and Luca = 12,000
b. William = 12,000 and Luca = 8,000
d. William = 4,000 and Luca = 16,000
l+w=20000
l=w+4000
w+4000+w=20000
2w+4000=20000
2w=16000
w=8000
l = 8000 +4000 = 12000
William scored 8000, Luca scored 12000