Answer:
Pr(the sum of the numbers rolled is either a multiple of 3 or an even number)=[tex]\frac{2}{3}[/tex]
Step-by-step explanation:
Let A be the event "sum of numbers is multiple of 3"
and B be the event "sum is an even number".
As our dice has six sides, so the sample space of two dices will be of 36 ordered pairs.
|sample space | = 36
Out of which 11 pairs have the sum multiple of 3 and 18 pairs having sum even.
So Pr(A)= [tex]\frac{11}{36}[/tex]
and Pr(B)= [tex]\frac{18}{36}[/tex]
and Pr(A∩B) = [tex]\frac{5}{36}[/tex], as 5 pairs are common between A and B.
So now Pr(A or B)= Pr(A∪B)
= Pr(A)+Pr(B) - Pr(A∩B)
= [tex]\frac{11}{36}[/tex] + [tex]\frac{18}{36}[/tex] - [tex]\frac{5}{36}[/tex]
= [tex]\frac{24}{36}[/tex]
= [tex]\frac{2}{3}[/tex]
Answer:
2/3 and NOT mutually exclusive
Step-by-step explanation:
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