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Sagot :
Using the probability concept, it is found that there is a
a) 0.2 = 20% probability that 3 is obtained.
b) 0.3333 = 33.33% probability that 6 is obtained.
c) 0.6667 = 66.67% probability of a number greater than 3.
d) 0.6667 = 66.67% probability of an even number.
A probability is the number of desired outcomes divided by the number of total outcomes.
Item a:
There are 5 numbers greater than 1, one of which is 3, hence:
[tex]p = \frac{1}{5} = 0.2[/tex]
0.2 = 20% probability that 3 is obtained.
Item b:
There are 3 numbers greater than 3, one of which is 6, hence:
[tex]p = \frac{1}{3} = 0.3333[/tex]
0.3333 = 33.33% probability that 6 is obtained.
Item c:
There are 3 even numbers, two of which are greater than 3, hence:
[tex]p = \frac{2}{3} = 0.6667[/tex]
0.6667 = 66.67% probability of a number greater than 3.
Item d:
There are 3 numbers greater than 3, two of which are even, hence:
[tex]p = \frac{2}{3} = 0.6667[/tex]
0.6667 = 66.67% probability of an even number.
A similar problem is given at https://brainly.com/question/25667645
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