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Find the probability that the sum of two randomly chosen numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is no greater than 10.

Sagot :

Using it's concept, it is found that there is a 0.64 = 64% probability that the sum of two randomly chosen numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is no greater than 10.

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem:

  • Two numbers that can be repeated chosen from a set of 10 numbers, thus, in total, there are [tex]10^2 = 100[/tex] outcomes.

For a sum no greater than 10, we have that:

  • 0 can be added with all the 10 numbers, as can 1.
  • 2 can be added with 9 of them, bar 9.
  • 3 can be added with 8 of them.
  • 4 can be added with 7 of them.
  • 5 can be added with 6 of them.
  • 6 can be added with 5 numbers, 7 with 4, 8 with 3, and 9 with 2.

Hence, the number of desired outcomes is:

[tex]D = 10(2) + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 64[/tex]

The probability is:

[tex]p = \frac{D}{T} = \frac{64}{100} = 0.64[/tex]

0.64 = 64% probability that the sum of two randomly chosen numbers from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} is no greater than 10.

A similar problem is given at https://brainly.com/question/25401798

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