In a certain soccer tournament you play 6 games. In every match you get 3 points if you win, 1 point for a draw and 0 points if you lose. For each match the probability you win is 0.5, the probability you draw is 0.1 and the probability you lose is 0.4, independently of the results of all other matches. What is the probability you Önish?

0.430625 = 0.431

Step-by-step explanation:

Let W represent winning, D represent a draw and L represent a loss.

12+ points can be garnered in each of the following ways.

6W 0D 0L

5W 1D 0L

5W 0D 1L

4W 2D 0L

4W 1D 1L

4W 0D 2L

3W 3D 0L

The probability of getting 12+ points is the sum of all these 7 probabilities.

Knowing that P(W) = 0.5

P(D) = 0.1

P(L) = 0.4

P(6W 0D 0L) = [6!/(6!0!0!)] 0.5⁶ 0.1⁰ 0.4⁰ = 0.015625

P(5W 1D 0L) = [6!/(5!1!0!)] 0.5⁵ 0.1¹ 0.4⁰ = 0.01875

P(5W 0D 1L) = [6!/(5!0!1!)] 0.5⁵ 0.1⁰ 0.4¹ = 0.075

P(4W 2D 0L) = [6!/(4!2!0!)] 0.5⁴ 0.1² 0.4⁰ = 0.09375

P(4W 1D 1L) = [6!/(4!1!1!)] 0.5⁴ 0.1¹ 0.4¹ = 0.075

P(4W 0D 2L) = [6!/(4!0!2!)] 0.5⁴ 0.1⁰ 0.4² = 0.15

P(3W 3D 0L) = [6!/(3!3!0!)] 0.5³ 0.1³ 0.4⁰ = 0.0025

The probability of getting 12+ points = 0.015625 + 0.01875 + 0.075 + 0.09375 + 0.075 + 0.15 + 0.0025 = 0.430625

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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