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Bas_tet [7]
3 years ago
5

Suppose P(E) = 81%; P(E and F) = 9%; and P(E or F) = 65%,

Mathematics
1 answer:
PIT_PIT [208]3 years ago
4 0

Using Venn probabilities, it is found that P(F) = -0.07, which is not a probability, as it is a negative value.

------------------

  • We have two events, E and F.
  • Their probabilities are: P(E) = 0.81, P(E \cap F) = 0.09, P(E \cup F) = 0.65
  • The probabilities are related by the following equation:

P(E \cap F) = P(E) + P(F) - P(E \cup F)

  • Replacing into the equation, we can find P(F).

0.09 = 0.81 + P(F) - 0.65

0.16 + P(F) = 0.09

P(F) = -0.07

P(F) = -0.07, which is not a probability, as it is a negative value.

A similar problem is given at brainly.com/question/21421475

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Answer:

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Step-by-step explanation:

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then integrated between

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∫∫f (y1, y2) dy1*dy2 =  ∫∫k(1 − y2) dy1*dy2 = k  ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k  ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)-  (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

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b)

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)=  9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

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