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kiruha [24]
3 years ago
11

A sample space consists of five simple events, E1, E2, E3, E4, and E5. a If P(E1) = P(E2) = 0.15, P(E3) = 0.4, and P(E4) = 2P(E5

), find the probabilities of E4 and E5.
b If P(E1) = 3P(E2) = 0.3, find the probabilities of the remaining simple events if you know that the remaining simple events are equally probable.
Mathematics
1 answer:
kvv77 [185]3 years ago
7 0

Answer:

Step-by-step explanation:

Given that

A sample space consists of five simple events, E1, E2, E3, E4, and E5.

a If P(E1) = P(E2) = 0.15,

P(E3) = 0.4, and P(E4) = 2P(E5),

We know that total probability =1

i.e. sum of probabilities of All Eis would be 1

0.15+0.15+0.4+2x+x=1

where x = P(E5)

Solving for x we get

x=0.30

So P(E4) = 0.2 and P(E5) = 0.1

-----

b) Let P(E3) =P(E4) = P(E5) = y

then we have

total probability = 0.3+0.1+3y =1\\3y =0.6\\y = 0.2

Probability of remain are 0.2 each.

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Ashley and Christopher want to camp on the other side of a lake at point B. They are both going to kayak to the campsite from th
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Applying the midsegment theorem, the distant Ashley will kayak more than Christopher is: 0.5 mi

<em><u>Recall:</u></em>

  • The midsegment of a triangle joins two sides of a triangle at their midpoint.
  • The third side is the base of the triangle.
  • Triangles have three midsegments.
  • Based on the midsegment theorem, the length of the midsegment = ½(third side).

The picture given shows a triangle with two midsegments: AB and BC

AB = the distance Ashley will kayak.

BC = the distance Christopher will kayak.

XY = 5 mi (base)

XZ = 6 mi (base)

Applying the midsegment theorem, find AB and BC.

AB = ½(XZ)

  • Substitute

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Thus, the distance Ashley will kayak more than Christopher = AB - BC

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