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One satellite is scheduled to be launched from Cape Canaveral in Florida, and another launching is scheduled for Vandenberg Air Force Base in California. Let A denote the event that the Vandenberg launch goes off on schedule, and let B represent the event that the Cape Canaveral launch goes off on schedule. If A and B are independent events with P(A) > P(B), P(A ∪B) = 0.626, and P(A ∩B) =0.144.

Required:
Determine the values of P(A) and P(B).

Respuesta :

MrRoyal

Answer:

[tex]P(A) = 0.45[/tex]

[tex]P(B) = 0.32[/tex]

Step-by-step explanation:

Given

[tex]P(A) > P(B)[/tex]

[tex]P(A\ u\ B) = 0.626[/tex]

[tex]P(A\ n\ B) = 0.144[/tex]

Required

Find P(A) and P(B)

We have that:

[tex]P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)[/tex] --- (1)

and

[tex]P(A\ n\ B) = P(A) * P(B)[/tex] --- (2)

The equations become:

[tex]P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)[/tex] --- (1)

[tex]0.626 = P(A) + P(B) - 0.144[/tex]

Collect like terms

[tex]P(A) + P(B) = 0.626 + 0.144[/tex]

[tex]P(A) + P(B) = 0.770[/tex]

Make P(A) the subject

[tex]P(A) = 0.770 - P(B)[/tex]

[tex]P(A\ n\ B) = P(A) * P(B)[/tex] --- (2)

[tex]0.144 = P(A) * P(B)[/tex]

[tex]P(A) * P(B) = 0.144[/tex]

Substitute: [tex]P(A) = 0.770 - P(B)[/tex]

[tex][0.770 - P(B)] * P(B) = 0.144[/tex]

Open bracket

[tex]0.770P(B) - P(B)^2 = 0.144[/tex]

Represent P(B) with x

[tex]0.770x - x^2 = 0.144[/tex]

Rewrite as:

[tex]x^2 - 0.770x + 0.144 = 0[/tex]

Expand

[tex]x^2 - 0.45x - 0.32x + 0.144 = 0[/tex]

Factorize:

[tex]x[x - 0.45] - 0.32[x - 0.45]= 0[/tex]

Factor out x - 0.45

[tex][x - 0.32][x - 0.45]= 0[/tex]

Split

[tex]x - 0.32= 0 \ or\ x - 0.45= 0[/tex]

Solve for x

[tex]x = 0.32\ or\ x = 0.45[/tex]

Recall that:

[tex]P(B) = x[/tex]

So, we have:

[tex]P(B) = 0.32 \ or \ P(B) = 0.45[/tex]

Recall that:

[tex]P(A) = 0.770 - P(B)[/tex]

So, we have:

[tex]P(A) = 0.770 - 0.32 \ or\ P(A) =0.770 - 0.45[/tex]

[tex]P(A) = 0.45 \ or\ P(A) =0.32[/tex]

Since:

[tex]P(A) > P(B)[/tex]

Then:

[tex]P(A) = 0.45[/tex]

[tex]P(B) = 0.32[/tex]

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