Question

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).

Answer 1

Answer:

$P(A) = 0.45$

$P(B) = 0.32$

Step-by-step explanation:

Given

$P(A) > P(B)$

$P(A\ u\ B) = 0.626$

$P(A\ n\ B) = 0.144$

Required

Find P(A) and P(B)

We have that:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$ --- (1)

and

$P(A\ n\ B) = P(A) * P(B)$ --- (2)

The equations become:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$ --- (1)

$0.626 = P(A) + P(B) - 0.144$

Collect like terms

$P(A) + P(B) = 0.626 + 0.144$

$P(A) + P(B) = 0.770$

Make P(A) the subject

$P(A) = 0.770 - P(B)$

$P(A\ n\ B) = P(A) * P(B)$ --- (2)

$0.144 = P(A) * P(B)$

$P(A) * P(B) = 0.144$

Substitute: $P(A) = 0.770 - P(B)$

$[0.770 - P(B)] * P(B) = 0.144$

Open bracket

$0.770P(B) - P(B)^2 = 0.144$

Represent P(B) with x

$0.770x - x^2 = 0.144$

Rewrite as:

$x^2 - 0.770x + 0.144 = 0$

Expand

$x^2 - 0.45x - 0.32x + 0.144 = 0$

Factorize:

$x[x - 0.45] - 0.32[x - 0.45]= 0$

Factor out x - 0.45

$[x - 0.32][x - 0.45]= 0$

Split

$x - 0.32= 0 \ or\ x - 0.45= 0$

Solve for x

$x = 0.32\ or\ x = 0.45$

Recall that:

$P(B) = x$

So, we have:

$P(B) = 0.32 \ or \ P(B) = 0.45$

Recall that:

$P(A) = 0.770 - P(B)$

So, we have:

$P(A) = 0.770 - 0.32 \ or\ P(A) =0.770 - 0.45$

$P(A) = 0.45 \ or\ P(A) =0.32$

Since:

$P(A) > P(B)$

Then:

$P(A) = 0.45$

$P(B) = 0.32$