Answer:
a) P(B'|A) = 0.042
b) P(B|A') = 0.625
Step-by-step explanation:
Given that:
80% of the light aircraft that disappear while in flight in a certain country are subsequently discovered
Of the aircraft that are discovered, 63% have an emergency locator,
whereas 89% of the aircraft not discovered do not have such a locator.
From the given information; it is suitable we define the events in order to calculate the probabilities.
So, Let :
A = Locator
B = Discovered
A' = No Locator
B' = No Discovered
So; P(B) = 0.8
P(B') = 1 - P(B)
P(B') = 1- 0.8
P(B') = 0.2
P(A|B) = 0.63
P(A'|B) = 1 - P(A|B)
P(A'|B) = 1- 0.63
P(A'|B) = 0.37
P(A'|B') = 0.89
P(A|B') = 1 - P(A'|B')
P(A|B') = 1 - 0.89
P(A|B') = 0.11
Also;
P(B ∩ A) = P(A|B) P(B)
P(B ∩ A) = 0.63 × 0.8
P(B ∩ A) = 0.504
P(B ∩ A') = P(A'|B) P(B)
P(B ∩ A') = 0.37 × 0.8
P(B ∩ A') = 0.296
P(B' ∩ A) = P(A|B') P(B')
P(B' ∩ A) = 0.11 × 0.2
P(B' ∩ A) = 0.022
P(B' ∩ A') = P(A'|B') P(B')
P(B' ∩ A') = 0.89 × 0.2
P(B' ∩ A') = 0.178
Similarly:
P(A) = P(B ∩ A ) + P(B' ∩ A)
P(A) = 0.504 + 0.022
P(A) = 0.526
P(A') = 1 - P(A)
P(A') = 1 - 0.526
P(A') = 0.474
The probability that it will not be discovered given that it has an emergency locator is,
P(B'|A) = P(B' ∩ A)/P(A)
P(B'|A) = 0.022/0.526
P(B'|A) = 0.042
(b) If it does not have an emergency locator, what is the probability that it will be discovered?
The probability that it will be discovered given that it does not have an emergency locator is:
P(B|A') = P(B ∩ A')/P(A')
P(B|A') = 0.296/0.474
P(B|A') = 0.625
The answer is 7 R 4! hoped this helps!
Whatever your color eyes are
hope this helps
<span>The answer should be 14t - 2m2 - 5/2 , if I am correct</span>
9.252.063 rounded to the nearest hundredth is 9.252.060
