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shyshy6184
22.02.2020 •
Mathematics
Let S be a sample space and E and F be events associated with S. Suppose that
Pr left parenthesis Upper E right parenthesis equals 0.6Pr(E)=0.6,
Pr left parenthesis Upper F right parenthesis equals 0.2Pr(F)=0.2
and
Pr left parenthesis Upper E intersect Upper F right parenthesis equals 0.1Pr(E∩F)=0.1.
Calculate the following probabilities.
(a) Pr left parenthesis E|F right parenthesisPr(E|F)
(b) Pr left parenthesis F|E right parenthesisPr(F|E)
(c) Pr left parenthesis E| Upper F prime right parenthesisPrE|F′
(d) Pr left parenthesis Upper E prime | Upper F prime right parenthesisPrE′|F′
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Ответ:
a) P(E|F) = 0.5
b) P(F|E) = 0.167
c) P(E|F') = 0.625
d) P(E′|F′) = 0.375
Step-by-step explanation:
P(E) = 0.6
P(F) = 0.2
P(E n F) = 0.1
a) P(E|F) = Probability of E occurring, given F has already occurred. It is given mathematically as
P(E|F) = [P(E n F)]/P(F) = 0.1/0.2 = 0.5
b) P(F|E) = Probability of F occurring, given E has already occurred. It is given mathematically as
P(F|E) = [P(E n F)]/P(E) = 0.1/0.6 = 0.167
c) P(E|F′) = Probability of E occurring, given F did not occur. It is given mathematically as
P(E|F') = [P(E n F')]/P(F')
But P(F') = 1 - P(F) = 1 - 0.2 = 0.8
P(E n F') = P(E) - P(E n F) = 0.6 - 0.1 = 0.5
P(E|F') = 0.5/0.8 = 0.625
d) P(E′|F′) = [P(E' n F')]/P(F')
P(F') = 0.8, P(Universal set) = P(U) = 1
P(E' n F') = P(U) - [P(E n F') + P(E' n F) + P(E n F) = 1 - (0.5 + 0.1 + 0.1) = 0.3
P(E′|F′) = 0.3/0.8 = 0.375
Ответ:
we are given expression:
and
.
in order to find it's greatest common factor, let us break it into smaller parts.
let us find greatest common factor of 54 and 27 first.
54 = 27 × 2
27 = 27 × 1
27 is the greatest common factor of 54 and 27.now, let us find the greatest common factor of m and m^3.
m is the greatest common factor of m and m^3.now, let us find the greatest common factor of n^4 and n^2.
n^2 is the greatest common factor of n^4 and n^2.therefore, following expressions are part of the greatest common factor of 54mn^4 and 27m^3n^2.
b) n2c) me) 27