Ams 310 Survey Of Probability And Statistics Worksheet - Midterm Exam I - Stony Brook University Page 4
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11(2). A random variable can have values of
.
0, 1, 2, 3, 4, 5, 6, 7
(a). It is know that
, nd the probability of
and
P (X
4) = 3P (X
5)
P (X
4)
.
P (X
5)
(b). If
and
, nd the probability for
.
P (X
6) = 0.9
P (X
2) = 0.8
P (2
X
6)
(c). If
,
, and
, fund
P (1
X
5) = 0.8
P (3
X
6) = 0.6
P (3
X
5) = 0.5
the probability for
.
P (1
X
6)
Solution
(a). Let
P (A) = P (X
4),
then
P ( ¯ A) = P (X
5),
we have
P (A) + P ( ¯ A) = 3P ( ¯ A) + P ( ¯ A) = 1,
4P ( ¯ A) = 1
hence
P ( ¯ A) = 0.25,
P (A) = 0.75
.
(b). Let
P (B) = (X
6),
P (C) = P (X
2),
P (2
X
6) = P (B
C)
Since
covers all values of the random variable, therefore
P (B
C)
P (B
C) = 1 = P (B) + P (C)
P (B
C)
hence
P (B
C) = P (B) + P (C)
1 = 0.9 + 0.8
1 = 0.7
(c). Let
P (D) = P (1
X
5),
P (E) = P (3
X
6),
we also have
P (D
E) = P (3
X
5),
hence
P (1
X
6) = P (D E) = P (D) + P (E) P (D E) = 0.8 + 0.6 0.5 = 0.9.
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