(I) Let A
and B are two events of the sample space such that P(A  B) =
1
3
1
, then
5
1- if B  A , the value of P(A) is equal to
1
2
1
1
(A)
(B)
(C)
(D)
3
5
5
2
2- if A and B are mutually exclusive ,the value of P(A) is equal to
1
2
1
2
(A)
(B)
(C)
(D)
5
15
15
3
(II) - A box contains three coins , one coin is fair , one coin is two headed, and
1
one coin is weighted so that the probability of heads appearing is . A coin is
3
selected at random and tossed .
3- The probability that heads appears is equal to
15
11
13
2
(A)
(B)
(C)
(D)
18
18
15
17
4- If heads appeared , the probability that it comes from the third coin
and P(B) =
1
2
1
5
(B)
(C)
(D)
3
11
5
11
(III) A random sample of 200 adults are classified according to sex and their
level of education in the following table
(A)
Education
Elementary
Secondary
College
Male
28
38
22
Female
50
45
17
If a person is selected at random from this group ,then
4- the probability that adult is female equal to
(A) 0.56
(B) 0.65
(C) 0.58
(D) 0.57
5- If the person has a college ,the probability that it is male equal to
(A) 0.465
(B) 0.564
(C) 0.324
(D)0.456
6- The probability that the person does not have a college given that the person is
female equal to
(A) 0.8482
(B) 0.8563
(C)0.0852
(D) 0.6583
2
(IV) A random variables X has a mean  =6 and a variance  = 4.
7. E(3X+4)=
(A) 24.0
8. Var(3X+5)=
(A) 41.0
(B) 14.0
(C) 36.0
(D) 25.0
(E) 22.0
 32X  5 =
(B) 17.0
9 By using Chebyshev,s theorem ,
(C) 14.0
P( -4<X<16)
(D)22.0
(E) 36.0
(A)  24/25
(B) < 24/25
(C) >1/5
(D)<1/5
(V) Let E(X) =  and Var (X) =  2
10- For any random variable X with mean  and variance  2 , we have
(A)  >  2
(B)  <  2
(C)  =  2
(D) Non of these
(11) For any random variable X having binomial distribution, we have
(A)  >  2
(B)  <  2
(C)  =  2
(D) Non of these
(12) For any random variable X having Poisson distribution, we have
(A)  >  2
(B)  <  2
(C)  =  2
(D) Non of these
(VI) If the random variable X has a continuous uniform distribution on the
interval (0,10),then
(13) P(X< 4) equals to
(A) 0.4
(B) 0.61
(C) 0.8
(D) 0.32
(14) The mean of X is
(A) 15
(B) 5
(C) 36.0
(D) 25.0
(VII) If the random variable X has normal distribution with mean  and
variance  2 ,then
(15)
P(X<  + 2  ) equals to
(A) 0.8772
(B) 0.4772
(C) 0.5772
(D) 0.9772
(16) If the random variable X has normal distribution with mean  and
variance = 4 ,then P( X> 1) = 0.9332, then  equals to
(A) 4
(B) 4.1
(C) 3
(D) 5
(VIII)Suppose that the percentage of females in a certain population is 20 % . A
sample of 3 people is selected at random from this population
(17) The probability that no females are selected is
(A)0.512
(B) 0.613
(C) 0.84
(D) 0.36
(18) The probability that at least 2 males are selected is
(A) 0.356
(B) 0.61
(C) 0.896
(D) 0.32
(19) The expected number of males equals to
(A) 2.4
(B) 2.54
(C) 3.5
(D) 1.5
(IX) Suppose that a family has 5 children, 3 of them are girls .A sample of 2
children is selected at random
(20) The probability that no girls are selected is
(A)0.1
(B) 0.41
(C) 0.3
(D)0. 5
(21) The expected number of boys is
(A)1.3
(B) 0.8
(C) 2.3
(D) 3.2
( X)The average life of manufacture batteries is 5 years, with stander deviation
of one year. Assuming the live of the battery follows approximately a normal
distribution.
(22) If a random sample of 5 batteries selected from a manufacture has a mean
of 3 years with a standard deviation of one year ,then the random variable
X has a mean  x equal to
(A) 5
(B) 2
(C) 3
(D) 7
2
(23) the variance  x is equal to
(A) 4
(B) 0.2
(C) 6
(D) 5
(24) the probability that the mean life of a random sample 16 of such batteries will
be less than 5.5 years is
(A)0.9772
(B) 0.0228
(C) 0.2297
(D) 0.7729
(25) If P ( X > k) = 0.9332, the value of k is
(A)5.4
(B) 4.5
(C) 3.4
(D) 4.3
(**) If the cumulative distribution function of random variable X is given by
 x

x0 

F(x) =  x  1
 , then
0
x  0 
(26) P(0 < X < 2) equals to
(A)0.6667
(B) 0.3334
(C) .2223
(D) 0.56667
(27) If P(X  k ) = 0.5 , the value of k is
(A) 1
(B) 4.1
(C) 3
(D) 2
(***) A secretary makes 2 errors per page, on average. Using Poisson distribution
,find
(28) The probability that on the next page he will make at least one error is
(A) 0.8647
(B) 0.4687
(C) 0.5687
(D) 0.6847
(29) The expected number of error in 3 pages is
(A)8
(B) 6
(C) 3
(D) 5
(****) Suppose that X has exponential distribution ,with
F(x) = 1- exp(- 0.25) ,then
(30) P(X< 4) is
(A)0.6321
(B) 0.3216
(C) 0.1326
(D) 0.6132
(31) The expected value of X ( E(X) ) is
(A)2.5
(B) 4
(C) 8
(D) 2.5