STAT 230 Exam 1 with Solution, Spring 2011
American University of Beirut · Introduction to Probability and Random Variables · 4 pages · Dr. Mohammad Kobeissi · with solution
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American University of Beirut 3. Chips are drawn at random, one at a time and
STAT 230 without replacement from an urn that contains
Introduction to Probability and Random Variables 5 pink chips and 7 white chips, until only those
Spring 2011 of the same color are left. Find the probability
that exactly 7 draws are needed.
quiz # 1
Name: solu bin (_aw_] or q Pay P |
ID:
Section: 12:30 PM 2PM
+ G C C4
. : Cc C gy *F
1. In how many ways can three different prizes be 4 _ >? ee
—3z + 6
awarded among nine contestants if no person ey C 6
is to receive more than one prize ?
Ax8xF 2 ate = 504 2
A, Four couples are to be seated in a row. Find
2. One of the five élévators in a building starts the probability that each husband sits next to
with seven passengers and stops at nine floors. his wife.
Assuming that it is equally likely that a pas-
senger gets off at any of these nine floors, find let C 2 CG, 5) Gq 2 G be
the probability that at least two of these pas- te 4 tou thes
sengers will get off at the same floor.
A = ; ab fees} twe pesserges
ger off ek be Some
floor } gix2 J
| ios
P(A) = l- P(A) 4! : pewmlatin of He
cl (<2 3] ¢%
9x8 x dnb Sx ae3 G@ cudes
18 4 et cel
3 2 : pamaia a o} ecc (
p coute xu
- } ss at = 0.96 3! : bokef nb. ef pemelile
5. Let A,B and C three independent events with 7. An absentminded professor wrote 15 letters and
probabilities 1/2,1/6,1/4. Find P((AN B)U
C).
P(& 0B) uc)= ARAB) + Plc) MArBnc)
ied LR) PCB) + Plc) - HA) PIB) PO)
1,I,4
= jefe ‘ ~ 2 6 4
27. 9 - 9 SbdS
= y a 14 .
6. The pdf of a random variable X is given in the
following table
cig | 2 1 4
P(X=2)| 1/8 2/8 5/8
Find E(2X — 1)
sealed them in envelopes before writing the ad-
dresses on the envelopes. Then he wrote the m ! ¢
addresses on the envelopes at random. What
is the probability that no letter was addressed
correctly ?
| aching forme la w iHrouk
reploceme
A. | ne feller wes 294. crreellyf
P(m) = 1- PA) |
er tate isi
8. A computer is instructed to generate a random
sequence using the digits 0 through 9; repeti-
tions are allowed. Find the shortest length the
sequence can be and still have at least 70%
probability of containing at least one 4.
Lek n be He lena Hh o}
Lhe sequre
Ae event os el con bashing
al Je ash one aS
Ak) 2 I- PIA)= I- (7)"
IB a
9. In a squash tournament between three play- 10.
ers A, B and C, each player plays the oth-
A plays B, A plays C and B
plays C). Assume the following probabilities:
P(A beats B) = 0.6; P(A beats C) = 0.7;
P(B beats C) = 0.6. Assuming independence
of the match results, calculate the probability
ers once (ie.
that A wins at least as many games as any
other player.
W -{ event A wias al hesk es
mend grees as “4
olta bps
ib means Lest Ss
Plw) = P Plass)n@ £9)
PALB)a B90 € $y)
4 PALIo (c48)0 RSA)
5.2 x0.6 + 0.bro$ xo- 3
—
ine.
+ 0.964 vo. 4
- ole olo§ + 0.112
- 0.64
Let A and B be two events with P(B) = 0.7
and P(A|B) = 0.4. Find P(AU B).
P(AuB)- PIA+ p(B) — Plan®)
P(A) +P 8) - (A) _Pisos)}
pla) + Plans)
_ p(B) « pals) x P68)
= ost 0.6 x04
=0.4d
11. Ona multiple-choice exam with four choices for 12. Marwan, Jad and Kate simultaneously toss coins.
each question, a student either knows the an-
swer to a question or marks it at random. Ifthe
probability that he or she knows the answers
is 2/3, what is the probability that an answer
that was marked correctly was not marked ran-
domly ?
R vestian answered 53
| astin nee
> eel
P(CIR) x PER) + PC (B) «tl ®)
ims)
“_—~
Aa
pages cole
[x &
2 g
a
~ 1, t tx 2 a
a ad
The coin tossed by each one turns up head
with probability 1/2, 1/4 and 1/3 respectively.
If one person gets an outcome different from
those of the other two, then he is the odd man
out. If there is no odd man out, the players
flip again and continue to do so until they get
an odd man out.
a) Find the probability that Marwan is out in
the first round.
l
m 1.3.2, hades
= 9°G*3 424 3
- +4
~ & 2h
- =f
24
b) What is the probability that Marwan will
be the odd man out ?
1. = | event florwen ovt ab He
4
i round §
P ( farwen ovt) = P(11,) + PI 1a) + Pig) + +>
1) 24] dy Zu] 24
2 Ore on
ria YP ere
ees ee 9 ons
(our>