Combinatorics
25 !
16 !
2. simple variation.
3. simple permutation 8! 4. simple permutation You can
13!
21!
paint the solution of the previous 8! Different ways, so the solution is féleképpen lehet beszínezni, tehát
1. simple variation
8!2
5. simple permutation 26! 6. permutation with repetition
26!
7. simple permutation. (7-1)!,
7 !19!
becasuse they can stand in 7! Ways in a queue. If they are standing in a ring, and dance forward, the order
remains the same. They can dance forward 7 different ways with the same order, because we divide 7!-
6!
. 9. simple permutation. 4! 10. simple
2
10 !
12. combination with repetitions.
7 ! 3 !
with 7. 8. See previous, but you can turn over the chain, so
permutation
10 !
11. simple combination.
3 ! 4 !
 64 
 
8
Probabilities
1. Can be divided by 5 or the last digit is 0. In this case since B  A , it is the A event itself. 2. Can be
divided by 5 and the last digit is 0. In this case since B  A , it is the B event itself. 3. The last digit
is 0, but it cannot be divided by 5. It is an impossible event. 4. It cannot be divided by 5. 5. It
cannot be divided by 5 or it’s last digit is 0. 6. It can be divided by 5 and the last digit is not 0. 7. It
cannot be divided by 5 and the last digit is not 0. In this case since B  A , it is the A event itself.
8. It is not true that that it can be divided by 5 and the last digit is 0. in this case it means that the
last digit is not 0. 9. It cannot be divided by 5 and the last digit is not 0. In this case since B  A , it
is the A event itself. 10. It can be divided by 5 and the last digit is not 0, or it cannot be divided by
5 and the last digit is 0. In this case since B  A , this is the AB event itself. 11. It can be divided
by 5 and the last digit is 0, or (it cannot be divided by 5 and the last digit is 0). In this case, the
resulting event: the last digit is not 5. 12. It cannot be divided by 5 and the last digit is not 0. In
this case since B  A , it is the B event itself. 13. It can be divided by 5, or it cannot be , but the
last digit is 0. In this case since B  A , this is the A event itself. 14. Theorem of Total Probability.
P(faulty)
=
P(faulty|morning)P(morning)
+
P(faulty|afternoon)P(afternoon)
+
P(faulty|night)P(night) = 0.01*0.33+0.015*0.33+0.05*0.33 = 0.025 = 2.5%. 15. Theorem of Total
Probability. P(positiv) = P(pos|first)P(first) + P(pos|second)P(second) = 0.25*0.4+0.3*0.6 = 0.28.
1 6 1 1
Pburns | pizza P pizza 
16. P =  +  = 31% 17. Bayes Theorem: P pizza | burns  

Pburns 
3 7
5 7
1
0 .2 
Pgood | Casio PCasio 
7 
9.2%
18.

PCasio | good  
0.31
Pgood 
0.99  0.1
Pgood | Casio PCasio 
 91.7%. 19.

0.99  0.1  0.01  0.9
Pgood | Casio PCasio   Pgood | imitation Pimitation 
P plays | classical Pclassical 

Pclassical | plays  
P plays 
0.1  0.3
P plays | classical Pclassical 
 6.4%. 20

0.1  0.3  0.6  0.7
P plays | classical Pclassical   P plays | other Pother 
Binomial distribution, because trials are independent, p is the same in each trials.
1
n  10, p   0.15, P X  8  0 21 Binomial distribution, because trials are independent, p is
6
the same in each trials. Sample size is big. Try to use Poisson with   np  16.7 . It is more than
10, so Poisson cannot be used. In this case normal distribution should be used with   np and
  np1  p  22 Binomial distribution, because trials are independent, p is the same in each
1
 0.02, P X  7   0 23 Binomial distribution, because trials are
32
independent, p is the same in each trials. Sample size is big. Try to use Poisson with
  np  3.125  3 . P X  70  0 24 Poisson, because estimated number proportional to the
time interval.   9, P X  0  0.0003 25 Poisson, because estimated number proportional to the
time
interval.
26.
  6, P X  15  0.0003  0.0001  0.0004
trials.
n  10, p 


 X   91  90 
  PU  3  0.1% 27.   2   , so [89.33, 90.66] 28.
P X  91  P

1 
 


3 

 X   10  8 
X 

P X  10  P

P X  30   P
 2   2.3%
29.
30.
  2.3%
1 
 
 

  1.64   , so [31.8, 48.2]