IB Math HL Y2
Mr. Jauk
1. (a) (i)
P(A  B) = P(A) + P(B) = 0.7
(ii)
(b)
Test Prep Probability
P(A  B) = P(A) + P(B) – P(A  B)
= P(A) + P(B) – P(A)P(B)
= 0.3 + 0.4 – 0.12 = 0.58
P(A  B) = P(A) + P(B) – P(A  B)
= 0.3 + 0.4 – 0.6 = 0.1
P( A  B)
P(A│B) =
P(B)
=
A1
(M1)
(M1)
A1
A1
(M1)
0.1
= 0.25
0.4
A1
[7]
Most candidates attempted this question and answered it well. A few
misconceptions were identified (eg P( A  B)  P( A)P(B) ). Many candidates
were unsure about the meaning of independent events.
2.
(a)
METHOD 1
 8  4 3 2 11 10 9 8 7
P(3 defective in first 8) =          
M1A1A1
 3  15 14 13 12 11 10 9 8
Note: Award M1 for multiplication of probabilities with decreasing
denominators.
Award A1 for multiplication of correct eight probabilities.
8
Award A1 for multiplying by   .
 3
=
56
195
A1
METHOD 2
 4 11
  
 3  5 
P(3 defective DVD players from 8) =
15 
 
8
M1A1
Note: Award M1 for an expression of this form containing three combinations.
IB Math HL Y2
Mr. Jauk
Test Prep Probability
4! 11!

3!1! 5!6!
=
15!
8!7!
56
= 195
(b)
M1
A1
P(9th selected is 4th defective player│3 defective in first 8) =
56 1

195 7
8
=
195
P(9th selected is 4th defective player) =
1
7
(A1)
M1
A1
[7]
There were two main methods used to complete this question, the most
common being a combinations approach. Those who did this coped well with
the factorial simplification. Many who did not manage the first part were able
to complete the second part successfully.
3.
 4
recognition of X ~ B  6, 
 7
3
3
 6  4   3  
4 3  33 

P(X = 3) =        20 
7 6 
 3  7   7  
2
4
 6  4   3  
4 2  34 

P(X = 2) =        15 
7 6 
 2  7   7  
P( X  3) 80  16 

 
P( X  2) 45  9 
(M1)
A1
A1
A1
[4]
Many correct answers were seen to this and the majority of candidates
recognised the need to use a Binomial distribution. A number of candidates,
although finding the correct expressions for PX = 3 and PX = 4, were unable
to perform the required simplification.
IB Math HL Y2
Mr. Jauk
4.
Test Prep Probability
(a)
A1A1
Note: Award A1 for a diagram with two intersecting regions and at
least the value of the intersection.
(b)
9
20
A1
(c)
9  3
 
12  4 
A1
[4]
Although this was the best done question on the paper, it was disappointing
that a significant number of candidates produced Venn diagrams with key
information missing.
5.
METHOD 1
(M1)
Let P(I) be the probability of flying IS Air, P(U) be the probability
flying UN Air and P(L) be the probability of luggage lost.
IB Math HL Y2
Mr. Jauk
P(I | L) =
Test Prep Probability
P I  L 
PL 


PL | I  P I 
 or Bayes' formula , PI | L  

PL | I  P I   PL |U  P U  

0.23 
=
=
(M1)
65
135
A1A1A1
70
65
0.18 
 0.23 
135
135
299
 0.543 , accept 0.542
551
A1
METHOD 2
Expected number of suitcases lost by UN Air is 0.18  70 = 12.6
M1A1
Expected number of suitcases lost by IS Air is 0.23  65 = 14.95
A1
P(I | L) =
14.95
12.6  14.95
M1A1
= 0.543
A1
[6
This question was well answered by the majority of candidates. Most
candidates used either tree diagrams or expected value methods.
6.
(a)
 30 
Total number of ways of selecting 4 from 30 =  
4
12 18 
Number of ways of choosing 2B 2G =   
 2  2 
(M1)
(M1)
12 18 
  
 2  2 
P(2B or 2G) =
= 0.368
 30 
 
4
(b)
12 
Number of ways of choosing 4B =   , choosing 4G =
4
12  18 
    
4 4
P(4B or 4G) =
 30 
 
4
= 0.130
A1
18 
 
4
N2
A1
(M1)
A1
N2
[6]