STAB52H3 Midterm Test
Solutions
October, 2009
Duration: one hour and fifty minutes
Please do NOT open this booklet until you are asked to do so
There are 7 pages including this page. Please check to see if you have all the pages.
Aids allowed: This test is open book. You are allowed to use a non-programmable calculator. No
other aids are allowed.
All your work must be presented clearly in order to get credit. Just an answer with no other work
shown will only qualify for zero credit. Show your work and answer in the space provided, in
ink. Pencil may be used, but then any re-grading will NOT be allowed.
You will get a mark out of 90 for this test.
Last name:____________________
First name:____________________
Other names:___________________
Student number: _________________
Circle your tutorial:
TUT001 (Mon 9:00 - 10:00)
TUT002 (Mon 10:00 - 11:00)
TUT003 (Mon 16:00 - 17:00)
TUT004 (Tue 15:00 - 16:00)
Page 1 of 7
Solutions
1. Events A, B, and C are such that P(A) = 0.7, P(B) = 0.6, P(C) = 0.5, P(AB) =0.4,
P(BC) = 0.3, P(AC) = 0.2, P(ABC) = 0.1.
[Recall that for any two events E1 and E2 , E1 \ E2 = E1 ∩ E2c .]
a) [6 points] Calculate P ( ( A \ B) ∩ (C \ B) ) .
Solution
P ( ( A \ B) ∩ (C \ B) ) = P ( A ∩ B c ) ∩ (C ∩ B c )
(
= P( A∩ B
c
∩C
)
)
= P ( AC \ B )
= P( AC ) − P( ABC )
= 0.2 − 0.1 = 0.1
a) [6 points] Calculate P ( ( A \ B) ∪ (C \ B) ) .
Sol
P ( ( A \ B) ∪ (C \ B) ) = P ( ( A \ B) ) + P ( (C \ B) ) − P ( ( A \ B) ∩ (C \ B) )
= P( A) − P( AB ) + P(C ) − P( BC ) − P ( ( A \ B) ∩ (C \ B ) )
= 0.7 − 0.4 + 0.5 − 0.3 − 0.1
= 0.4
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2) [6 points] A professor has 20 books written by 20 different authors. Five of these books are
Probability books, five Statistics books, five Calculus books and five Algebra books. He wants to
distribute these books at random among four students Adam, Bob, Clara and David so that each
student receives five books. Find the probability that one of these students will receive all five
Statistics books.
Sol
( Sheldon Ross)
20


There are 
 possible divisions of these books among these four students. As there
 5, 5, 5, 5 
 15 
are 
 possible divisions of the books leading to a particular student having all 5
 5, 5, 5 
Statistics books, it follows that the desired probability is given by
 15

4×

 5, 5, 5  = 0.000257997936 ■
20




 5, 5, 5, 5 
3)[6 points] Suppose that the two friends Adam and Bob takes turns in tossing a biased coin
which lands heads with probability 0.4. Suppose that Adam tosses first (and Bob next and then
Adam and so on). They continue tossing this coin until a head comes up and the person who
tosses the first head wins the game. Find the probability that Adam wins the game.
Sol
Let H be the event of getting a head. Then P(H) = 0.4.
Let X = be the number of tails before the first H. Then X~Geo(0.4).
The required probability is P(X = 0) + P(X = 2) + P(X = 4) +…
P ( X = 0) + P ( X = 2) + P ( X = 4) + ⋯ = q0 p + q2 p + q4 p + ⋯
p
1− q2
0.4
=
1 − 0.62
= 0.625 □
=
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4) A box contains two biased coins. One shows heads with probability 0.1 when spun. The other
shows heads with probability 0.5. Suppose you pick one of these two coins at random and spin it
twice. (Pitman)
a)[5 points] Find the probability of getting a head on the first spin.
Sol
P( H1 ) = P(0.1 coin) P( H1 | 0.1 coin) + P(0.5 coin) P( H1 | 0.5 coin)
= 0.5 × 0.1 + 0.5 × 0.5
= 0.3 □
b) [3 points] Find the probability of getting a head on the second spin.
Sol
P ( H1 ) = P ( H 2 ) = 0.3 □
c) [5 points] Find the probability of getting a heads on both spins.
Sol
P( H1 H 2 ) = P(0.1 coin) P( H1 H 2 | 0.1 coin) + P(0.5 coin) P( H1 H 2 | 0.5 coin)
= 0.5 × 0.1× 0.1 + 0.5 × 0.5 × 0.5
= 0.13 □
d) [5 points] Given that heads came up on both spins, find the conditional probability that the
coin selected was the one that shows heads with probability 0.5.
Sol
P(0.5 coin | H1 H 2 ) =
P( H1H 2 | 0.5 coin) P (0.5 coin)
P (0.1 coin) P( H1 H 2 | 0.1 coin) + P(0.5 coin) P( H1H 2 | 0.5 coin)
0.5 × 0.5 × 0.5
0.5 × 0.1× 0.1 + 0.5 × 0.5 × 0.5
25
=
= 0.9615384615 □
26
=
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5) [6 points] The manufacturer of a low-calorie dairy drink wishes to compare the taste appeal of
a new formula (formula B) with that of the standard formula (formula A). Each of four judges is
given three glasses in random order, two containing formula A and the other containing formula
B. Each judge is asked to state which glass he or she most enjoyed. Suppose that the two
formulas are equally attractive. Find the probability that at least three (i.e. three or more) of the
four judges will choose the new formula as the most enjoyed?
Sol
Let Y be the number of judges stating a preference for the new formula.
6) [6 points] Let X~ Poisson(λ). If P ( X ≥ 1) = 0.5 , find that value of λ.
Sol
P ( X ≥ 1) = 0.5 ⇒ 1 − e − λ = 0.5
⇒ e − λ = 0.5
⇒ λ = − ln(0.5) = 0.6931471806 □
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7) One-third of the people in a population have blood type O + . If people are tested one by one
for their blood type, calculate the probability that
a) [3 points] the first person to have blood type O + is the fourth person tested.
3
 2 1 8
Ans =     = = 0.0987654321 □
 3   3  81
b) [5 points] the third person with blood type O + is the tenth person tested.
2
7
 9   1   2   1  36 × 27
       = 10 = 0.07803688462 □
3
 2 3   3   3 
8) The cumulative distribution function of a random variable X is give by:
 16
if x ≥ 4
1 −
F ( x) =  x 2
 0
if x < 4.
a) [5 points] Find f, the probability density function of X.
Sol
 32
if x ≥ 4

f ( x) =  x3
 0 if x < 4.
b) [4 points] Calculate P (5 ≤ X ≤ 7) .
Ans P (5 ≤ X ≤ 7) = F (7) − F (5) = [1 − 16 / 49] − [1 − 16 / 25] = 0.313
Page 6 of 7
9) The probability density function of a continuous random variable X is give by:
1 − 18 ( x −3)2
e
8π
f ( x) =
x ∈ R.
[6 points] Calculate P ( −1 ≤ X ≤ 5) . Give your answer to the nearest three decimal places.
Sol
1  x −3 

2 
− 
1 − 18 ( x −3)2
1
Note that f ( x) =
e
=
e 2
8π
2 2π
2
x ∈ R. and X ~ N ( µ = 3, σ 2 = 4) .
P (−1 ≤ X ≤ 5) = P ((−1 − 3) / 2 ≤ Z ≤ (5 − 3) / 2)
= P (−2 ≤ Z ≤ 1)
= (1 − 0.1587) − 0.0228
= 0.8185 □
[10 points] 10) Let X be a continuous random variable with probability density function:
1
, x ∈ R.
π (1 + x 2 )
Find the probability density function of Y = arctan(X). Identify the distribution of Y (i.e. give the
name of the distribution of Y).
f ( x) =
(Ghahramani)
y= h(x) = arctan(x) and h −1 ( y ) = tan y
Note that as x varies from - ∞ to ∞, y varies from -π/2 to π/2
h′( x) =
1
1
1
and h′(h −1 ( y )) =
=
2
2
1+ x
1 + tan y sec 2 y

1
1


2
f X (h ( y )) π  1 + tan y  1
fY ( y ) =
=
= ,
π


1
h′(h −1 ( y )


2
 1 + tan y 
−1
 π π
y ∈  − ,. 
 2 2
 π π
i.e. Y~ Uniform  − ,.  □
 2 2
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