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MATH 264 QUIZ #2
December 8, 2004
QUESTION
The government of Turkey sponsors a national lottery called “SAYISAL LOTO 6/49”. To play, you pick
six numbers between one and forty-nine. Once a week, 6 numbers are randomly selected. The winning
ticket belongs to people with all six correct numbers. Lesser prizes are given to tickets with five, four, or
three correct choices. In one play of the game,
(a) find the probability of winning the grand prize (express your result as a fractional number),
(b) find the probability of guessing 4 numbers correctly (express your result as a fractional number).
Bonus.
(a) Find the probability of winning a prize (that is, the probability of guessing at least 3 of the 6
numbers correctly) in one play of the game.
(b) Find the probability of winning a prize in 5 play of the game (In this part of the question, you do
not need to simplify your answer).
ANSWER
Let A denote the set of winning numbers and B denote the remaining numbers. Then a, the number of
objects of type A is equal to 6, and b, the number of objects of type B is equal to 43. We will choose 6
number among 49 numbers, without replacement. Let X be the number of matches between our selection
and set A. Thus, X is a hypergeometric random variable, and its distribution is given by the formula:
6
43
x 6−x
f (x) = P (X = 6) =
49
6
(a) The desired probability is P (X = 6), and by hypergeometric formula, it is equal to
6 43
1
1
1
6
0
=
.
P (X = 6) = = =
49 · 48 · 47 · 46 · 45 · 44
49
49
13983816
6·5·4·3·2·1
6
6
(b) The desired probability is P (X = 4), and by hypergeometric formula, it is equal to
6 43
6 · 5 · 4 · 3 43 · 42
·
15 · 903
13545
645
4
2
P (X = 4) = = 4 · 3 · 2 · 1 2 · 1 =
=
=
.
49
13983816
13983816
13983816
665896
6
Bonus.
(a) The desired probability is P (X ≥ 3), and by hypergeometric formula, it is equal to
P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)
6 43
6 43
6 43
6 43
+
+
+
3
3
4
2
5
1
6
0
=
49
6
6 · 5 · 4 43 · 42 · 41
6·5·4·3·2
·
+ 13545 +
· 43 + 1
3·2·1
5·4·3·2·1
= 3·2·1
13983816
20 · 12341 + 13545 + 6 · 43 + 1
=
13983816
246820 + 13545 + 258 + 1
260624
=
=
≈ 0.02.
13983816
13983816
(b) Assuming independence of trials, and using the result of part (a) of bonus and the binomial formula
(with n = 5 and p = 0.02), we see that the desired probability is
5
f (1) + f (2) + f (3) + f (4) + f (5) = 1 − f (0) = 1 −
(0.02)0 (0.98)5 ≈ 0.10.
0
MATH 264 QUIZ #2
December 8, 2004
QUESTION
The government of Turkey sponsors a national lottery called “SAYISAL LOTO 6/49”. To play, you pick
six numbers between one and forty-nine. Once a week, 6 numbers are randomly selected. The winning
ticket belongs to people with all six correct numbers. Lesser prizes are given to tickets with five, four, or
three correct choices. In one play of the game,
(a) find the probability of winning the grand prize (express your result as a fractional number),
(b) find the probability of guessing 3 numbers correctly (express your result as a fractional number).
Bonus.
(a) Find the probability of winning a prize (that is, the probability of guessing at least 3 of the 6
numbers correctly) in one play of the game.
(b) Find the probability of winning a prize in 5 play of the game (In this part of the question, you do
not need to simplify your answer).
ANSWER
Let A denote the set of winning numbers and B denote the remaining numbers. Then a, the number of
objects of type A is equal to 6, and b, the number of objects of type B is equal to 43. We will choose 6
number among 49 numbers, without replacement. Let X be the number of matches between our selection
and set A. Thus, X is a hypergeometric random variable, and its distribution is given by the formula:
6
43
x 6−x
f (x) = P (X = 6) =
49
6
(a) The desired probability is P (X = 6), and by hypergeometric formula, it is equal to
6 43
1
1
1
6
0
=
.
P (X = 6) = = =
49 · 48 · 47 · 46 · 45 · 44
49
49
13983816
6·5·4·3·2·1
6
6
(b) The desired probability is P (X = 3), and by hypergeometric formula, it is equal to
6 43
6 · 5 · 4 43 · 42 · 41
·
3
3
3 · 2 · 1 = 20 · 12341 = 246820 .
P (X = 3) = = 3 · 2 · 1
49
13983816
13983816
13983816
6
Bonus.
(a) The desired probability is P (X ≥ 3), and by hypergeometric formula, it is equal to
P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)
6 43
6 43
6 43
6 43
+
+
+
3
3
4
2
5
1
6
0
=
49
6
6 · 5 · 4 · 3 43 · 42 6 · 5 · 4 · 3 · 2
246820 +
·
+
· 43 + 1
4·3·2·1 2·1
5·4·3·2·1
=
13983816
246820 + 15 · 903 + 6 · 43 + 1
=
13983816
246820 + 13545 + 258 + 1
260624
=
=
≈ 0.02.
13983816
13983816
(b) Assuming independence of trials, and using the result of part (a) of bonus and the binomial formula
(with n = 5 and p = 0.02), we see that the desired probability is
5
f (1) + f (2) + f (3) + f (4) + f (5) = 1 − f (0) = 1 −
(0.02)0 (0.98)5 ≈ 0.10.
0