September 30, 2011
Hypergeometric Distribution (Continued) - Solutions
Example 4
Determine the probability of winning the jackpot in lotto 5/35
Solution
P(X = 5) =
5 C5
×30 C0
1
=
= 3.08 × 10−6
324632
35 C5
Example 5
A quality control engineer inspects a random sample of 3 batteries from each lot of 24 car batteries
ready to be shipped. If such a lot contains 6 batteries with slight defects, what are the probabilities that
the inspector’s sample will contain
A. None of the batteries with defects?
B. Only one of the batteries with defects?
C. At least two of the batteries with defects?
Solution
A. 𝑁 = 24
𝑀=6
𝑛=3
𝑋=0
𝑃(𝑋 = 0) =
B. 𝑁 = 24
𝑀=6
𝑛=3
𝑀=6
𝑛=3
×18 C3
816
=
= 0.4032
2024
24 C3
𝑋=1
𝑃(𝑋 = 1) =
C. 𝑁 = 24
6 C0
6 C1
×18 C2
918
=
= 0.4536
2024
24 C3
𝑋≥2
𝑃(𝑋 ≥ 2) = 1 − [P(X = 0) + P(X = 1)] = 1 − [0.4032 + 0.4536] = 0.1432
Example 6
Among 16 cities that a professional society if considering for its next 3 annual conventions, 7 are in the
western part of the United States. To avoid arguments, the selection is left to chance. If none of the
cities can be chosen more than once, what are the probabilities that
A. None of the conventions will be held in the western part of the United States?
B. All of the conventions will be held in the western part of the United States?
Solution
A. 𝑁 = 16
𝑀=7
𝑛=3
𝑋=0
𝑃(𝑋 = 0) =
B. 𝑁 = 24
𝑀=6
𝑛=3
7 C0
× 9 C3
84
=
= 0.15
560
16 C3
𝑋=3
𝑃(𝑋 = 3) =
7 C3
× 9 C0
35
=
= 0.0625
560
16 C3
Example 7
If 6 of the 18 new buildings in a city violate the building code, what is the probability that a building
inspector who randomly selects 4 of the new building for inspection will catch
A.
B.
C.
D.
None of the buildings that violate the building code?
1 of the new buildings that violate the building code?
2 of the new buildings that violate the building code?
At least 3 of the new buildings that violate the building code?
Solution
A. 𝑁 = 18
𝑀=6
𝑛=4
𝑋=0
𝑃(𝑋 = 0) =
B. 𝑁 = 18
𝑀=6
𝑛=4
6 C0
×12 C4
495
=
= 0.1618
3036
18 C4
𝑋=1
𝑃(𝑋 = 1) =
6 C1
×12 C3 1320
=
= 0.4314
3036
18 C4
C. 𝑁 = 18
𝑀=6
𝑛=4
𝑋=2
𝑃(𝑋 = 2) =
D. 𝑁 = 18
𝑀=6
𝑛=4
6 C2
×12 C2
990
=
= 0.3235
3036
18 C4
𝑋≥3
𝑃(𝑋 ≥ 3) = 1 − [P(X = 0) + P(X = 1) + P(X = 2)]
= 1 − [0.1618 + 0.4314 + 0.3235]
= 0.0833
Example 8
An agricultural cooperative claims that 90% of the watermelons shipped out are ripe and ready to eat.
Find the probabilities that among 18 watermelons shipped out,
A. All 18 are ripe and ready to eat.
B. At least 16 are ripe and ready to eat.
C. At most 14 are ripe and ready to eat.
Solution
A. 𝑝 = 0.9
𝑛 = 18
𝑋 = 18
18
𝑃(𝑋 = 18) = ( ) (0.9)18 (0.1)0 = 0.1501
18
B. 𝑝 = 0.9
𝑛 = 18
𝑋 ≥ 16
𝑃(𝑋 ≥ 16) = P(X = 16) + P(X = 17) + P(X = 18)
18
18
18
= ( ) (0.9)16 (0.1)2 + ( ) (0.9)17 (0.1)1 + ( ) (0.9)18 (0.1)0
16
17
18
= 0.2835 + 0.3 + 0.1501
= 0.7336
C. 𝑝 = 0.9
𝑛 = 18
𝑋 ≤ 14
𝑃(𝑋 ≤ 14) = 1 − [P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)]
18
= 1 − [( ) (0.9)15 (0.1)3 + 0.7336] = 0.0984
15