Probability and Independence: Worksheet 11 for MATH 10B, Summaries of Discrete Mathematics

Worksheet 11. MATH 10B. Tu 2/28/19. 1. Concept of Independence. • What is the probability that two people chosen at random were born during ...

Typology: Summaries

2022/2023

Uploaded on 05/11/2023

strawberry3
strawberry3 🇺🇸

4.6

(39)

387 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Week 5 Worksheet 11 MATH 10B
Tu 2/28/19
1. Concept of Independence
What is the probability that two people chosen at random were born during the
same month of the year?
Solution: 12
11
12
1
12 =1
12
What is the probability that in a group of n people chosen at random, there are
at least two born in the same month of the year?
Solution: If n > 12, by the pigeonhole principle, at least two people were
born in the same month. Otherwise, P(at least two born in the same month) =
1P(all born in different month) = 1 12
12 ×11
12 ×10
12 . . . 12n+1
12 .
How many people chosen at random are needed to make the probability greater
than 1/2 that there are at least two people born in the same month of the year?
Solution: In other words, we want to find the minimum nsuch that 1
P(all born in different month) = 1 12
12 ×11
12 ×10
12 . . . 12n+1
12 >1
2. When
n= 4, P (at least two born in the same month) 0.482, and When n=
5, P (at least two born in the same month) 0.618. Thus, n5.
2. Two dies were rolled. Are the events that the first die rolled is a 1 and that the sum
of the two dice is a 7 independent?
Solution: Let Abe the event that first die rolled is a 1, and Bbe the event that
the sum of the two dice is a 7. Then,
P(AB) = 1
36 (1)
P(A) = 1
6(2)
P(B) = 6
36 (3)
Therefore, P(AB) = P(A)P(B).
3. Find the smallest number of people you need to choose at random so that the proba-
bility that at least two of them were both born on April 1st exceeds 1/2.
1
pf3

Partial preview of the text

Download Probability and Independence: Worksheet 11 for MATH 10B and more Summaries Discrete Mathematics in PDF only on Docsity!

Week 5 Worksheet 11

Tu 2/28/

  1. Concept of Independence
    • What is the probability that two people chosen at random were born during the same month of the year?

Solution:

1

12

1 12 =^

1 12

  • What is the probability that in a group of n people chosen at random, there are at least two born in the same month of the year?

Solution: If n > 12, by the pigeonhole principle, at least two people were born in the same month. Otherwise, P (at least two born in the same month) = 1 − P (all born in different month) = 1 − 1212 × 1112 × 1012... 12 − 12 n +1.

  • How many people chosen at random are needed to make the probability greater than 1/2 that there are at least two people born in the same month of the year?

Solution: In other words, we want to find the minimum n such that 1 − P (all born in different month) = 1 − 1212 × 1112 × 1012... 12 − 12 n+1 > 12. When n = 4 , P (at least two born in the same month) ≈ 0 .482, and When n = 5 , P (at least two born in the same month) ≈ 0 .618. Thus, n ≥ 5.

  1. Two dies were rolled. Are the events that the first die rolled is a 1 and that the sum of the two dice is a 7 independent?

Solution: Let A be the event that first die rolled is a 1, and B be the event that the sum of the two dice is a 7. Then,

P (A ∩ B) =

P (A) =

P (B) =

Therefore, P (A ∩ B) = P (A)P (B).

  1. Find the smallest number of people you need to choose at random so that the proba- bility that at least two of them were both born on April 1st exceeds 1/2.

Week 5 Worksheet 11

Tu 2/28/

Solution: P (at least two of them were both born on April 1st) = 1 − P (none of them were both born on April 1st) −P (exactly one of them were both born on April 1st) = 1−(^365366 )n−

(n 1

( 3661 )(^365366 )(n−1)

  1. Solving the equation, we have n > 614.
  1. Assume that the probability a child is a boy is 0.51 and that the sexes of children born into a family are independent. What is the probability that a family of five children has - exactly three boys?

Solution:

3

(0.51)^3 (0.49)^2

  • at least one boy?

Solution: 1 − 0. 495

  • at least one girl?

Solution: 1 − 0. 515

  • all children of the same sex?

Solution: 0. 515 + 0. 495

  1. Let E be the event that a randomly generated bit string of length three contains an odd number of 1s, and let F be the event that the string starts with 1. Are E and F independent?

Solution: P (E) =

P (F ) =

P (E ∩ F ) =

= P (E)P (F )

Therefore, independent.