- #1

- 242

- 25

- Homework Statement
- Hello, I very much struggle with probability problems no matter how much I revise. Nonetheless, i am determined to hone my comprehension of the topic and to develop my overall dexterity. Consequently while revising below I came across the following questions, would anyone be able to advise whether the methods I have used to solve them would be correct or if I can improve upon my workings in any manner? I think I stumbled more so on question 2 to be perfectly candid.

Question 1; The integers 1 to 12 are placed in one or more of the following categories;

S: square numbers

P: prime numbers (where 1 is not a prime number)

E:even numbers

T: multiples of 3

If one of the numbers is chosen at random find the probability of it falling within the following sets:

a) T'

b) S ∪ E

c)P⋂T

Question 2; A shortlist of 3 parliamentary candidates is to be selected from a group of 5 men and 6 women. What is the probability that the shortlist consists only of men?

Assume that each selection of 3 candidates is equally likely

- Relevant Equations
- P(A ∪ B)=P(A)+P(B)-P(A⋂B)

P(A⋂B) = P(A)P(B)

Question 1:

a) T' is the complementary event of T

Therefore, T'=1-T

In set T = {3,6,9,12}

P(T)=4/12 =1/3

P(T')=1-1/3=2/3

b) The addition rule states; P(A ∪ B)=P(A)+P(B)-P(A⋂B)

Therefore, P(S ∪ E) = P(S)+P(E)-P(S⋂E)

S={1,4,9}

P(S)=3/12=1/4

E={2,4,6,8,10,12}

P(E)=6/12=1/2

(S⋂E)={4}

P(S⋂E)=1/12

P(S ∪ E) =3/12+6/12-1/12=8/12=2/3

c)P(A⋂B) = P(A)P(B)

Therefore, P(P⋂T)=P(P)P(T)

P={2,3,5,7,11}

P(P)=5/12

T={3,6,9,12}

P(T)=4/12

P(P⋂T)=5/12*4/12=20/144=5/36

Question 2;

This is a combination problem so the order of selection does not matter.

The total number of ways the constituency can be formed;

5+6=11 in total

3 of 11 = 11 C 3 = 11!/3!*8!=165 ways

The number of ways a constituency of 3 men can be formed;

5 men in total

3 of 5 = 5 C 3 =5!/3!*2!=10

The probability that the constituency is only formed of men is 10/165=2/33

Would this be correct, I feel that there is perhaps more to this problem which I have neglected? Thank you to anyone who replies

a) T' is the complementary event of T

Therefore, T'=1-T

In set T = {3,6,9,12}

P(T)=4/12 =1/3

P(T')=1-1/3=2/3

b) The addition rule states; P(A ∪ B)=P(A)+P(B)-P(A⋂B)

Therefore, P(S ∪ E) = P(S)+P(E)-P(S⋂E)

S={1,4,9}

P(S)=3/12=1/4

E={2,4,6,8,10,12}

P(E)=6/12=1/2

(S⋂E)={4}

P(S⋂E)=1/12

P(S ∪ E) =3/12+6/12-1/12=8/12=2/3

c)P(A⋂B) = P(A)P(B)

Therefore, P(P⋂T)=P(P)P(T)

P={2,3,5,7,11}

P(P)=5/12

T={3,6,9,12}

P(T)=4/12

P(P⋂T)=5/12*4/12=20/144=5/36

Question 2;

This is a combination problem so the order of selection does not matter.

The total number of ways the constituency can be formed;

5+6=11 in total

3 of 11 = 11 C 3 = 11!/3!*8!=165 ways

The number of ways a constituency of 3 men can be formed;

5 men in total

3 of 5 = 5 C 3 =5!/3!*2!=10

The probability that the constituency is only formed of men is 10/165=2/33

Would this be correct, I feel that there is perhaps more to this problem which I have neglected? Thank you to anyone who replies

Last edited: