Question

5.1 Suppose we have the following sample space S ={ 1;2;3;4;7;8;9;10;11;12;15; and we have events: A = {1;3;11;15;} and B = {2;7;9;10;11;15}
5.1.1 Find the probability of A ∩ B.

Answer 1

Answer:

4/11

Step-by-step explanation:

we have the following sample space S ={ 1;2;3;4;7;8;9;10;11;12;15;

events: A = {1;3;11;15;} and B = {2;7;9;10;11;15}

A ∩ B = { 11 , 15 }  = 2 Values

S ={ 1;2;3;4;7;8;9;10;11;12;15}  =  11

A = {1;3;11;15;} = 4 Values

B = {2;7;9;10;11;15}  = 6 Values

Probability of event A = 4/11

Probability of event B = 6/11

probability of A ∩ B  = (4/11)(2/4) + (6/11)(2/6)

= 2/11 + 2/11

= 4/11

probability of A ∩ B  = 4/11

Answer 2

Answer:

S ={ 1,2,3,4,7,8,9,10,11,12,15}=11 elements

A={1,3,11,15}= 4 elements

B={2,7,9,10,11,15}= 6 elements

A∩ B={11,15}=2 elements

→n(A∪B)=n(A)+n(B)-n(A∩B)

            =6+4-2

   =8 elements

P(A∩ B)

     $=\frac{n(A\bigcap B)}{n(A\bigcup B)}\\\\=\frac{2}{8}\\\\=\frac{1}{4}$