Respuesta :
Answer:
Yes, The events A and B are mutually exclusive event.
Step-by-step explanation:
Two chips are drawn at random and without replacement from a bag containing two blue chips and two red chips.
Total chips (n ) = 4
Two chips are chooses from 4 chips then the total possibilities are = [tex]\binom{4}{2}[/tex]
total outcomes = [tex]\frac{4!}{(2!)(2!)}[/tex] = 6
Let [tex]b_{1}[/tex] and [tex]b_{2}[/tex] denote blue chips and [tex]r_{1}[/tex] and [tex]r_{2}[/tex] denote red chips.
A: {At least one of the chips is blue} = { [tex]b_{1}[/tex][tex]b_{2[/tex] , [tex]b_{1}r_{1}[/tex] , [tex]b_{1}r_{2}[/tex] , [tex]b_{2}r_{1}[/tex], [tex]b_{2}r_{2}[/tex] }
P( A) = [tex]\frac{5}{6}[/tex]
B: {Both chips are red} = { [tex]r_{1}r_{2}[/tex] }
P( B) = [tex]\frac{1}{6}[/tex]
[tex](A\bigcup B)[/tex] = { [tex]b_{2}b_{1}[/tex] , [tex]b_{1}r_{1} , b_{1}r_{2} , b_{2}r_{1} , b_{2}r_{2} ,r_{1}r_{2}[/tex] }
[tex]P(A\bigcup B)[/tex] = [tex]\frac{6}{6}[/tex]
If A and B are mutuall exclusive
[tex]P(A\bigcup B)[/tex] = P( A) + P( B)
P( A) + P( B) = [tex]\frac{5}{6} + \frac{1}{6}[/tex] = 1
[tex]P(A\bigcup B)[/tex] = [tex]\frac{6}{6}[/tex] = 1
Hence
[tex]P(A\bigcup B)[/tex] = P( A) + P( B)
Then A and B are mutuall exclusive
