Respuesta :
Answer and explanation:
Given : A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement.
To find : Define the following events and the following conditional probabilities ?
Solution :
Number of yellow balls = 1
Number of red balls = 2
Number of green balls = 3
Let A be the event that 'One of the balls is yellow'
So, [tex]P(A)=\frac{1}{6}+\frac{5}{6}\times \frac{1}{5}=\frac{1}{3}[/tex]
Let B be the event that 'At least one ball is red'
So, [tex]P(B)=1-\frac{4}{6}\times \frac{3}{5}=\frac{3}{5}[/tex]
Let C be the event that 'Both balls are green'
So, [tex]P(C)=\frac{3}{6}\times \frac{2}{5}=\frac{1}{5}[/tex]
Let D be the event that 'Both balls are of the same color'
So, [tex]P(D)=\frac{2}{6}\times \frac{1}{5}+\frac{3}{6}\times \frac{2}{5}=\frac{4}{15}[/tex]
Conditional probabilities,
a) [tex]P(B|A)[/tex]
[tex]P(B|A)=\frac{P(B\cap A)}{P(A)}[/tex]
[tex]P(B\cap A)[/tex] means one ball is yellow and one is red,
[tex]P(B|A)=\frac{\frac{1}{6}\times \frac{2}{5}}{\frac{1}{3}}[/tex]
[tex]P(B|A)=\frac{\frac{1}{15}}{\frac{1}{3}}[/tex]
[tex]P(B|A)=\frac{3}{15}[/tex]
[tex]P(B|A)=\frac{1}{5}[/tex]
b) [tex]P(D|B)[/tex]
[tex]P(D|B)=\frac{P(D\cap B)}{P(B)}[/tex]
[tex]P(D\cap B)[/tex] means both ball are red,
[tex]P(D|B)=\frac{\frac{2}{6}\times \frac{1}{5}}{\frac{3}{5}}[/tex]
[tex]P(D|B)=\frac{\frac{1}{15}}{\frac{3}{5}}[/tex]
[tex]P(D|B)=\frac{5}{15\times 3}[/tex]
[tex]P(D|B)=\frac{1}{9}[/tex]
c) [tex]P(C|D)[/tex]
[tex]P(C|D)=\frac{P(C\cap D)}{P(D)}[/tex]
[tex]P(C\cap D)[/tex] means both ball are green,
[tex]P(C|D)=\frac{\frac{3}{6}\times \frac{2}{5}}{\frac{4}{15}}[/tex]
[tex]P(C|D)=\frac{\frac{1}{5}}{\frac{4}{15}}[/tex]
[tex]P(C|D)=\frac{3}{4}[/tex]
