Answer:0.6
Explanation:
Given
Bowl contains 4 red and 2 Green balls
Probability of selecting two red balls given atleast one of the ball is red + Probability of selecting two red balls given that no ball is red =1
Probability of selecting two red balls given that no ball is red [tex]=\frac{Probability \ of\ selecting\ two\ red\ balls\ out\ of\ 4\ balls}{probability\ of\ selecting\ 2\ balls\ out\ of\ 6 balls}[/tex]
[tex]P=\frac{^{4}C_2}{^{6}C_2}[/tex]
[tex]P=\frac{2}{5}[/tex]
Required Probability[tex]=1-\frac{2}{5}=\frac{3}{5}[/tex]