Three boxes contain red and green balls. Box 1 has 5 red balls* and 5 green balls*, Box 2 has 7 red balls* and 3 green balls* and Box 3 contains 6 red balls* and 4 green balls*. The respective probabilities of choosing a box are 1/4, 1/2, 1/4. What is the probability that the ball chosen is green?
Accepted Solution
A:
Answer: 0.375Step-by-step explanation:Let G be the event that the chosen marble is green. Let [tex]B_i[/tex] be the event that we choose Box i.The probabilities of choosing a box are :[tex]P(B_1)=\dfrac{1}{40}=0.25\\\\P(B_2)=\dfrac{1}{2}=0.5\\\\P(B_3)=\dfrac{1}{4}=0.25[/tex]Now, [tex]P(G|B_1)=\dfrac{5}{10}=0.5[/tex][tex]P(G|B_2)=\dfrac{3}{10}=0.3[/tex][tex]P(G|B_1)=\dfrac{4}{10}=0.4[/tex]By using the law of total probability, we have[tex]P(G)=P(G|B_1)P(B_1)+P(G|B_2)P(B_2)+P(G|B_3)P(B_3)\\\\\Rightarrow\ P(G)=0.25\cdot0.5+0.5\cdot0.3+0.25\cdot0.4=0.375[/tex]