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Urn 1 contains 5 red balls and 3 black balls. Urn 2 contains 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If an urn is selected at random and a ball is drawn, find the probability it will be red.

Answer :

caylus
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Choice urn_1: 1/3
    choice R: 5/8        ==>1/3*5/8=5/24
Choice urn_2: 1/3
   Choice R: 3/4        ==> 1/3*3/4=1/4
Choice urn_3: 1/3
   Choice R:4/6=2/3  ==> 1/3*2/3=2/9

Total: 5/24+1/4+2/9=15/72+18/72+16/72=49/72

Answer:

The probability of getting red balls is 0.68.

Step-by-step explanation:

Given : Urn 1 contains 5 red balls and 3 black balls. Urn 2 contains 3 red balls and 1 black ball. Urn 3 contains 4 red balls and 2 black balls. If an urn is selected at random and a ball is drawn.

To find : The probability it will be red ?

Solution :

Total urn = 3

If an urn is selected at random then the probability is [tex]P(U)=\frac{1}{3}[/tex]

In urn 1 - 5 red balls + 3 black balls

Probability of getting red ball from urn 1 - [tex]P(R_1)=\frac{5}{8}[/tex]

In urn 2 - 3 red balls + 1 black balls

Probability of getting red ball from urn 2 - [tex]P(R_2)=\frac{3}{4}[/tex]

In urn 3 - 4 red balls + 2 black balls

Probability of getting red ball from urn 3 - [tex]P(R_3)=\frac{4}{6}[/tex]

Choosing red ball from urn is [tex]P(R)=P(R_1)+P(R_2)+P(R_3)[/tex]

[tex]P(R)=\frac{5}{8}+\frac{3}{4}+\frac{4}{6}[/tex]

[tex]P(R)=\frac{15+18+16}{24}[/tex]

[tex]P(R)=\frac{49}{24}[/tex]

The probability of getting red balls is

[tex]P=P(U)\times P(R)[/tex]

[tex]P=\frac{1}{3}\times \frac{49}{24}[/tex]

[tex]P=\frac{49}{72}[/tex]

[tex]P=0.68[/tex]

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