Question

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 1

Hello,

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 2

The probability of getting red balls is 0.68.

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
`P(U)=(1)/(3)`

In urn 1 - 5 red balls + 3 black balls

Probability of getting red ball from urn 1 -
`P(R_1)=(5)/(8)`

In urn 2 - 3 red balls + 1 black balls

Probability of getting red ball from urn 2 -
`P(R_2)=(3)/(4)`

In urn 3 - 4 red balls + 2 black balls

Probability of getting red ball from urn 3 -
`P(R_3)=(4)/(6)`

Choosing red ball from urn is
`P(R)=P(R_1)+P(R_2)+P(R_3)`

`P(R)=(5)/(8)+(3)/(4)+(4)/(6)`

`P(R)=(15+18+16)/(24)`

`P(R)=(49)/(24)`

The probability of getting red balls is

`P=P(U)* P(R)`

`P=(1)/(3)* (49)/(24)`

`P=(49)/(72)`

`P=0.68`