Question

As shown in Figure 02, an urn contains 12 red balls and 4 green balls. The red balls are numbered from 1 to 12, and the green balls are numbered from 1 to 4. One ball is randomly drawn from the urn. Which of the following answers is correct? (Let: R = red; G = green; and E = even.)

1. P(G ∪ R) = 0.000.

2. P(R|E) = 0.375.

3. P(G ∪ E) = 0.625.

4. P(G|R) = 0.500.

Please provide a walkthrough explanation on each answer given.

Answer 1

Solution:

Given:

An urn contains 12 red balls and 4 green balls.

The red balls are numbered from 1 to 12, and the green balls are numbered from 1 to 4.

Let  R = red; G = green; and E = even.

A ball is randomly drawn from the urn.

i) P(G ∪ R) = 0.000.

Lets check if this is correct or incorrect.

Using addition rule probability:

P( G U R)= P(G) + P(R) - P( G ∩ R)

P( G U R)= ( 4/16) + ( 12/16) - 0

( Here Green and Red balls are mutually exclusive events thus P( G ∩ R) = 0 )

P( G U R)= ( 4 + 12 ) / 16

P( G U R)= 16 / 16

P( G U R)= 1.000

Thus first answer is incorrect.

ii) P(R|E) = 0.375.

Using conditional rule of probability:

Red and Even: we have 12 red balls , so we have 6 red balls which are even.

Also we have 6 red even + 2 Green even = 8 Even balls.

Thus

Thus second answer is also incorrect.

iii) P(G ∪ E) = 0.625.

Using addition rule probability:

P(G ∪ E) = P(G ) + P(E ) - P( G ∩ E)

P(G ∪ E) = 4/16 + 8/16 - 2/16

( Since we have 2 Green even balls , P( G ∩ E) = 2/16 )

Thus

P(G ∪ E) = ( 4 + 8 - 2 ) / 16

P(G ∪ E) = 10 / 16

P(G ∪ E) = 0.625

Thus third answer is correct.

iv) P(G|R) = 0.500.

Using conditional rule of probability:

Since Green and Red are mutually exclusive events,

Thus

Thus fourth is incorrect.