Four balls are selected at random

Question

Four balls are selected at random WITHOUT replacement from an urn with 3 WHITE and 5 BLUE balls. Calculate the probability that:

 

Two balls will be white and 2 balls will be blue

 

All four balls will be the same color

 

EXACTLY THREE of the balls will be of one color

Answer 1

Possible draws

wwwb-ng3

wwbw-ng 3

wwbb

wbww-ng 3

wbwb

wbbw

wbbb-ng 3

bwww- ng 3

bwwb

bwbw

bwbb-ng 3

bbww

bbwb-ng3

bbbw-ng 3

bbbb-ng

 

These are good 

PwPwPbPb=3/8(2/7)(5/6)(4/5)
=1/14

PwPbPwPb=3/8(5/7)(2/6)(4/5)
=1/14

PwPbPbPw=3/8(5/7)(4/6)(2/5)
=1/14

PbPwPwPb=5/8(3/7)(2/6)(4/5)
=1/14

PbPwPbPw=5/8(3/7)(4/6)(2/5)
=1/14

PbPbPwPw=5/8(4/7)(3/6)(2/5)

=1/14

 

Sum all these up

The probability of drawing two whites and two blues is 6/14

=42%

 

 

all Four balls will be the same color. Since there are only three white balls these must be four blue balls

Pbbbb=(5/8)(4/7)(3/6)(2/5)

=(1/7)(3/6)=1/14

 

=7.1%

 

 

Possible draws with three colors
wwwb=3/8(2/7)(1/6)(5/5)

=1/56
wwbw=3/8(2/7)(5/6)(1/5)

=1/56
wbww=3/8(5/7)(2/6)(1/5)

=1/56
wbbb=3/8(5/7)(4/6)(3/5)

=3/28
bwww=5/8(3/7)(2/6)(1/5)

=1/56

bwbb=5/8(3/7)(4/6)(3/5)

=3/28
bbwb=5/8(4/7)(3/6)(3/5)

=3/28
bbbw=5/8(4/7)(3/6)(3/5)

=3/28

p3=4/56+12/28=1/14+6/14
=7/14=1/2

50%