Some Notes

Homework 8, Qn. 4

Make a list, in steps:
First Baby posseblities

B
G

Possibilities with the Second Baby:

BB
BG
GB
GG

Possibilities with the Third Baby

BBB
BGB
GBB
GGB
BBG
BGG
GBG
GGG

Possibilities with the Fourth Baby
BBBB
BGBB
GBBB
GGBB
BBGB
BGGB
GBGB
GGGB
BBBG
BGBG
GBBG
GGBG
BBGG
BGGG
GBGG
GGGG

The Complete list

Possibilities with the FIFTH BABY:
BBBBB
BGBBB
GBBBB
GGBBB*
BBGBB
BGGBB*
GBGBB*
GGGBB
BBBGB
BGBGB*
GBBGB*
GGBGB
BBGGB*
BGGGB
GBGGB
GGGGB

BBBBG
BGBBG*
GBBBG*
GGBBG
BBGBG*
BGGBG
GBGBG
GGGBG
BBBGG*
BGBGG
GBBGG
GGBGG
BBGGG
BGGGG
GBGGG
GGGGG

Total number of outcomes in the sample space S, is

n(S) = 32

E be the event that there are exactly 3 boys and 2 girls. Outcomes in E are marked *.
The number of outcomes in E is

n(E) = 10.

P(E) = n(E)/n(S) = 10/32 = .3125

Homework 8, Qn. 8

Three members of your family (two parents and yourself) went a buy three cell phones. The phones come in three colors: red, blue and black. You all pick them randomly. What is the probability that all three phones will be red or blue?

Solution:

Use the noraion R = red, B=blue and A = Black

Make a list by steps:
First member's choices:

R
B
A

Choices of first and second member together

RR
BR
AR
RB
BB
AB
RA
BA
AA

Final List of Choices

Choices of all three members together

RRR
* BRR
* ARR
RBR
* BBR
* ABR
RAR
BAR
AAR
RRB
* BRB
* ARB
RBB
* BBB
* ABB
RAB
BAB
AAB
RRA
BRA
ARA
RBA
BBA
ABA
RAA
BAA
AAA

The sample space S consists of the above complete list.

So, n(S) = 27

Let E be the event that all three phones will be red or blue
Outcomes in E are marked by *.
So, n(E) = 8

So, P(E) = n(E)/n(S) = 8/27