Probability

Question

Toss a coin 5 times. What is the probability that there are exactly 2 Heads and 3 tails?

 

Five men and three women applied for 2 scholarships. What is the probability that both the winners
will be men? (Hint:Name the applicants as Bob, Bill, David, Jeff, Steve, Laura, Mary and Kerrie.
List all the possible outcomes. List all the 28 outcomes.)

 

Five men and three women applied for 2 scholarships. What is the probability that the winners will
be a man and a woman?

Answer 1

5!/2!3! = 5*4/2 = 10

 

HHHHH

HHHHT                                      

HHHTH                                     

HHHTT

HHTHH

HHTHT

HHTTH

HHTTT  <------ 1

HTHHH 

HTHHT

HTHTH

HTHTT <---- 2

HTTHH

HTTHT <--- 3

HTTTH <--- 4

HTTTT

THHHH

THHHT

THHTH

THHTT <--- 5

THTHH

THTHT <----6

THTTH <--- 7

THTTT

TTHHH

TTHHT <--- 8

TTHTH <--- 9

TTHTT

TTTHH <--- 10

TTTHT

TTTTH

TTTTT

 

10/64 = 5/32 = 0.15625 = 15.625%

 

 

 

---------------------------------------

 

(8 choose 2) = 8*7/2 = 28 possible permutations

R is for Robert, not Bob; B is for Bill

 

(5 choose 2) = 5*4/2 = 10 men

 

RB    <--- 1

RD   <--- 2

RJ  <---3

RS  <--- 4

RL *

RM *

RK *

BD  <-- 5

BJ  <-- 6

BS <-- 7

BL *

BM *

BK *

DJ <-- 8

DS <--- 9

DL  *

DM *

DK *

JS <--- 10

JL  *

JM *

JK *

SL  *

Sm *

SK *

LM

LK

MK  

 

10/28 = 5/14 = 35 and 5/7 %

 

one man and one woman

 

(5 choose 1)(3 choose 1) = 15

 

15/28 = 53 and 4/7 % <---- marked with a star asterisk *

Answer 2

Five men and three women applied for 2 scholarships. What is the probability that the winners will
be a man and a woman?

 

a man can be chosen in 5C1 ways and a woman can be chosen in 3C1 ways; one of each can occur, therefore, in 5(3)=15 ways.

two persons can be chosen in 8C2 ways, i.e. 28 ways.

 

Therefore P(1M,1W) = 15/28