Question

Need help question's screenshot given

Answer 1

http://snag.gy/CffXa.jpg

Answer 2

@ganeshie8
@UnkleRhaukus
@zepdrix
@paki

Answer 3

probability = (# of favorable outcomes )/( total # of outcomes)

Answer 4

start by finding the total number of outcomes in the sample space

Answer 5

\(A\) has \(n+1\) coins
that means there are \(n+2\) possibilities for number of heads : {0,1,2,...,n+1}

similarly for \(B\) there are \(n+1\) possibilities for number of heads

together the sample space contains \((n+1)(n+2)\) different outcomes

Answer 6

fine so far ? :)

Answer 7

total number of outcomes of a is 2^n and for b is 2^n+1

Answer 8

am i right????

Answer 9

ohhh i understood but we need the sample space as the total outcome of n+1 and n coins, then y taking s as outcome of heads ???

Answer 10

nope, our event is "counting number of heads in a toss"

Answer 11

because later we are going to compare the heads count between A and B

Answer 12

ok i understand then should we use p(a>B)

Answer 13

kindof.. but its going to be simpler than that... we just count the number of favorable outcomes and take ratio

Answer 14

so do we agree that the total number of outcomes equal (n+2)(n+1) ?

Answer 15

yes

Answer 16

good, lets go ahead and find the outcomes in favor

Answer 17

we want A to have more heads than B

Answer 18

suppose A has 1 head, then B can have only 0 heads. yes ?

Answer 19

lets create a quick table

Answer 20

yes

Answer 21

number of heads in A toss | possible number of heads in B toss
1 | {0} : `1` way
2 | {0,1} : `2` ways
3 | {0,1,2} : `3` ways
....
n+1 | {0, 1,2,...,n} : `n+1` ways

Answer 22

adding them up you get :
1 + 2 + 3 + ... + (n+1)

thats the sum of first n+1 positive integers ?

Answer 23

that equals (n+1)(n+2)/2

Answer 24

so number of favorable outcomes = (n+1)(n+2)/2

take the ratio for the probability

Answer 25

then the ratio is
p(a>b) = ((n+1)(n+2)/2)/(n+1)(n+2) =1/2

Answer 26

am i right ??/

Answer 27

Looks good!

Answer 28

thanks