Suppose you and a friend each choose at random an integer between 1 and 8.?

For example, some possibilities are (3,7), (7,3), (4,4), (8,1), where your number is written first and your friends number second. Find


(a) p(you pick 5 and your friend picks 8)


(b) p(sum of the two numbers picked is %26lt;4)


(c) p(both numbers match)


(d) p(the sum of the two numbers is a prime)


(e) p(your number is greater than your friends number)

Suppose you and a friend each choose at random an integer between 1 and 8.?
The probability of an event is the number of equally likely ways it can happen divided by the total number of equally likely things that can happen.





In this case, there are 64 different outcomes. Now, just count the ones that count as "Hits"





a) you pick 5 and he picks 8 can only happen 1 way. 1/64


b) sum %26lt;4. (1,1) (1,2) (2,1). 3 ways or 3/64


c) match (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) (7,7) (8,8) or 8/64


d) primes between 2 and 16 are 2, 3, 5, 7, 11, 13


2 = (1,1)


3 = (1,2) (2,1)


5 = (1,4) (2,3) (3,2) (4,1)


7 = (1,6) (2,5) (3,4) (4,3) (5,2) (6,1)


11 = (3,8) (4,7) (5,6) (6,5) (7,4) (8, 3)


13 = (5,8) (6,7) (7,6)


or 22/64


e)7 ways if he picks 1, 6 ways if he picks 2, 5 if he picks 3, 4,3,2,1 = 28 ways or 28/64
Reply:I'll just do c).





Suppose both friends choose 1.


P(first friend choosing 1) = 1/8


P(second choosing 1) = 1/8


P(both choosing 1) = 1/8* 1/8


Now there are 8 ways they can choose the same number.


So P(same number) = 8* 1/8* 1/8 = 1/8
Reply:(a)


1/8 * 1/8 = 1/ 64





(b)


The numbers could be: (1,1) (2,1) (1,2).


That's 3 sets out of 64: = 3/ 64.





(c)


This could happen eight times.


8/ 64 = 1/ 8.





(d)


Consider all of the sets of numbers.


(1+1)(1+2)(1+3)(1+4)(1+5)(1+6)(1+7)(1+...


(2+1)(2+2)(2+3)(2+4)(2+5)(2+6)(2+7)(2+...


(3+1)......... →





↓ ↓





(8+1)(8+2)(8+3)(8+4)(8+5)(8+6)(8+7)(8+...


Now when the numbers are added, the values range between 2 and 16.


So 5, 7, 11 and 13 are the prime numbers possible.


4/ 64 = 1/ 16








(e)


Consider all of the sets of numbers.


(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,...


(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(2,7)(2,...


(3,1)......... →





↓ ↓





(8,1)(8,2)(8,3)(8,4)(8,5)(8,6)(8,7)(8,...





It can be seen that the total number of times your number is greater then your friends is:


1 + 2 + 3 + 4 + 5 + 6 + 7 = 28


28/ 64 = 7/ 16
Reply:sosy
Reply:hmmm i am confused an what your asking

quince