Suppose that you have a possibly biased coin that gives heads with probability p , where p is unknown (

The answer to this question is  p(head)=p(tail)=1/2 .

Step-by-step explanation:

Let's flip the coin two times. When we have the respective possibilities.

Equation:

P(HH)=p2;        (i) //two time head.

P(HT)=p(1-p);    (ii)    //one head on tail

P(TH)=(1-p)p;     (iii)      //one tail on head.

P(TT)=(1-p)2.      (iv)  //two time tail.

In equation (ii) and (iii) that is P(HT)=P(TH) having the same value that is one head and one tail so we consider that both equations are the same.

So we consider  HT as a head and TH as a tail. both have equal chances of getting head or tail.

If we see HH or TT as results, we assume that this never occurred and replicate the action to make another set, renew until we get an HT or a TH, i.e. an unbiased toss. The probability of getting Head or Tail prevails the same for every step and we will be holding p(head)=p(tail)=1/2 out from the given process.

So, the process is to flip the coin double,

End if we receive HT or TH, repeat flipping unless.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

a) 7x=7

step-by-step explanation: