This may prove to be somewhat tricky.) Suppose that we have 10 coins, which are weighted so thatwhen flipped theith coin shows heads with probabilityp=i/10 (i= 1, . . . ,10). You randomly pick onecoin, you flip it, and it shows heads. Withoutputting this coin back, you pick a second coin. What is theprobability that the second coin shows heads, given that the first coin showed heads?

2/10

Step-by-step explanation:

Probability of getting head on i'th coin = i/10

Probability Pr of getting head on 2nd coin (Event A) = 2 / 10

Probability Pr of getting head on 1st coin (Event B) = 1/10

Probability A given B = Pr (A/B) = Pr (A∩B) / Pr B ;

where A∩B = Pr (A & B) = Pr A X Pr B

Putting in above formula :

Pr (A/B) = [(1/10)x(2/10)]  / 1/10

= 2/10

center= (-2,0)

radius= sqrt2/2

step-by-step explanation:

b

step-by-step explanation:

a difference is the result of subtraction. while each of these expressions has subtraction in it, a difference will be the result when subtraction is the last operation applied in the order of operations for the expression. this means subtraction must be outside of the parenthesis. only b has this.