Suppose that on each play of a game, a gambler either wins 1 with probability p or loses 1 with probability 1–p (or q). The gambler keeps betting until she or he is either up a total of n or down a total of m. What is the probability the gambler will quit an overall winner? You must consider both cases when p = 0.5 and when p ≠ 0.5

Step-by-step explanation:

From the given information,

Considering both cases when p = 0.5 and when p ≠ 0.5

the probability that the gambler will quit an overall winner is:

where ;

N.k = n  and k  = m

Hence, the probability changes to:

is  p ≠ 0.5   and k/N =  is P = 0.5

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option