If henry plays both games conservatively (cc), find the probability that henry will earn
a. 2 points
b. 1 1 /2 points
c. 1 point
d.1 / 2 points
e. 0 points

a) for 2 points,

Pa = Pww = 0

b) for 1.5 point,

Pb = Pwt + Ptw = 0 + 0 = 0

c) for 1 point

Pc = Pwl + Ptt + Plw = 0 + 0.64 + 0 = 0.64

d) for 0.5 point

Pd = Ptl + Plt = 0.16 + 0.16 = 0.32

e) for 0 point

Pe = Pll = 0.04

Step-by-step explanation:

The remaining part of the question is attached.

Given;

According to the rules of the game

Win = 1 point

Tie = 1/2 point

Lose= 0 point

If henry play the game conservatively, the probability of

Win = 0

Tie = 0.8

Lose = 0.2

Playing the game twice and conservatively, the following outcomes are possible with the corresponding points and probabilities.

ww = 2 points Pww = 0 × 0 = 0

wt = 1.5 point Pwt = 0 × 0.8 = 0

wl = 1 point Pwl = 0 × 0.2 = 0

tw = 1.5 point Ptw = 0.2 × 0 = 0

tt = 1 point Ptt = 0.8 × 0.8 = 0.64

tl = 0.5 point Ptl = 0.8 × 0.2 = 0.16

lw = 1 point Plw = 0.2 × 0 = 0

lt = 0.5 point Plt = 0.2 × 0.8 = 0.16

ll = 0 point Pll = 0.2 × 0.2 = 0.04

Where w = win , t = tie and l = lose.

a) for 2 points,

Pa = Pww = 0

b) for 1.5 point,

Pb = Pwt + Ptw = 0 + 0 = 0

c) for 1 point

Pc = Pwl + Ptt + Plw = 0 + 0.64 + 0 = 0.64

d) for 0.5 point

Pd = Ptl + Plt = 0.16 + 0.16 = 0.32

e) for 0 point

Pe = Pll = 0.04

answer: slight chance that it will happen

step-by-step explanation: