Daniel plays two games in the casino. The first game he believes he has 60% chance to win. If he wins the first game, he will win the second game with 25% chance, if he loses the first game, he will win the second game with 75% chance. The winning prices for the first and second game are $1 and $2 respectively A. What is the probability that he wins the second game?
B. If he wins the second game, what is the probability that he also won the first game?
C. What is the expected value for the price he wins from the two games? What is the variance of it?

1) the probability that he wins the second game is 0.45

2) the probability he wins the first game given that he won the second game is 0.333  

3) Expected Value is $ 1.5 and the variance is 0.75

Step-by-step explanation:

Given the data in the question;

Let Win1 represent first game victory

and Loss1 represent first game loss

Let Win2 denote second game victory

and Loss2 denote second game loss

now from the Tree diagram of Daniel;

P ( Win1 ) = 0.6

P ( Loss1 ) = 0.4

P ( Win2 | Win1 ) = 0.25

P ( Loss2 | Win1 ) = 0.75

P ( Win2 | Loss1 ) = 0.75

P ( Loss2 | Loss1 ) = 0.25

1) probability that he wins the second game will be;

P ( Win2 )  =  [P ( Win1 ) × P ( Win2 | Win1 )] + [P ( Loss1 ) × P ( Win2 | Loss1 ) ]

we substitute

= [0.6 × 0.25] + [0.4 * 0.75] = 0.45

Therefore the probability that he wins the second game is 0.45

2)

If he wins the second game, what is the probability that he also won the first game  will be;

{From Bayes' Theorem}

P ( Win1 | Win2 )  =  P ( Win1 ) × P ( Win2 | Win1 ) / P ( Win2 )

= 0.6 × 0.25 / 0.45 = 0.333  

Therefore the probability he wins the first game given that he won the second game is 0.333  

c)

winnings       probability

$0                  0.4 × 0.25 = 0.1

$1                   0.6 × 0.75 = 0.45

$2                  0.4 × 0.75 = 0.3

$3                  0.6 × 0.25 = 0.15

so Expected Value = ∑( winnings × individual win probability)

=  (0 × 0.1) + (1 × 0.45) + (2 × 0.3) + (3 × 0.15)

= 0 + 0.45 + 0.6 + 0.45

= $ 1.5

Variance = ∑( X² ) - (∑(X))²

∑( X² ) = Summation of ( Winnings² × Probability of Each Win )

so

Variance = 0 × 0 × 0.1 + 1 × 1 × 0.45 + 2 × 2 × 0.3 + 3 × 3 × 0.15 - ( 1.5 )²

Variance = 0 + 0.45 + 1.2 + 1.35 - 2.25

Variance = 3 - 2.25

Variance = 0.75

Therefore Expected Value is $ 1.5 and the variance is 0.75

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

it would open into a square depending on the type of envelope i’m not sure about the second part