Miguel is playing a game in which a box contains four chips with numbers written on them. two of the chips have the number 1, one chip has the number 3, and the other chip has the number 5. miguel must choose two chips, and if both chips have the same number, he wins $2. if the two chips he chooses have different numbers, he loses $1 (–$1). let x = the amount of money miguel will receive or owe. fill out the missing values in the table. (hint: the total possible outcomes are six because there are four chips and you are choosing two of them.) xi 2 –1 p(xi) what is miguel’s expected value from playing the game? based on the expected value in the previous step, how much money should miguel expect to win or lose each time he plays? what value should be assigned to choosing two chips with the number 1 to make the game fair? explain your answer using a complete sentence and/or an equation

Using the competed discrete probability table, the expected value of the equation would be -$0.50

Number of possible outcomes = 4

The probability distribution :

Expected value = Σ(X × P(x)) = (2 × 0.16667) + (-1 × 0.8333)

Expected value = - $0.50

The amount of money Miguel should expect to win or lose :

Since the expected value is negative, Hence, Miguel should expect to lose -$0.5 each time he plays.

To make the game fair :

Expected value should be equal to 0 ;

Let the value = v

(0.16667v) + (-0.8333) = 0

0.16667v = 0.8333

v = (0.8333 ÷ 0.16667)

v = $4.998

v = $5.0

Hence, to make the game fair, the winning value should be $5.0

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the answer is 45 i just did it on oddesy ware

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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