Question

(1 point) A game of chance involves rolling an unevenly balanced 4-sided die. The probability that a roll comes up 1 is 0.21, the probability that a roll comes up 1 or 2 is 0.44, and the probability that a roll comes up 2 or 3 is 0.55 . If you win the amount that appears on the die, what is your expected winnings? (Note that the die has 4 sides.)

Answer 1

Here we have given the 4 sided dice 1 , 2 , 3 , 4

Consider X : be the number occurs on dice

P ( X= 1 ) = 0.21

P ( X = 1 or X=2 ) = 0.44

P (X= 2 or X = 3 ) = 0.55

In dice all the events are mutually exclusive i,e there is no common between any two event

Addition rule of probability

P ( A or B ) = P ( A ) + P ( B ) - P ( A and B )

Because of mutually exclusive property

P ( A and B ) = 0

So we get formula as

P ( A or B ) = P ( A ) + P ( B )

Now

P ( X = 1 or X = 2 ) = P ( X = 1 ) + P ( X = 2 )

We have given P ( X = 1 ) = 0.21   and P ( X = 1 or X = 2 ) = 0.44

we plug the value in above formula

0.44 = 0.21 + P ( X = 2 )

P( X =2 ) = 0.44 - 0.21 = 0.23

P ( X = 2 ) = 0.23

Now to find P (X = 3 )

P ( X = 2 or X = 3 ) = P ( X = 2 ) + P ( X = 3 )

We have calculated above P ( X = 2 ) = 0.23

P ( X= 2 or X = 3 ) = 0.55 ( given )

P ( X = 2 or X = 3 ) = P ( X = 2 ) + P ( X = 3 )

0.55 = 0.23 + P ( X = 3 )

P ( X = 3 ) = 0.55 - 0.23 = 0.32

P ( X = 3 ) = 0.32

Now we calculate

P (X =4 )

We know that sum of probability is always 1

So we get

P ( X = 1 ) + P ( X = 2 ) +P ( X = 3 ) + P ( X = 4 )= 1

P( X= 1 ) = 0.21 , P( X = 2 ) = 0.23,   P( X = 3 ) = 0.32 ,

0.21 + 0.23 + 0.32 + P(X=4)= 1

0.76 + P (X=4) = 1

P (X=4) = 1- 0.76 = 0.24

P(X=4) = 0.24

All calculated probabilities

P( X= 1 ) = 0.21 ,

P( X = 2 ) = 0.23,  

P( X = 3 ) = 0.32 ,

P(X=4) = 0.24

Formula for expected value

x	P(X) = probability	x*p(X)
1	0.21	=1*0.21 =0.21
2	0.23	=2*0.23=0.46
3	0.32	=3*0.32 = 0.96
4	0.24	=4*0.24=0.96

Expected winning = 2.59

Final answer :

Expected winning = 2.59

I hope this will help you :)