Respuesta :
Answer:
C.$1
Step-by-step explanation:
Since, when a die is rolled,
Then the total possible outcomes = 6 ( i.e. 1, 2, 3, 4, 5 or 6 )
Outcomes of getting number less than 3 = 2 ( i.e. 1 or 2 )
Outcomes of getting 3 or 4 = 2
Outcomes of getting 5 = 1,
Outcomes of getting 6 = 1,
[tex]\because \text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}[/tex]
Thus, the probability of getting number less than 3 = [tex]\frac{2}{6}=\frac{1}{3}[/tex]
The probability of getting 3 or 4 = [tex]\frac{2}{6}=\frac{1}{3}[/tex]
The probability of getting 5 = [tex]\frac{1}{6}[/tex]
The probability of getting 6 = [tex]\frac{1}{6}[/tex]
∵ If the die shows a number less than 3 then nothing will get; if the die shows a 3 or 4, we will get x dollars; if the die shows a 5, we will get $1; and if the die shows a 6, we will get $3
So, the expected value = [tex]x\times \frac{1}{3}+0\times \frac{1}{3}+1\times \frac{1}{6}+3\times \frac{1}{6}[/tex]
[tex]=\frac{x}{3}+\frac{1}{6}+\frac{1}{2}[/tex]
[tex]=\frac{2x+1+3}{6}[/tex]
[tex]=\frac{2x+4}{6}[/tex]
The game is fair if,
[tex]\frac{2x+4}{6}=x[/tex]
[tex]2x + 4 = 6x[/tex]
[tex]4 = 4x[/tex]
[tex]\implies x = 1[/tex]
Hence, the value of x would be $ 1.
The value of [tex]x[/tex] would be [tex]\$ 1[/tex]
Total possible outcomes when a die is rolled [tex]=6 (i.e, 1,2,3,4,5\; or \; 6)[/tex]
Outcomes of getting number less than [tex]3=2 ( i.e, 1\; or \;2)[/tex]
Outcomes of getting [tex]3\; or\; 4=2[/tex],
Outcomes of getting [tex]5=1,[/tex]
Outcomes of getting [tex]6=1,[/tex]
[tex]Probability=\dfrac{Favourable\; outcomes}{Total\; number\; of\; outcomes}[/tex]
Thus, the probability of getting number less than [tex]3 = \dfrac{2}{6}[/tex]
[tex]=\dfrac{1}{3}[/tex]
The probability of getting [tex]3\; or \;4 = \dfrac{2}{6}[/tex]
[tex]=\dfrac{1}{3}[/tex]
The probability of [tex]5=\dfrac{1}{6}[/tex]
The probability of [tex]6=\dfrac{1}{6}[/tex]
∴ According to question If the die show a number less than [tex]3[/tex] then nothing will be paid, if the die shows a [tex]3\;or\;4[/tex], we get [tex]x[/tex] dollars; if the die shows a [tex]5[/tex] , we get [tex]\$ 1[/tex]; and if the die shows a [tex]6[/tex], we will get [tex]\$ 3[/tex]
So Expected value [tex]=x\times \dfrac{1}{3} + 0\times \dfrac{1}{3} + 1\times \dfrac{1}{6} +3\times \dfrac{1}{6} \\[/tex]
[tex]=x\times \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{2}[/tex]
[tex]= \dfrac{2x+1+3}{6}[/tex]
The game is fair if,
[tex]\dfrac{2x+4}{6}=x[/tex]
[tex]2x+4=6x[/tex]
[tex]4=4x[/tex]
[tex]x=1[/tex]
Therefore , the value of [tex]x[/tex] would be [tex]\$ 1[/tex]
Hence option c is correct.
Learn more about probability concept here;
https://brainly.com/question/23044118?referrer=searchResults
