Answer: Our required probability is 0.83.
Step-by-step explanation:
Since we have given that
Number of dices = 2
Number of fair dice = 1
Probability of getting a fair dice P(E₁) = 
Number of unfair dice = 1
Probability of getting a unfair dice P(E₂) = 
Probability of getting a 3 for the fair dice P(A|E₁)= 
Probability of getting a 3 for the unfair dice P(A|E₂) = 
So, we need to find the probability that the die he rolled is fair given that the outcome is 3.
So, we will use "Bayes theorem":

Hence, our required probability is 0.83.
Y coordinate on solving both equations comes out to be 0
-2x+3y=-6
3y = -6+2x
put the value of 3y in equation 2nd
5x-2(-6+2x) =15
5x+12-4x = 15
x=3
put value of X in 3y = -6+2x
3y = -6+2*3 = -6+6 = 0
thus y = 0
Answer:
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
Indicates leg lengths of 1 and√3 and hypotenuse 2, the desired ratio is √3/2