Probability of biased die

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bandula

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Please give correct method for find probability.

Rall a biased die . The probability that even number appears is twice as that of an odd number appears. Let A be the event that number less than 3 appers.
find probability of A.
A ={ 1,2} 1,2 are odd and even 3 , 4 & 5,6 also other odd and even set there are 3 sets hence P(A) =1/3 is that P(A) correct ? plese tell definit method for find P(A)
 
An ordinary die has six numbers and the probability of any one showing after a toss is 16\dfrac{1}{6}
So we could modal that as six numbered identical cubes in a box
Thus with your bias die we have nine faces {1,2,2,3,4,4,5,6,6}\{1,2,2,3,4,4,5,6,6\} drawing an even number is twice as likely rolling an odd.
What is the probability of rolling a number less than three?
 
bandula said:
Please give correct method for find probability.

Rall a biased die . The probability that even number appears is twice as that of an odd number appears. Let A be the event that number less than 3 appers.
find probability of A.
A ={ 1,2} 1,2 are odd and even 3 , 4 & 5,6 also other odd and even set there are 3 sets hence P(A) =1/3 is that P(A) correct ? plese tell definit method for find P(A)
Click to expand...
Yes, that is completely valid. Here is a litte more detailed, a little harder, method but gives a check:

Let p be the probabillity that a specific odd number, 1, 3, or 5, is rolled. Then the probability that 2, 4, or 6 is rolled is 2p each. So we must have p+ 2p+ p+ 2p+ p+ 2p= 9p= 1. p= 1/9.

The probability a 1, odd, is rolled is 1/9 and the probability a 2, even, is rolled is 2/9. The probability 1 or 2 is rolled is 1/9+ 2/9=3/9= 1/3.
 
Roll a biased die. The probability that an even number appears is twice that of an odd number appearing.
P({2,4,6}) = 2 [MATH]\times[/MATH] P({1,3,5})
P({1,2})=?
 
HallsofIvy said:
Yes, that is completely valid. Here is a litte more detailed, a little harder, method but gives a check:

Let p be the probabillity that a specific odd number, 1, 3, or 5, is rolled. Then the probability that 2, 4, or 6 is rolled is 2p each. So we must have p+ 2p+ p+ 2p+ p+ 2p= 9p= 1. p= 1/9.

The probability a 1, odd, is rolled is 1/9 and the probability a 2, even, is rolled is 2/9. The probability 1 or 2 is rolled is 1/9+ 2/9=3/9= 1/3.
Click to expand...
thanks a lot
 
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