A two fair dice is thrown . find the probability distribution of even number?
Answers
Explanation:
If the dice is six sided, only 50% of the first dice fulfill the requirement of being even (2,4,6). If the other dice only needs to be even (not identical to the first) we need one of the three even numers to pop up on the second dice from the six available ones. So here again the chances are 3/6 or 1/2. So just getting two even dice (not necessarily equal), has a 1/2 ×1/2 = 1/4 (25%) chance.
Lets answer the question slightly differently now. What about two identical even numbers?
Now what if the second dice needs to be identical to the first (which was even), making it only a 1 in 6 or 1/6 chance of getting the specific even number of the first dice.
As both criteria need to be fulfilled symultaneously we need to multiply 1/2 by 1/6 making the answer (1×1)/(2×6) or 1/12. So the chance would be 1/2 (about 8%). But this is if you throw the dice one after the other so the second can target the outcome of the first.
Strangely enough this does not feel correct, You would expect a lower chance. Just thinknabout it. There is a 1 in 36 (1/36) chance 1/6 × 1/6= 1/36 chance of getting identical dice if you throw them together. Now we only want even identicals. As we only identify even or uneven pairs, we have a chance of 1 in 2 (1/2) that our 1/36 dubbles is actually an even dubbel. So only half of 1 in 36, or 1 in 72 (1/72) will fulfill the requirements. So the chance of throwing 2 identical even dice is in this calculation 1,4%.
But in this last case you could argue that in the 1/36 above (where we got identical dice, even or uneven) already half of them are even, making the chance of getting an even and identical set of two dice not 1/72 but 1/18! The sentence above could have read ”As we only identify even or uneven pairs, we have a chance of 1 in 2 (1/2) that our 1/36 dubbles is actually an even dubbel. So only half of 1 in 36, or 1 in 18 (1/18) will fulfill the requirements.