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Old 07-29-2007   #1
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Default probability?

Two fair dice are tossed. Find the probability that the sum is odd, given that the sum is 5 or less.please explain your reasoning so I can understand...
Old 07-29-2007   #2
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You are given that the sum is 5 or less, so we should only consider those sums. If you toss a pair of dice, there are 10 ways to get a sum of 5 or less:1,1 1,2 1,3 1,42,1 2,2 2,33,1 3,24,1Remember that the order of the dice matters, so 1 on first die and 4 on second is different from 4 on first die and 1 on second.Looking at the sums that come up from these, you get2 3 4 53 4 54 55So, of these 10 sums, 6 are odd. The probability is 6/10 = 0.6.edit: Those of you who said that the probability is 6/36, you are not reading carefully enough. That is the probability of sum is 5 or less and sum is odd. That was not what was asked for. We are given that the sum is 5 or less, so we only look at those 10 possibilities.
Old 07-29-2007   #3
Kiwi Maths Teacher
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If you throw two dice together then the probabilities of interest are theseP(1)=0P(2)=1/36P(3)=2/36P(4)=3/36P(5)=4/36The answer to your question is[P(3)+P(5)]/[P(2)+P(3)+P(4)+P(5)]=[6/36]/[10/36]=3/5 or 0.6Here is the reasoningThe only outcomes of interest are if the total is 3 or 5hence the numeratorGiven that the sum is 5 or less then the denominator is the sum of the probabilities that the total is 2 or 3 or 4 or 5.Hope this helps.

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