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The sample space is 62 or 36. The easier way to do this problem is to find the probability of doubles or 10 and subtract from 1. The probability of a double is 6/36 (1,1; 2,2; 3,3; 4,4; 5,5; 6,6). There are 3 ways to get a 10; 4,6; 6,4; & 5,5. We already counted 5,5 in the double so there are 2/36 we add to the 6/36 to obtain 8/36. So, 1-8/36 = 28/36 or 7/9 or 0.7778 probability that neither double nor a total of 10 will appear.

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Q: Two unbiased dice are thrown Find the probability that neither double nor a total of 10 will appear?
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Two events A and B with probability 0.5 and 0.7 respectively have joint probability of 0.4 The probability that neither A nor B happens is?

Let me denote -A as the event that A does not happen. So we want Pr[-(A and B)] Now, the event that neither A nor B occurs is the opposite of either A occurring, or B occurring or both occurring. So Pr[-(A and B)] = 1 - Pr(A or B)= 1 - [Pr(A) + Pr(B) - Pr(A and B)] (since A+B is double counted)= 1 - (0.5 + 0.7 - 0.4)= 1 - 0.8= 0.2


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Related questions

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Two events A and B with probability 0.5 and 0.7 respectively have joint probability of 0.4 The probability that neither A nor B happens is?

Let me denote -A as the event that A does not happen. So we want Pr[-(A and B)] Now, the event that neither A nor B occurs is the opposite of either A occurring, or B occurring or both occurring. So Pr[-(A and B)] = 1 - Pr(A or B)= 1 - [Pr(A) + Pr(B) - Pr(A and B)] (since A+B is double counted)= 1 - (0.5 + 0.7 - 0.4)= 1 - 0.8= 0.2


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