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The remake of The Taking of Pelham 1 2 3 starring John Travolta and Denzel Washington hits theaters today, and while we've posted lists in the past that tied into the week's premieres, I felt like this was the perfect opportunity to spotlight one ...
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Some 61 positions were lost "primarily through attrition" (a hiring freeze, in place since Jan. . 1. ). In all, these actions will pare the Met's full-and part-time work force to about 2200 in the fiscal year beginning July . 1. . ...
What is the probability that at least 1 of 2 will occur?
Q. I am doing a problem for statistics class that deals with the following question: What is the probability that at least 1 of 2 randomly selected batteries will be rejected by plan A? I have already determined that the probability of 1 battery being rejected is 2.5%, so would would it be for at least 1 of 2? Would you just to 2.5%^2, or is there a different procedure for "at least"? Thanks. @SV... So basically I would add the probability of those two instances? The probability of 1 would be 2.5%, and the probability of 2 would be 2.5%^2, so the answer would come out to: 8.75% Is this correct? I recommend everyone check out this answer, it helped me the most, especially zantu3's solution:
Asked by tdude51 - Wed Oct 3 18:33:51 2007 - - 2 Answers - 0 Comments
A. The probability that at least one is rejected is: Pr[first is rejected] + Pr[second is rejected] - Pr[both are rejected] 0.025 + 0.025 - 0.025^2 = 0.049375 = 4.9375% The reason we have to subtract off Pr[both are rejected] is because it is counted in Pr[first is rejected] and in Pr[second is rejected].
Answered by Phineas Bogg - Wed Oct 3 18:52:32 2007
Q. I am doing a problem for statistics class that deals with the following question: What is the probability that at least 1 of 2 randomly selected batteries will be rejected by plan A? I have already determined that the probability of 1 battery being rejected is 2.5%, so would would it be for at least 1 of 2? Would you just to 2.5%^2, or is there a different procedure for "at least"? Thanks. @SV... So basically I would add the probability of those two instances? The probability of 1 would be 2.5%, and the probability of 2 would be 2.5%^2, so the answer would come out to: 8.75% Is this correct? I recommend everyone check out this answer, it helped me the most, especially zantu3's solution:
Asked by tdude51 - Wed Oct 3 18:33:51 2007 - - 2 Answers - 0 Comments
A. The probability that at least one is rejected is: Pr[first is rejected] + Pr[second is rejected] - Pr[both are rejected] 0.025 + 0.025 - 0.025^2 = 0.049375 = 4.9375% The reason we have to subtract off Pr[both are rejected] is because it is counted in Pr[first is rejected] and in Pr[second is rejected].
Answered by Phineas Bogg - Wed Oct 3 18:52:32 2007
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