A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2? I find: p = 1/6 But the answer is: 0,4129
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Which side of 17,5 do you want? 1 - 0,587064 = 0,412936 That looks pretty close.
Originally Posted by Apprentice123 A given honest is launched 100 times in a row. What is the probability that in 18 or more of these releases occur side 2? I find: p = 1/6 But the answer is: 0,4129 You have calculated the probability of fewer than 18 not 18 or more. CB
Not is P(X >= x) = (P X >= x-0,5) ???
Originally Posted by Apprentice123 Not is P(X >= x) = (P X >= x-0,5) ??? What does that mean? (I'm betting whatever you intend it to mean the answer is no) CB
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