SOLUTION: The probability of an event A occurring in each of a series of independent trials is 0.666 Find the distribution function of the number of occurrences of in 9 trials

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Question 372037: The probability of an event A occurring in each of a series of independent trials
is 0.666 Find the distribution function of the number of occurrences of in 9 trials
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
9 occurences: P%289%29=1%2A%280.666%29%5E%289-0%29%2A%281-0.666%29%5E%280%29=0.0258
8 occurences: P%288%29=9%2A%280.666%29%5E%289-1%29%2A%281-0.666%29%5E%281%29=0.1164
7 occurences: P%287%29=36%2A%280.666%29%5E%289-2%29%2A%281-0.666%29%5E%282%29=0.2334
6 occurences: P%286%29=84%2A%280.666%29%5E%289-3%29%2A%281-0.666%29%5E%283%29=0.2731
5 occurences: P%285%29=126%2A%280.666%29%5E%289-4%29%2A%281-0.666%29%5E%284%29=0.2055
4 occurences: P%284%29=126%2A%280.666%29%5E%289-5%29%2A%281-0.666%29%5E%285%29=0.1030
3 occurences: P%283%29=84%2A%280.666%29%5E%289-6%29%2A%281-0.666%29%5E%286%29=0.0344
2 occurences: P%282%29=36%2A%280.666%29%5E%289-7%29%2A%281-0.666%29%5E%287%29=0.0074
1 occurences: P%281%29=9%2A%280.666%29%5E%289-8%29%2A%281-0.666%29%5E%288%29=0.0009
0 occurences: P%280%29=1%2A%280.666%29%5E%289-9%29%2A%281-0.666%29%5E%289%29=0.0001