Probability of N unrelated events, each with different probabilities, what is the chance X number of outcomes occur Given the probability of N unrelated events, each with different probabilities, what is the chance X number of outcomes occur?
Said specifically there are 8 unrelated contracts, what is the chance a company wins 1,2,3,4,5,6,7, and 8 of them?
P(Winning Contract1) = 70%
P (Winning Contract 2) = 45%
P(C3)= 30%
P(C4)=50%
P(C5)=55%
P(C6)=610%
P(C7)=2%
P(C8)= 2%
Since there are (I think?) 8 choose 4 or 70 possible ways to win 4 contracts, and you have to sum those probabilities.
P(Winning one contract; not necessarily contract 1, but either 1, or 2, or 3, 4, or..) = ?
P (Winning two contracts) =?
P (Winning three contracts) = ?
P (Winning four contracts) = ?
…
P (Winning 8 contracts)?
Having a bit of trouble with this, help would be appreciated.

These calculations are time consuming. I give you an examplary calculation. Let´s   calculate the probability of winning 7 contracts. We have 8 different situations.


*

*Not winning contract 1 but winning the other contracts


The probability is $(1-0.7)\cdot 0.45 \cdot  0.3\cdot 0.5 \cdot 0.55 \cdot 0.61\cdot 0.02\cdot 0.02$


*

*Not winning contract 2 but winning the other contracts


The probability is $0.7\cdot (1-0.45) \cdot  0.3\cdot 0.5 \cdot 0.55 \cdot 0.61\cdot 0.02\cdot 0.02$


*

*Not winning contract 3 but winning the other contracts


The probability is $0.7\cdot 0.45 \cdot  (1-0.3)\cdot 0.5 \cdot 0.55 \cdot 0.61\cdot 0.02\cdot 0.02$


*

*Not winning contract 4 but winning the other contracts


The probability is $0.7\cdot 0.45 \cdot  0.3\cdot (1-0.5) \cdot 0.55 \cdot 0.61\cdot 0.02\cdot 0.02$


*

*Not winning contract 5 but winning the other contracts


The probability is $0.7\cdot 0.45 \cdot  0.3\cdot 0.5 \cdot (1-0.55) \cdot 0.61\cdot 0.02\cdot 0.02$


*

*Not winning contract 6 but winning the other contracts


The probability is $0.7\cdot 0.45 \cdot  0.3\cdot 0.5 \cdot 0.55 \cdot (1-0.61)\cdot 0.02\cdot 0.02$


*

*Not winning contract 7 but winning the other contracts


The probability is $0.7\cdot 0.45 \cdot  0.3\cdot 0.5 \cdot 0.55 \cdot 0.61\cdot (1-0.02)\cdot 0.02$


*

*Not winning contract 8 but winning the other contracts


The probability is $0.7\cdot 0.45 \cdot  0.3\cdot 0.5 \cdot 0.55 \cdot 0.61\cdot 0.02\cdot (1-0.02)$
Now all the probabilities have to be summed up to get the probability of winning 7 contracts.
In the case of not winning 4 contracts but winning 4 contracts ${8 \choose 4}=70$ situations have to be calculated.
