# Thread: Need help in finding the probability of events occuring?

1. ## Need help in finding the probability of events occuring?

Question:
i have a total of 16 events whereby the probability of 2 events occuring is 0.3, the probability of 6 events occuring is 0.003 and the probability of 8 events happening is 0.0003.

How can i find out the probability of 1 event occuring out of the 16 events?
How can i find out the probability of 5 events occuring out of the 16 events?
How can i find out the probability of 10 events occuring out of the 16 events?

Assumptions:
the 16 events occuring are considered to be independent events.

Note:
i cannot use binomial distribution to solve as there are 3 different probabilties for the 3 cases. is this question considered to be multinomial distribution?

2. Originally Posted by spidermman
Question:
i have a total of 16 events whereby the probability of 2 events occuring is 0.3, the probability of 6 events occuring is 0.003 and the probability of 8 events happening is 0.0003.

How can i find out the probability of 1 event occuring out of the 16 events?
How can i find out the probability of 5 events occuring out of the 16 events?
How can i find out the probability of 10 events occuring out of the 16 events?

Assumptions:
the 16 events occuring are considered to be independent events.

Note:
i cannot use binomial distribution to solve as there are 3 different probabilties for the 3 cases. is this question considered to be multinomial distribution?

I don't think there's enough information. Is this the entire question?

3. To make the question clearer, i meant that the 2 events have individual probability of 0.3 occuring, 6 events have individual probability of 0.003 occuring and 8 events have individual probability of 0.0003. Hence, there are a total of 16 individual events.

Assume that the 16 events will occur and they are independent,
i would like to find out that out of this 16 events, what is the probability that 5 events occured while 11 events does not occur?

and i would also like to find out the probability of number (from 0 to 16) of events occuring?

4. Originally Posted by spidermman
To make the question clearer, i meant that the 2 events have individual probability of 0.3 occuring, 6 events have individual probability of 0.003 occuring and 8 events have individual probability of 0.0003. Hence, there are a total of 16 individual events.

Assume that the 16 events will occur and they are independent,
i would like to find out that out of this 16 events, what is the probability that 5 events occured while 11 events does not occur?

and i would also like to find out the probability of number (from 0 to 16) of events occuring?
This question still makes no sense to me. Not for the least reason being that (2)(0.3) + (6)(0.003) + (8)(0.0003) = 0.6204 so one of the 16 events is not certain to occur. And since the events have different probabilities it makes no sense to ask about the probability of a particular number of events occuring.

Is this exactly how the question is stated?

5. The question goes like this: Probability of occuring
Event #1: 0.0003
Event #2: 0.3
Event #3: 0.003
Event #4: 0.003
Event #5: 0.0003
Event #6: 0.0003
Event #7: 0.003
Event #8: 0.0003
Event #9: 0.0003
Event #10: 0.0003
Event #11: 0.003
Event #12: 0.003
Event #13: 0.003
Event #14: 0.0003
Event #15: 0.3
Event #16: 0.0003

So lets call X{i} to be a bernoulli random variable that is 1 when Event #i happens. And Y = sum{Xi} for 1 to 16.

The question is what (i) P (Y=5), (ii) P(Y=10), (iii) P(Y=15) ?

6. Originally Posted by spidermman
The question goes like this: Probability of occuring
Event #1: 0.0003
Event #2: 0.3
Event #3: 0.003
Event #4: 0.003
Event #5: 0.0003
Event #6: 0.0003
Event #7: 0.003
Event #8: 0.0003
Event #9: 0.0003
Event #10: 0.0003
Event #11: 0.003
Event #12: 0.003
Event #13: 0.003
Event #14: 0.0003
Event #15: 0.3
Event #16: 0.0003

So lets call X{i} to be a bernoulli random variable that is 1 when Event #i happens. And Y = sum{Xi} for 1 to 16.

The question is what (i) P (Y=5), (ii) P(Y=10), (iii) P(Y=15) ?
Looks to me like $\Pr(Y = 5) = \sum_{a,b,c} \frac{5!}{a! \, b! \, c!} (0.3)^a (0.003)^b (0.0003)^c$

where a + b + c = 5 subject to the restrictions $0 \leq a \leq 2$, $0 \leq b \leq 6$ and $0 \leq c \leq 8$.

etc.

There might be a clever way of avoiding the tedium of working out the required combinations of a, b and c but I don't see it right now.