Independent events

P

Paneczko

New member
Joined
Aug 4, 2020
Messages
3
I'm new to probability and i'm reading DeGroot's probability and statistatistics, so i have a problem in exercise 1.11.11

Suppose that the probability that any particle emmited by a radioactive material will penetrate a certain shield is 0.01. If 10 particles are emmited what is the probability that at least one will penetrate de shield?

I think about the sommatory from i=1 to n=10 of ((0.01)^i)((0.99)^(10-i))(11-i)

But i'm not sure, can someone help me please?
 
Paneczko said:
I'm new to probability and i'm reading DeGroot's probability and statistatistics, so i have a problem in exercise 1.11.11

Suppose that the probability that any particle emmited by a radioactive material will penetrate a certain shield is 0.01. If 10 particles are emmited what is the probability that at least one will penetrate de shield?

I think about the sommatory from i=1 to n=10 of ((0.01)^i)((0.99)^(10-i))(11-i)

But i'm not sure, can someone help me please?
Click to expand...
You can do a summation similar to that; but what is the (11-i) for?

There is an easier way, though: first find the probability that NONE of the 10 particles will penetrate.
 
Dr.Peterson said:
You can do a summation similar to that; but what is the (11-i) for?

There is an easier way, though: first find the probability that NONE of the 10 particles will penetrate.
Click to expand...

Thank you! I used the (11-i) in the final of summation because i taught that the probability that exactly one particle penatrate is (0.01)((0.99)^9)10 (this one i know is right)

And for exactly two particles to penatrete i taught that the probability is:

((0.01)^2)((0.99)^8)9! (Like, 9! arrangements)

And so on! But i will try what you said! Thank you so much! I already spot some errors in my first resolution
 
Have you learned about the binomial distribution? That is what you need to use, if you don't use the much quicker alternative I suggested.
 
You must log in or register to reply here.
Top