# Thread: The Binomial Distribution

1. ## The Binomial Distribution

can anyone help?

A circuit has 6 breakers. The probability that each breaker will fail is
0.1. If the circuit is activated, .find the probability that exactly two
breakers will fail. Each breaker is independent of any other breaker.

2. Originally Posted by jickjoker
can anyone help?
A circuit has 6 breakers. The probability that each breaker will fail is
0.1. If the circuit is activated, .find the probability that exactly two
breakers will fail. Each breaker is independent of any other breaker.
You should understand that this is not a tutorial service.
You can look it up.

3. Originally Posted by jickjoker
can anyone help?

A circuit has 6 breakers. The probability that each breaker will fail is
0.1. If the circuit is activated, .find the probability that exactly two
breakers will fail. Each breaker is independent of any other breaker.
What have you tried? Where are you stuck?

4. just trying to verify that i have worked it out correctly, i like to do brainteazers in my spare time and this
one come up , i think the answer is

6c2 , 0.1 *2 , 0.9 *4 = 0.098415
9.8% chance
i think thats correct

5. Originally Posted by jickjoker
just trying to verify that i have worked it out correctly, i like to do brainteazers in my spare time and this
one come up , i think the answer is
6c2 , 0.1 *2 , 0.9 *4 = 0.098415
9.8% chance
i think thats correct
That is correct: $\binom{6}{2}(.1)^2(.9)^4=0.098415$