Question

How can i know how to calculate the odds of a current passing in an electrical circuit?

each switch is closed in 0.7 odds(and then electricity can be passed). also, each switch is working independently of others.

1)what are the chances that a current(electrical current) is passing from A to B?

2)if a current is not passing, what are the chances that 3 switches are closed?

Answer 1

`"Part 1) 0.80164"`
`"Part 2) 0.31125"`

Explanation:

`"There are 5 switches than can be open or closed."`
`"Hence there are at most "2^5 = 32" cases to investigate."`
`"We can take a few shortcuts though :"`
`"If both 1 & 4 are open OR both 2 & 5 are open, current"`
`"cannot pass."`
`"So (1 OR 4) AND (2 OR 5) must be closed."`
`"But there are additional criteria :"`
`"If (4 & 2) are open, 3 must be closed."`
`"If (1 & 5) are open, 3 must be closed."`
`"So if we note (O,C,O,C,C) as 1,and 3 open and 2,4,5 closed,"`
`"we have following cases only, that can work : "`

(C,C,&,&,&)
(C,&,C,&,C)
(&,C,C,C,&)
(&,&,&,C,C)

`"Note that there is overlap with the & notation that indicates"`
`"that a gate might be either open or closed."`
`"So we must be careful in extracting all cases from this."`

`"The first case has 8 possibilities because of the 3 stars."`
`"The second only 2 additional possibilities as if the first star is"`
`"equal to C, we are in case 1."`
`"The third has also 2 additional possibilities for the same reason."`
`"The last one has 4 additional possibilities :"`

(O,O,&,C,C), and
(C,O,O,C,C), (O,C,O,C,C)

`"The odds for case 1 are "0.7^2 = 0.49"`
`"The odds for the additional possibilities in case 2 are "0.7^3*0.3`
`"Same for case 3."`
`"Case 4 : "0.3^2*0.7^2 + 0.7^3*0.3^2 + 0.7^3*0.3^2`

`"So we have in total :"`

`"0.49 + 0.1029 + 0.1029 + 0.0441 + 0.03087 + 0.03087"`
`"= 0.80164"`

`"Part 2 is only the case if 1&4 both open or 2&5 both open,"`
`"and the rest closed. The odds for that are"`

`0.3^2*0.7^3 + 0.3^2*0.7^3 = 0.06174`

`=> 0.06174 / (1 - 0.80164) = 0.31125`