Problem Solving

I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be
0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?

A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?

(A) 0.06
(B) (0.06)^10
(C) 1 - (0.06)^10
(D) (0.94)^10
(E) 1 - (0.94)^10

OA E

massi2884 wrote:I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be
0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?

A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?

(A) 0.06
(B) (0.06)^10
(C) 1 - (0.06)^10
(D) (0.94)^10
(E) 1 - (0.94)^10

OA E

Probability of String does not fail = (1-0.06)^10; All 10 bulbs are glowing.

Hence, to fail; prob. = 1- prob. of success = 1 - (1-0.06)^10 = 1-0.94^10.

Yes you should always ask such question to gain a better understanding of all concepts.

Let the event of the first light bulb failing be E1, for the 2nd be E2 and so on.
OK what we need is the Union of the events E1, E2... E10 i.e., P(E1 OR E2 OR E3 OR .... OR E10)
From sets we know that n(A union B) = n(A) + n(B) - n(A intersection B)..
Extending to probability P(E1 OR E2) = P(E1) + P(E2) - P(E1 AND E2)
Such a formula can also be derived for 10 events. But the Key here is that its not simply equal to the sum of n(A) and n(B). When you simply add the probability you are also adding the occurance of events like {E1 AND E2} etc. twice.
Now if you want to find this union, one way is to derive such a formula for 10 events and then calculate the probability using it which is very tedious.

So a better way is to find the probability of the P(NOT E1 AND NOT E2...... AND NOT E10) and subtract it from 1 as aptly demonstrated by others.
P(E1 OR E2 OR E3 OR .... OR E10) = 1-P(NOT E1 AND NOT E2...... AND NOT E10)
= 1-P(NOT E1)*P(NOT E2)*....*P(NOT E10) ---- since independent events
= 1-[(1-P(E1))* (1-P(E2)) *(1-P(E3))* ..... *(1-P(E10))]