# probability of sequential events

something has to travel from A to B. between these points there are two stages, where at each one it can fail with a probability of 0.3.

what is the total probability that the thing successfully travels from A to B?

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Note that it can fail either right away at point 1 (you know the probability of that) or (with probability that it does not fail at point 1) it can fail at point 2 with a probability that you know. Add these to get total probability of failure, F, then 1-F is the probability of success. –  Mr. F Mar 20 '12 at 15:16

If you assume that two stages are independent, then probability of success is multiplication of success of each stage. Hence $P = 0.7*0.7 = 0.49.$

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they are independent, but they are sequential. so if it fails at the first one, it doesnt even get to the second one. –  clamp Mar 20 '12 at 15:18
@clamp thats right. What I have mentioned is short cut. You can also find probability of failure which is $0.3*0.3$(fail in both stages) + $2*0.3*0.7$(fail in one stage but pass in other). Now subtract it by 1. You will get $1 - (0.09 + 0.42) = 0.49.$ –  quartz Mar 20 '12 at 15:22

I'll call your "it" a "car" and I'll assume that if the car fails at stage 1, then the car does not reach stage 2.

Presumably, the probability that the car fails at the second stage given that the car did not fail at the first stage is $0.3$. (Note then that if $S_1$ is the event "the car fails at stage 1" and $S_2$ is the event "the car fails at stage 2", then $S_1$ and $S_2$ are not independent.)

Let

$\ \ \ \ \ \ \ \ S_1$ be the event that the car fails at stage 1,

and let

$\ \ \ \ \ \ \ \ S_2$ be the event that the car fails at stage 2.

You can use the multiplication rule to find the probability that the car makes the journey: $$P(S_1^C\cap S_2^C)=P(S_1^C)P(S_2^C\mid S_1^C)=(0.7)(0.7)=.49.$$

Or, you can argue as follows:

Noting that $S_1\cap S_2=\emptyset$ and that $S_2\subset S_1^C$, the probability that the journey is unsuccessful is \eqalign{ P(S_1 \cup S_2)&=P(S_1)+P(S_2)\cr &=P(S_1)+P(S_2\cap S_1^C )\cr &= P(S_1)+P(S_2|S_1^C )P(S_1^C)\cr &=(0.3)+(0.3)(0.7)\cr &=0.51. } So the probability that the car makes the journey is $1-0.51=0.49$.

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