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A smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $0.7, 0.5$ and $0.3$ respectively. What is the probability that he successfully transfers his goods?

  • A] $0.105$
  • B] $0.5$
  • C] $0.245$
  • D] $0.045$

Is it as simple as calculating success in all three scenarios: $0.3 * 0.5 * 0.7 = 0.105$. Is A correct answer ?

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    $\begingroup$ Definitely looks correct to me $\endgroup$ – gt6989b Nov 25 '18 at 16:41
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    $\begingroup$ Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent. $\endgroup$ – BlackMath Nov 25 '18 at 18:17
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    $\begingroup$ It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included. $\endgroup$ – Teepeemm Nov 25 '18 at 19:37
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Let $A_i$ be an event $i$-th police stop him. We are interested in $$P(A_1'\cap A_2'\cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 \cdot 0.5 \cdot 0.7 = 0.105$$

(since $A_1, A_2, A_3$ are independant so are $A_1', A_2', A_3'$ ) so your answer is correct.

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You can think this in two ways. first way Not being caught on first check post and not being caught on second check post and not being caught on third post. i.e. $(1-P(A1))×(1-P(A2))×(1-P(A3) = 0.3×0.5×0.7 = 0.105$

second way Find P(being caught)

Caught on first check post Or Not being caught on first post and caught on second post Or Not being caught on first post and Not being caught on second post and caught on third post.

$P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3) =0.7 + 0.3×0.5 + 0.3×0.5×0.3 =0.895$

Now, P(not being caught) = 1- P(being caught) = 1- 0.895 =0.105

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