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Target gets hit by three missiles.

The probability of hitting the target with the
- First missile is 0.4;
- Second missile is 0.5;
- Third missile is 0.7;

The target has a probability of 0.2 getting destroyed with one missile, 0.6 by two missiles and for sure (1) by three missiles.

What's the probability of the target being destroyed?

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This is the exact question on the book, teacher told us that the result is 0.458 but didn't describe the method of solving, except saying that it's not easy.

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    $\begingroup$ you need to start by getting the probability of exactly 1,2 and 3 hits separately - being hit by 3 missiles is simplest, all missiles have to hit - can you do that calculation? Two missile hits can happen in 3 different ways 1,2 or 3 has to miss - can you work out the probabilities of the 3 distinct events and add them together - 1 hit is similar to 2 hits, 1 hits and 2 misses - you have to consider the probability of hits and misses in your calculations - $\endgroup$
    – Cato
    Commented Jan 31, 2017 at 10:26
  • $\begingroup$ I see. So, let's say: D -> target being destroyed; M1,M2,M3 -> Probabilities of missiles; H1,H2,H3 -> the three scenarios you mentioned, where P(H3) is the easiest to find (0.4*0.5*0.7), since all the targets should hit. So I have to basically find P(D) which is P(D/H1)*P(H1) + P(D/H2)*P(H2) + P(D/H3)*P(H3)? @Cato $\endgroup$
    – bashbin
    Commented Jan 31, 2017 at 10:39
  • $\begingroup$ yes I would imagine it would end up in that form - an example of 1 missile hit is that just 1 hits, P=.4*.5*.3 - but that can happen in 3 ways, but the 3 ways exclude each other $\endgroup$
    – Cato
    Commented Jan 31, 2017 at 10:41
  • $\begingroup$ Thanks!! I was having problems finding P(H1) and P(H2) earlier, but thanks to you ringing me the bell that there can be 3 ways for these two scenarios, the problem is solved. @Cato $\endgroup$
    – bashbin
    Commented Jan 31, 2017 at 10:50

2 Answers 2

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First: hit and destroy, Second :no hit, Third: no hit

$0.4\times 0.2 \times 0.5 \times 0.3=0.012$

First: no hit, Second :hit and destroy, Third: no hit

$0.6\times 0.5 \times 0.2 \times 0.3=0.018$

First: no hit, Second :no hit, Third: hit and destroy

$0.6\times 0.5 \times 0.7 \times 0.2=0.042$

First: no hit, (Second and Third) :hit and destroy

$0.6\times 0.5 \times 0.7 \times 0.6=0.126$

second: no hit, (first and Third) :hit and destroy

$0.5 \times 0.4 \times 0.7 \times 0.6=0.084$

third: no hit, (first and second) :hit and destroy

$0.3 \times 0.4 \times 0.5 \times 0.6=0.036$

(First, Second and Third) :hit and destroy

$0.4\times 0.5 \times 0.7 =0.14$

Add all, we get required probability as $0.458$

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    $\begingroup$ Good Answer, Kk...k...Kiran... $\endgroup$ Commented Jan 31, 2017 at 11:42
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$$\begin{align*} p =& P(A \lor B \lor C) \\=& 1-P(\neg A)P(\neg B)P(\neg C)\\=&1-(1-P(A))(1 - P(B))(1-P(C)) \end{align*}$$

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