2
$\begingroup$

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?

--

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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
    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
    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
    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
    Jan 31, 2017 at 10:50

2 Answers 2

1
$\begingroup$

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 $0.458$

$\endgroup$
1
  • 1
    $\begingroup$ Good Answer, Kk...k...Kiran... $\endgroup$ Jan 31, 2017 at 11:42
0
$\begingroup$

$$\begin{align*} p =& P(A \lor B \lor C) \\=& 1-P(\neg A)P(\neg B)P(\neg C)\\=&1-(1-P(A))(1 - P(B))(1-P(C)) \end{align*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.