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Three projectiles are fired at an airplane. Probability that it will be hit with the first projectile is $0.5$, for the second is $0.6$, and for the third is $0.8$. Probability that an airplane will be shot down with only one projectile is $0.3$, with two projectiles $0.6$, and with all three projectiles is $1$. Find the probability that an airplane is shot down.

Event definitions:

$H_i$: Airplane is hit by $i$ projectile, $1\le i\le 3$

$A:$ Airplane is shot down.

$P(H_1)=0.5,P(H_2)=0.6,P(H_3)=0.8$

What are equations for probabilities that an airplane will be shot down with only one, two and three projectiles?

What is the equation for $P(A)$?

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    You have it wrong for the probabilities of $H_i$ : those are the probabilities that airplane is hit by projectile $i$. So you have to compute the probabilities of one hit, two hits and three hits, and use law of total probabilities with the given conditional probabilities of $A$. – 2017-01-10
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    @Nicolas FRANCOIS Could you show the equation for $P(A)$ with probabilities of one, two and three hits? – 2017-01-10
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    Just make a tree, with stage $i$ representing the result of projectile $i$, and compute the number of hits for each leaf of the tree. I have too much work for the moment to typeset it... – 2017-01-10

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A simple way to do it, is by enumerating the possibilities:

1) No projectile hits

2) first projectile hits, second and third miss

3) Second projectile hits, first and third miss

...

8)All three projectiles hit

Every case has an propability to occur (multiply hit and miss propabilites) and secondy every case has an chance of knocking the airplane down. Multiply those and add them up.

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    If we define event space $\Omega=\{000,001,010,011,100,101,110,111\}$, and without using any given probability, we get $P(A)=7/8$. How do we exactly need to use given data and sum them? – 2017-01-10
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    Each event has a different propability to occur and each has a propability for the airplane to go down. Just counting won't do the trick here. For example: 001 occurs with propability of $\frac{4}{25}$ and will send the airplane go crashing in $30\%$, so multiply that to get $\frac{6}{125}$. Second is event 010 which occurs with propability of $\frac{3}{50}$ and will send the airplane down again in $30\%$, so another $\frac{9}{500}$ chance. Do that with all events and summ them up. – 2017-01-10