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January 12th, 2015, 06:58 AM   #1
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Shooter hitting a target twice

If a shooter has four bullets and the probability that he hits the target is 0.8, what is the probability that he will hit the target twice if he shoots four times.

I think that it is:
0.8*0.8*(1-0.2)*(1-0.2)*6 (it is multiplied by 6 because there are 6 possible ways for him to hit the target exactly 6 times).

Is my reasoning correct?
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January 12th, 2015, 08:57 AM   #2
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I think your reasoning might be correct, but what you have written has many errors. Would you like to try again?
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January 12th, 2015, 09:02 AM   #3
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Hello, Chemist@!

Quote:
If a shooter has four bullets and the probability that he hits the target is 0.8,
what is the probability that he will hit the target twice if he shoots four times?

I think that it is: $\:0.8\cdot0.8\cdot\color{red}{(1-0.2)}\cdot\color{red}{(1-0.2)}\cdot6 \quad\color{red}{??}$
(it is multiplied by 6 because there are 6 possible ways for him
to hit the target exactly 2 times).

Is my reasoning correct?$\quad \color{blue}{\text{Sorry, no.}}$

You have: $\:(0.8)(0.8)(0.8)(0.8)(6) \;\;$ . . . something about "four hits".

You want "2 hits and 2 misses": $\:{4\choose2}(0.8)^2(0.2)^2$

Thanks from Chemist@
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