My Math Forum Shooter hitting a target twice

 Probability and Statistics Basic Probability and Statistics Math Forum

 January 12th, 2015, 05:58 AM #1 Member   Joined: Sep 2012 Posts: 61 Thanks: 0 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?
 January 12th, 2015, 07:57 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,396 Thanks: 2477 Math Focus: Mainly analysis and algebra I think your reasoning might be correct, but what you have written has many errors. Would you like to try again?
January 12th, 2015, 08:02 AM   #3
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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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": $\:{4\choose2}(0.8)^2(0.2)^2$

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