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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*(10.2)*(10.2)*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: 7,313 Thanks: 2447 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 Thanks: 407  Hello, Chemist@! Quote:
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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