Bullseye -- II

Probability Level 3

Four men go to play darts. The following are the chances of them hitting the bullseye:

"Deadshot" Ash -- 70%

"Cleanshot" Bill -- 60 %

"Splitshot" Carl -- 50%

"Strayshot" Dean -- 40%

What is the percentage probability (1 decimal place) that AT LEAST one of them MISSES hitting the bullseye?


The answer is 91.6.

Satyen Nabar by Satyen Nabar , 48, India

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2 solutions

Satyen Nabar
Jun 6, 2014

The combined chances of them all hitting bullseye is multiplication of 0.7, 0.6, 0.5 and 0.4. = 0.084

1- 0.084 = 0.916 (91.6%) chance of at least one of them missing.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Got this answer correct ,yet no BULLSEYE!!. Ientered 0.916

Raman Gupta - 7 years ago
Math Man
Jun 16, 2014

with inclusion exculsion , we get 1-(7/10)(6/10)(5/10)(4/10)=0.916=91.6%

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