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Erik has .94 chance of being chosen, Bailey has .85 chance and Bert has .8 chance. What's the probability Bert would be chosen and Erik and Bailey would loose? What's the probability at least one will be chosen?

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  • $\begingroup$ How many total are being chosen? What have you tried so far? $\endgroup$ – brogrenkp May 6 '16 at 1:42
  • $\begingroup$ The problem doesn't state how many are being chosen. You may assume only one can be chosen. I don't understand the question at all. $\endgroup$ – user54272 May 6 '16 at 1:44
  • $\begingroup$ They all can be chosen, but only with a specific probability. Bert can be chosen with the probability of $0.8$. And Erik will be not chosen with a probability of $1-0.94=0.06$. Does it help to anser the first question ? $\endgroup$ – callculus May 6 '16 at 1:51
  • $\begingroup$ I figured it out Thanks $\endgroup$ – user54272 May 6 '16 at 1:54
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Erik has .94 chance of being chosen, Bailey has .85 chance and Bert has .8 chance.

What's the probability Bert would be chosen and Erik and Bailey would loose?

Ans: 0.8*0.06*0.15

What's the probability at least one will be chosen?

Ans: 1 - probability none will be chosen = 1-[0.06*0.15*0.20]

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    $\begingroup$ That´s right. Congrats. $\endgroup$ – callculus May 6 '16 at 2:01
  • $\begingroup$ It's funny how similar the equation is to what the problem question is. 0.8 is the probability Bert is chosen times the probability the others are not. The question spells out the answer. Now why would we know to multiply? $\endgroup$ – Breedly May 6 '16 at 2:38

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