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An airline serves a dinner to all the passengers on an airplane. They get their choice of steak or fish. Three steak meals and three fish meals are set aside for the six-member crew. If the meals are distributed to the crew members randomly, what is the probability that both pilots get the fish?

Guest Apr 6, 2015

Best Answer 

 #1
avatar+80935 
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The probability that the first pilot gets fish = 1/2 = 50

The probabilty that the second pilot gets fish = 2/5 = .40

So 

P(A and B)  = .50 * .40  =  .20  = 20%

 

 

  

CPhill  Apr 6, 2015
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1+0 Answers

 #1
avatar+80935 
+8
Best Answer

The probability that the first pilot gets fish = 1/2 = 50

The probabilty that the second pilot gets fish = 2/5 = .40

So 

P(A and B)  = .50 * .40  =  .20  = 20%

 

 

  

CPhill  Apr 6, 2015

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