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Misha Larkins [42]

3 years ago

6

Three red and three blue flags are arranged randomly along

2 answers:

laiz [17]3 years ago

3 0

$|\Omega|=\dfrac{6!}{3!3!}=\dfrac{4\cdot5\cdot6}{2\cdot3}=20\\A=\{RBRBRB,BRBRBR\}\\|A|=2\\\\P(A)=\dfrac{2}{20}=\dfrac{1}{10}$

Taya2010 [7]3 years ago

3 0

The answer would be B

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