3 years ago

Help with number 4 and number 7 please

1 answer:

6 0

Answer is in the picture

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ratelena [41]

$|\Omega|=_{12}C_3=\dfrac{12!}{3!9!}=\dfrac{10\cdot11\cdot12}{2\cdot3}=220\\|A|=4\cdot3\cdot5=60\\\\P(A)=\dfrac{60}{220}=\dfrac{3}{11}\approx0,273$

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3 years ago