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Find the probability of throwing at teast 3 sixes in 5 throws of a dia. |
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Answer» Solution :In one throw of a die p(getting a 60=`1/6` p(getting a NON 6 )=`5/6` In 5 THROWS of a die p(atleast 3 SIXES)=p(3 sixer)+p(4 sixer)+p(5 sixer) = `"^5C_3(1/6)^3(5/6)^2+^5C_4(1/6)^4(5/6)+("^5C_5)(1/6)^5=250/6^5+25/6^5+1/6^5=276/6^5` |
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