You flip two fair six−sidded dice and add the number of spots. what probability of getting a number divisible by 2 but not by 5 ?


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Question Number 131062 by benjo_mathlover last updated on 01/Feb/21

$You flip two fair six - sidded dice and add$ $the number of spots . what probability of getting$ $a number divisible by 2 but not by 5 ?$
	Answered by EDWIN88 last updated on 01/Feb/21

	$W r i t e L_{n} f o r e v e n t h e n u m b e r i s$ $d i v i s i b l e b y n . N o w P (L_{2}) = 1 / 2 .$ $(c o u n t t h e c a s e; o r m o r e e l e g a n t l y n o t i c e$ $t h a t e a c h d i e h a s t h e s a m e n u m b e r o f$ $o d d a n d e v e n f a c e s a n d w o r k f r o m t h e r e)$ $N o w P (L_{2} - L_{5}) = P (L_{2}) - P (L_{2} \cap L_{5})$ $b u t L_{2} \cap L_{5} c o n t a i n s o n l y t h r e e$ $o u t c o m e s (6, 4), (5, 5) a n d (4, 6) s o$ $P (L_{2} - L_{5}) = \frac{1}{2} - \frac{3}{36} = \frac{5}{12}$
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