Question

when a die is thrown twice what is the probability that 5 will not come up either time

Answer 1

we have
fevourable events =10(1,1,2,2,3,3,4,4,6,6)
total event=36
p(e)= 10/36
=5/18

Answer 2

Hey there!!!

Here When 2 times a die is thrown then there will be following favourable outcomes

( 1 ,1 ) , ( 1, 2) , ( 1,3) ,( 1 ,4) , ( 1, 5) , ( 1,6)

(2,1) , (2,2) , ( 2,3),. (2,4) , (2,5) , ( 2,6)

(3 ,1 ) , (3 ,1 ), (3 ,1 ) , (3 ,1 ) ,(3 ,1 ) ,(3 ,1 )

(4 , 1 )(4 , 2 )(4 , 3 )(4 , 4)(4 , 5)(4 , 6 )

(5, 1). (5, 2). (5, 3). (5, 4 ) (5, 5) (5, 6)

(6,1). (6,2). (6,3). (6,4). (6,5). (6,6)

So ,

$P ( 5 \: will \: not \: Come \: up \: Either time ) \: \: = \frac{25}{36}$
∵ (Total number of Desired Outcomes That , 5 will come up either time is 11
so , 5 will not come up either time = 36 - 11 = 25 )
$so \: \: P \: (5will \: not \: \: come \: up \: either \: time \: ) \: = \frac{25}{36}$


➪ 0.96

Hope this would help you !!