Question

A die is thrown twice. What is the probability that
i) 5 will come up atleast once ?
ii) 5 will not come up either time?​

Answer 1

Answer:

I hope it will help you

Step-by-step explanation:

When a die is thrown twice, the possible outcomes =

Total number of possible outcomes = 

The outcomes when 5 comes up at least once =

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

Number of such favourable outcomes = 11 

Therefore, the probability that 5 comes at least once is 



Answer 2

Sample space :

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

( 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 )

Total no of cases = 36

$P(E) \: = \frac{no \: of \:favourable \: outcomes}{total \: no \: of \: outcomes}$

( i ) E1 = 5 will come up at least one

$p(E1) = \frac{11}{36}$

( ii ) E2 = 5 will no come up either time

$p(E2) = 1 - p(e1)$

$p(E2) = 1 - \frac{11}{36}$

$p(E2) = \frac{25}{36}$