Question

A die is thrown twice what is the probability that 5 will not come up rither time

Answer 1

Hi there!
Here's the answer:

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Let S be sample space
n(S) - No. of total outcomes when a dice is rolled two times
n(S) = 6^2 = 36

Let E be event that 5 will show up on dice both time
& E' be the Event that 5 will show up at least once

[^^^ Atleast once => 1 to Maximum
Maximum is 2, as dice is rolled twice, only two 5's can occur]

P(E) + P(E') = 1
(°•° Sum of probabilities = 1 )
=> P(E) = 1 - P(E')

E' = {(5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (1,5) , (2,5) , (3,5) , (4,5) , (6,5)}
n(E') - No. of possible cases for event E'
n(E') = 11

Now find P(E')
P(E') = n(E') / n(S)
P(E') = 11 / 36

•°• P(E) = 1 - (11/36)
=> P(E) = 25/36

•°• Required probability is 25/36.

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Hope it helps