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Question

A die is thrown twice. What is the probability that 5 will not come up either time?


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Solution

Probability of an event =numberoffavorableoutcomestotalnumberofoutcomes

We know that,

If a die is rolled twice, the total number of outcomes is 36.

Sample space is defined by (S)

S={(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)}

So, n(S)=36

Step 1: Calculating the probability of getting 5 on either time

Let A be the event of getting 5 either time.

The outcomes where 5 turns up ether time are ={(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5)}

So, n(A)=11

Therefore, the probability of getting 5 on either time, P(A)=n(A)n(S)

P(A)=1136

Step 2: Calculating the probability of not getting 5 on either time

So, the probability that 5 will not come up on either time =1P(A)

=11136=2536

Hence, the probability that 5 will not come up either time is 2536.


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