Question

A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?
[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment].

Answer 1

i) p(5 will not come up either time)= 36-11/36=25/36
ii)p(5 will come at least once)= 11/36

Answer 2

$\huge{\underline{\underline{\tt{\green{Question}}}}}\:$

A dice is thrown twice. What is the probability that

(i) 5 will not come up either time ?

(ii) 5 will come up at least once ?

$\huge{\underline{\underline{\tt{\pink{Answer}}}}}\:$

Probability of getting 5 on throwing the dice once .

Sample space $\begin{cases}(5,1)\\(5,2)\\(5,3)\\(5,4)\\(5,5)\\(5,6)\end{cases}$

For tossing it twice we have 11 outcomes

Number of outcomes we get on throwing a dice once = 18

So , hence total outcomes on throwing it twice = 18 × 2 = 36

Number of times we can get a 5 = 11

i ) Number of times we won't get a 5

$\hookrightarrow\:36\:-\:11\:=\:25$

Event 1 = Probability of not getting a 5 either time

Event 1 = $\frac{Favorable\: outcomes}{Total\: number\:of\: outcomes}$

Event 1 = $\dfrac{25}{36}$

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ii ) Number of times we will get 5 atleast once = 11

Event 2 = Probability of getting a 5 atleast one time

Event 2 = $\frac{Favorable\: outcomes}{Total\: number\:of\: outcomes}$

Event 2 = $\dfrac{11}{36}$

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