1.

Q => 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?​

Answer»

align="absmiddle" alt="\huge\star\:\:{\orange{\UNDERLINE{\purple{\mathbf{Question}}}}}" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%5Chuge%5Cstar%5C%3A%5C%3A%7B%5Corange%7B%5Cunderline%7B%5Cpurple%7B%5Cmathbf%7BQuestion%7D%7D%7D%7D%7D" TITLE="\huge\star\:\:{\orange{\underline{\purple{\mathbf{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\star\:\:{\orange{\underline{\red{\mathbf{Answer}}}}}

Probability of getting 5 on throwing the dice once .

Sample SPACE \begin{cases}(5,<klux>1</klux>)\\(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}

_____________________________________

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}

_______________________________________



Discussion

No Comment Found

Related InterviewSolutions