Two different dice are tossed together find the probability that the product is 6

Answer

Nội dung chính Show

Related Videos
What is the probability of rolling the product of 6 with 2 dice?
When two dice are tossed together then what is the probability of getting doublets?
When two dice are tossed what is the probability that the total score is a prime number?

Verified

Hint- Write samples of cases of total favorable and that of event to reach up to the probability.
If we toss two different dice than, total favorable cases are
$S = \left[ \begin{gathered}
  \left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right),\left( {1,6} \right) \\
  \left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {2,6} \right) \\
  ................................................... \\
  ...................................\left( {6,5} \right),\left( {6,6} \right) \\
\end{gathered} \right]$
Thus $n(S) = 36$
Now the favorable outcome to get product of 6 are
$\left[ {\left( {1,6} \right),\left( {6,1} \right),\left( {2,3} \right),\left( {3,2} \right)} \right]$
Thus let E be an event of getting a product 6 on toss of two dice then $n(E) = 4$
Now $P(E) = \dfrac{{Favorable{\text{ cases of event E}}}}{{Total{\text{ possible sample cases}}}} = \dfrac{{n(E)}}{{n(S)}}$……………………………. (1)
Using equation (1)
$P(E) = \dfrac{4}{{36}} = \dfrac{1}{9}$
Hence probability of getting a product 6 while toss of two different dice is $\dfrac{1}{9}$
Note- Whenever we have to solve such type of problems always write down the set of all possible sample cases and then the cases corresponding to that particular event, this helps in reducing the chances of leaving any sample case .Then use the probability basics of $P(E) = \dfrac{{Favorable{\text{ cases of event E}}}}{{Total{\text{ possible sample cases}}}} = \dfrac{{n(E)}}{{n(S)}}$to reach to the solution.

Given: Two different dice are tossed together.

To do: To find the probability of the product of two numbers on the top of the dice is 6.

Solution:

Two dice are tossed

$S=$[$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)$,

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$,

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)$,

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)$,

$(5,1),(5,2),(5,3),(5,4), (5,5),(5,6)$,

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$]

Total number of possiblen outcomes when two dice are tossed$=6\times6$

Events of getting product as$( 6)=( 1,\ 6),\ ( 2,\ 3),\ ( 3,\ 2),\ ( 6,\ 1)$

Number of favourbalble outcomes$=6$

Probability of getting product as $( 6)=\frac{Number\ of\ possible\ outcomes}{Total\ nuumber\ of\ possible\ outcomes}$

$=\frac{6}{36}$

$=\frac{1}{6}$

Therefore the probability of getting a product of 6 as a result when two dice are rolled is $\frac{1}{6}$.

Getting Image
Please Wait...

Course

NCERT

Class 12Class 11Class 10Class 9Class 8Class 7Class 6

IIT JEE

Exam

JEE MAINSJEE ADVANCEDX BOARDSXII BOARDS

NEET

Neet Previous Year (Year Wise)Physics Previous YearChemistry Previous YearBiology Previous YearNeet All Sample PapersSample Papers BiologySample Papers PhysicsSample Papers Chemistry

Download PDF's

Class 12Class 11Class 10Class 9Class 8Class 7Class 6

Exam CornerOnline ClassQuizAsk Doubt on WhatsappSearch DoubtnutEnglish DictionaryToppers TalkBlogJEE Crash CourseAbout UsCareerDownloadGet AppTechnothlon-2019

Logout

+91

Updated On: 27-06-2022

UPLOAD PHOTO AND GET THE ANSWER NOW!

Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

15051

9.0 K